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Class 11 Mathematics - Permutations and Combinations - Derivations of formulas and their connections Expert Quiz

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पहचान \(\sum_{k=0}^{r}{}^{m}C_k{}^{n}C_{r-k}={}^{m+n}C_r\) का सबसे सही संयोजनात्मक आधार क्या है?

What is the most correct combinatorial basis of the identity \(\sum_{k=0}^{r}{}^{m}C_k{}^{n}C_{r-k}={}^{m+n}C_r\)?

Explanation opens after your attempt
Correct Answer

A. दो अलग समूहों से कुल (r) वस्तुएं चुननाChoosing total (r) objects from two separate groups

Explanation

Simple Explanation

पहले समूह से (k) और दूसरे से (r-k) चुनकर सभी cases जुड़ते हैं। परीक्षा में दो समूहों वाली selection में वेंडरमोंड पहचानें। / Choose (k) from the first group and (r-k) from the second, then add all cases. In exams identify Vandermonde in two-group selection.

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\(\sum_{r=0}^{n}{}^{n}C_r^2={}^{2n}C_n\) को सिद्ध करने में कौन-सा विचार सबसे उपयुक्त है?

Which idea is most suitable to prove \(\sum_{r=0}^{n}{}^{n}C_r^2={}^{2n}C_n\)?

Explanation opens after your attempt
Correct Answer

B. दो (n)-आकार के समूहों से कुल (n) वस्तुएं चुननाChoosing total (n) objects from two groups of size (n)

Explanation

Simple Explanation

दूसरे समूह से (n-r) चुनना \({}^{n}C_{n-r}={}^{n}C_r\) के बराबर है। परीक्षा में square sum को दो समान समूहों से जोड़ें। / Choosing (n-r) from the second group equals \({}^{n}C_{n-r}={}^{n}C_r\). In exams connect square sums with two equal groups.

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({}^{n}C_r\cdot r!(n-r)!) हमेशा किसके बराबर होता है?

What is ({}^{n}C_r\cdot r!(n-r)!) always equal to?

Explanation opens after your attempt
Correct Answer

B. (n!)

Explanation

Simple Explanation

पहले (r) वस्तुएं चुनें, फिर चुनी और न चुनी दोनों सूचियों को क्रम में रखें। परीक्षा में factorial cancellation से भी यही तुरंत मिलता है। / Choose (r) objects first, then arrange both chosen and unchosen lists. In exams factorial cancellation also gives this quickly.

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\(r{}^{n}C_r=n{}^{n-1}C_{r-1}\) की double counting व्याख्या क्या है?

What is the double-counting interpretation of \(r{}^{n}C_r=n{}^{n-1}C_{r-1}\)?

Explanation opens after your attempt
Correct Answer

A. एक चुने हुए (r)-समूह में एक सदस्य को चिह्नित करनाMarking one member in a chosen (r)-group

Explanation

Simple Explanation

बाईं ओर समूह चुनकर mark करते हैं, दाईं ओर marked member पहले चुनते हैं। परीक्षा में (r) factor को marked choice मानें। / The left side chooses a group and marks one member, while the right side chooses the marked member first. In exams treat the factor (r) as a marked choice.

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(r(r-1){}^{n}C_r=n(n-1){}^{n-2}C_{r-2}) में (r(r-1)) किसे गिनता है?

What does (r(r-1)) count in (r(r-1){}^{n}C_r=n(n-1){}^{n-2}C_{r-2})?

Explanation opens after your attempt
Correct Answer

A. चुने हुए समूह में ordered दो marked सदस्यOrdered two marked members in the selected group

Explanation

Simple Explanation

(r)-समूह में पहला और दूसरा marked member क्रम सहित चुने जाते हैं। परीक्षा में (r(r-1)) देखकर ordered marking सोचें। / In an (r)-group, the first and second marked members are chosen with order. In exams think of ordered marking when you see (r(r-1)).

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\(\sum_{r=0}^{n}r{}^{n}C_r=n2^{n-1}\) में दाईं ओर \(2^{n-1}\) क्यों आता है?

Why does \(2^{n-1}\) appear on the right side of \(\sum_{r=0}^{n}r{}^{n}C_r=n2^{n-1}\)?

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Correct Answer

A. marked member चुनने के बाद बाकी (n-1) members स्वतंत्र रूप से चुने या छोड़े जाते हैंAfter choosing the marked member, the remaining (n-1) members are freely chosen or left

Explanation

Simple Explanation

पहले marked member के (n) choices हैं और बाकी पर two choices हैं। परीक्षा में binomial sum में extra (r) को marking से जोड़ें। / There are (n) choices for the marked member and two choices for each remaining member. In exams connect the extra (r) in binomial sums with marking.

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(\sum_{r=0}^{n}r(r-1){}^{n}C_r) का सही सरल रूप कौन-सा है?

What is the correct simplified form of (\sum_{r=0}^{n}r(r-1){}^{n}C_r)?

Explanation opens after your attempt
Correct Answer

B. (n(n-1)2^{n-2})

Explanation

Simple Explanation

Ordered दो marked members चुनने के (n(n-1)) ways हैं और बाकी (n-2) freely चुने जाते हैं। परीक्षा में दो marks हों तो \(2^{n-2}\) आता है। / There are (n(n-1)) ways to choose two ordered marked members and the remaining (n-2) are chosen freely. In exams two marks lead to \(2^{n-2}\).

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\(\sum_{r=0}^{n}r^2{}^{n}C_r\) का सही रूप क्या है?

What is the correct form of \(\sum_{r=0}^{n}r^2{}^{n}C_r\)?

Explanation opens after your attempt
Correct Answer

C. (n(n+1)2^{n-2})

Explanation

Simple Explanation

(r-2=r(r-1)+r) लिखकर दो standard sums जोड़ते हैं। परीक्षा में \(r^2\) को split करना तेज तरीका है। / Write (r-2=r(r-1)+r) and add two standard sums. In exams splitting \(r^2\) is the fastest method.

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\(\sum_{r=0}^{n}{}^{n}C_r{}^{r}C_k\) का simplified form कौन-सा है?

What is the simplified form of \(\sum_{r=0}^{n}{}^{n}C_r{}^{r}C_k\)?

Explanation opens after your attempt
Correct Answer

A. \(^{n}C_k2^{n-k}\)

Explanation

Simple Explanation

पहले (k) marked members चुनें, फिर बाकी (n-k) members subset में आएं या न आएं। परीक्षा में nested selection में marked set पहले चुनें। / Choose the (k) marked members first, then each of the remaining (n-k) members may or may not enter the subset. In exams choose the marked set first in nested selection.

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\({}^{n}C_r{}^{r}C_s{}^{s}C_t\) को सही क्रम बदलकर किस रूप में लिखा जा सकता है?

By changing the order of selection, \({}^{n}C_r{}^{r}C_s{}^{s}C_t\) can be correctly written as which form?

Explanation opens after your attempt
Correct Answer

A. \(^{n}C_t{}^{n-t}C_{s-t}{}^{n-s}C_{r-s}\)

Explanation

Simple Explanation

पहले सबसे अंदर के (t) members चुनें, फिर (s-t), फिर (r-s) members जोड़ें। परीक्षा में nested choices को layers में उल्टा गिनें। / First choose the innermost (t) members, then add (s-t), then (r-s) members. In exams count nested choices backward by layers.

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यदि (n) अलग वस्तुओं को (a,b,c) आकार के labelled समूहों में बांटना हो और (a+b+c=n), तो सूत्र कौन-सा है?

If (n) distinct objects are divided into labelled groups of sizes (a,b,c) and (a+b+c=n), which formula is correct?

Explanation opens after your attempt
Correct Answer

B. \(\frac{n!}{a!b!c!}\)

Explanation

Simple Explanation

Labelled groups fixed हैं और हर group के अंदर order नहीं गिना जाता। परीक्षा में multinomial denominator में group-size factorials रखें। / Labelled groups are fixed and order inside each group is not counted. In exams put group-size factorials in the multinomial denominator.

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(12) अलग वस्तुओं को (4,4,4) के unlabelled समूहों में बांटने का formula क्या होगा?

What is the formula for dividing (12) distinct objects into unlabelled groups of (4,4,4)?

Explanation opens after your attempt
Correct Answer

B. \(\frac{12!}{4!4!4!3!}\)

Explanation

Simple Explanation

तीनों groups equal size और unlabelled हैं, इसलिए groups की (3!) अदला-बदली भी हटती है। परीक्षा में समान unlabelled groups पर extra factorial divide करें। / All three groups have equal size and are unlabelled, so the (3!) interchange of groups is also removed. In exams divide extra by factorial for equal unlabelled groups.

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\(\frac{n!}{a!b!c!}\) को \({}^{n}C_a{}^{n-a}C_b\) से जोड़ने की शर्त क्या है?

What condition connects \(\frac{n!}{a!b!c!}\) with \({}^{n}C_a{}^{n-a}C_b\)?

Explanation opens after your attempt
Correct Answer

A. (a+b+c=n)

Explanation

Simple Explanation

पहले (a) चुनें, फिर बचे में से (b), और (c) अपने आप तय होता है। परीक्षा में sequential selection को factorial form से मिलाएं। / Choose (a) first, then (b) from the remaining, and (c) is automatically fixed. In exams match sequential selection with factorial form.

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(n) distinct objects को (r) distinct boxes में onto तरीके से भेजने में inclusion-exclusion का सही सूत्र कौन-सा है?

Which inclusion-exclusion formula counts onto mappings from (n) distinct objects to (r) distinct boxes?

Explanation opens after your attempt
Correct Answer

A. (\sum_{k=0}^{r}(-1)^k{}^{r}C_k(r-k)^n)

Explanation

Simple Explanation

Total functions से empty boxes वाले cases inclusion-exclusion द्वारा हटते हैं। परीक्षा में onto का अर्थ हर box non-empty समझें। / Cases with empty boxes are removed from total functions by inclusion-exclusion. In exams read onto as every box being non-empty.

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(n) distinct objects को (3) distinct boxes में onto distribute करने की संख्या क्या है?

What is the number of onto distributions of (n) distinct objects into (3) distinct boxes?

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Correct Answer

A. \(3^n-3\cdot2^n+3\)

Explanation

Simple Explanation

तीन empty choices घटती हैं, फिर दो empty boxes के over-subtraction को जोड़ा जाता है। परीक्षा में onto distribution में inclusion-exclusion सावधानी से लगाएं। / Three empty-box choices are subtracted, then over-subtraction of two empty boxes is added. In exams apply inclusion-exclusion carefully for onto distribution.

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(n) distinct objects में से कोई object छोड़े बिना उन्हें (2) non-empty labelled boxes में बांटने की संख्या क्या है?

What is the number of ways to distribute (n) distinct objects into (2) non-empty labelled boxes without leaving any object?

Explanation opens after your attempt
Correct Answer

A. \(2^n-2\)

Explanation

Simple Explanation

Total \(2^n\) assignments में दोनों all-in-one empty-box cases invalid हैं। परीक्षा में labelled boxes हों तो empty cases अलग से घटाएं। / From total \(2^n\) assignments, the two all-in-one empty-box cases are invalid. In exams subtract empty cases separately for labelled boxes.

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(n) distinct objects को (2) non-empty unlabelled groups में बांटने की संख्या कौन-सी है?

What is the number of ways to divide (n) distinct objects into (2) non-empty unlabelled groups?

Explanation opens after your attempt
Correct Answer

A. \(\frac{2^n-2}{2}\)

Explanation

Simple Explanation

Labelled count में दोनों groups की अदला-बदली दो बार गिनती है। परीक्षा में unlabelled दो groups के लिए (2) से divide करें। / In the labelled count, interchanging the two groups counts each division twice. In exams divide by (2) for two unlabelled groups.

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(n) distinct objects से (r) objects repetition के साथ बिना order चुने जाएं, तो formula किससे आता है?

If (r) objects are chosen from (n) distinct objects with repetition and without order, which idea gives the formula?

Explanation opens after your attempt
Correct Answer

A. Stars and bars

Explanation

Simple Explanation

यह multiset selection है और answer \({}^{n+r-1}C_r\) होता है। परीक्षा में repetition with no order को bars method से करें। / This is multiset selection and the answer is \({}^{n+r-1}C_r\). In exams handle repetition with no order by the bars method.

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\(x_1+x_2+x_3+x_4=25\) और \(x_i\geq2\) हो, तो solutions की संख्या कौन-सी है?

If \(x_1+x_2+x_3+x_4=25\) and \(x_i\geq2\), what is the number of solutions?

Explanation opens after your attempt
Correct Answer

B. \(^{20}C_3\)

Explanation

Simple Explanation

चार variables को पहले (2) देने पर (17) बचता है। परीक्षा में lower bound घटाकर non-negative stars and bars लगाएं। / Give (2) first to the four variables, leaving (17). In exams subtract the lower bound and apply non-negative stars and bars.

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\(x_1+x_2+x_3=20\) में \(x_1\geq3\), \(x_2\geq4\), \(x_3\geq5\) हो, तो count क्या है?

In \(x_1+x_2+x_3=20\), if \(x_1\geq3\), \(x_2\geq4\), \(x_3\geq5\), what is the count?

Explanation opens after your attempt
Correct Answer

A. \(^{10}C_2\)

Explanation

Simple Explanation

Minimum (3+4+5=12) हटाने पर (8) बचता है। परीक्षा में unequal lower bounds को पहले shift करें। / Removing the minimum (3+4+5=12) leaves (8). In exams shift unequal lower bounds first.

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\(x_1+x_2+x_3+x_4=18\) में exactly (2) variables zero हों, तो count क्या है?

In \(x_1+x_2+x_3+x_4=18\), if exactly (2) variables are zero, what is the count?

Explanation opens after your attempt
Correct Answer

A. \(^{4}C_2{}^{17}C_1\)

Explanation

Simple Explanation

पहले zero variables चुनें, फिर बाकी दो variables positive sum (18) बनाते हैं। परीक्षा में exactly zero cases में positive distribution लगाएं। / Choose the zero variables first, then the remaining two variables form a positive sum of (18). In exams use positive distribution for exactly-zero cases.

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\(x_1+x_2+x_3=15\) में \(0\leq x_i\leq6\) हो, तो inclusion-exclusion में कौन-सा expression सही है?

In \(x_1+x_2+x_3=15\) with \(0\leq x_i\leq6\), which expression is correct by inclusion-exclusion?

Explanation opens after your attempt
Correct Answer

A. \(^{17}C_2-3{}^{10}C_2+3{}^{3}C_2\)

Explanation

Simple Explanation

Upper bound तोड़ने पर \(x_i\geq7\) के cases घटते और double violations जुड़ते हैं। परीक्षा में bounded solutions में (7) shift करें। / Cases with \(x_i\geq7\) are subtracted and double violations are added. In exams shift by (7) for bounded solutions.

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\(x_1+x_2+x_3+x_4=10\) में हर \(x_i\leq4\) हो, तो valid count किस expression से मिलेगा?

If \(x_1+x_2+x_3+x_4=10\) and every \(x_i\leq4\), which expression gives the valid count?

Explanation opens after your attempt
Correct Answer

A. \(^{13}C_3-4{}^{8}C_3+6{}^{3}C_3\)

Explanation

Simple Explanation

Violation \(x_i\geq5\) से शुरू होती है और inclusion-exclusion लागू होता है। परीक्षा में upper limit (4) हो तो (5) subtract shift लें। / A violation begins with \(x_i\geq5\), so inclusion-exclusion applies. In exams use a subtract shift of (5) for upper limit (4).

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(n) letters की derangement संख्या \(D_n\) के लिए recurrence (D_n=(n-1)\(D_{n-1}+D_{n-2}\)) में (n-1) factor क्यों आता है?

In the derangement recurrence (D_n=(n-1)\(D_{n-1}+D_{n-2}\)), why does the factor (n-1) appear?

Explanation opens after your attempt
Correct Answer

A. पहले letter की गलत position चुनने के लिएTo choose the wrong position of the first letter

Explanation

Simple Explanation

पहला letter अपनी original position छोड़कर (n-1) positions में जा सकता है। परीक्षा में derangement recurrence में first object की गलत choice देखें। / The first letter can go to (n-1) positions other than its original position. In exams watch the first object's wrong choice in derangement recurrence.

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(D_n=n!\left(1-\frac{1}{1!}+\frac{1}{2!}-\cdots+(-1)^n\frac{1}{n!}\right)) किस principle से आता है?

The formula (D_n=n!\left(1-\frac{1}{1!}+\frac{1}{2!}-\cdots+(-1)^n\frac{1}{n!}\right)) comes from which principle?

Explanation opens after your attempt
Correct Answer

A. Inclusion-exclusion

Explanation

Simple Explanation

Fixed points वाले permutations को alternating तरीके से घटाया और जोड़ा जाता है। परीक्षा में कोई object सही स्थान पर न हो तो inclusion-exclusion सोचें। / Permutations with fixed points are subtracted and added alternately. In exams think of inclusion-exclusion when no object is in its correct place.

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(5) letters को envelopes में इस प्रकार रखना है कि कोई letter सही envelope में न जाए। Count कौन-सी है?

(5) letters are placed into envelopes so that no letter goes into its correct envelope. Which count is correct?

Explanation opens after your attempt
Correct Answer

A. \(D_5=44\)

Explanation

Simple Explanation

यह (5) objects का derangement है और \(D_5=44\) होता है। परीक्षा में letters-envelopes mismatch को derangement pattern मानें। / This is the derangement of (5) objects and \(D_5=44\). In exams treat letters-envelope mismatch as a derangement pattern.

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(n) people को (n) seats पर इस तरह बैठाना कि exactly (k) people सही seats पर हों, count क्या है?

What is the count for seating (n) people in (n) seats so that exactly (k) people are in their correct seats?

Explanation opens after your attempt
Correct Answer

A. \(^{n}C_kD_{n-k}\)

Explanation

Simple Explanation

पहले सही बैठे (k) people चुनें, फिर बाकी (n-k) का derangement करें। परीक्षा में exactly fixed points के लिए choose fixed plus derange rest करें। / First choose the (k) correctly seated people, then derange the remaining (n-k). In exams use choose fixed plus derange rest for exactly fixed points.

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(n) distinct objects की circular arrangements में ((n-1)!) को (n!) से derive करने का सही कारण क्या है?

What is the correct reason for deriving ((n-1)!) from (n!) for circular arrangements of (n) distinct objects?

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Correct Answer

A. हर circular arrangement (n) rotations से linear arrangements में गिनी जाती हैEach circular arrangement is counted as (n) rotations in linear arrangements

Explanation

Simple Explanation

Linear (n!) count में rotations duplicate होते हैं। परीक्षा में circular arrangement में rotational overcount को (n) से divide करें। / Rotations are duplicates in the linear count (n!). In exams divide rotational overcount by (n) in circular arrangements.

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(n) distinct beads की bracelet arrangements में (\frac{(n-1)!}{2}) कब सही है?

When is (\frac{(n-1)!}{2}) correct for bracelet arrangements of (n) distinct beads?

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Correct Answer

A. जब rotations और reflections दोनों same माने जाएंWhen both rotations and reflections are considered the same

Explanation

Simple Explanation

Bracelet में mirror images same मानी जाती हैं। परीक्षा में reflection same होने पर circular count को (2) से divide करें। / In a bracelet, mirror images are considered the same. In exams divide the circular count by (2) when reflection is the same.

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(8) people को round table पर बैठाना है और (A), (B), (C) साथ रहें। Count कौन-सी है?

(8) people are seated around a round table and (A), (B), (C) stay together. Which count is correct?

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Correct Answer

A. \(5!\cdot3!\)

Explanation

Simple Explanation

तीन लोगों का one block और बाकी (5) people मिलकर (6) circular objects बनाते हैं, इसलिए (5!) arrangements हैं। परीक्षा में circular block count में objects minus one factorial लें। / The block of three people and the remaining (5) people form (6) circular objects, giving (5!) arrangements. In exams use one less factorial for circular block objects.

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(n) people को round table पर बैठाने में (A) और (B) adjacent न हों, तो count कौन-सी है?

If (n) people are seated around a round table and (A) and (B) are not adjacent, which count is correct?

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Correct Answer

A. ((n-1)!-2(n-2)!)

Explanation

Simple Explanation

Total circular arrangements से (A,B) adjacent block के arrangements घटते हैं। परीक्षा में circular not adjacent को complement से करें। / Subtract adjacent block arrangements of (A,B) from total circular arrangements. In exams handle circular not-adjacent by complement.

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(2n) people में (n) men और (n) women हैं। Round table पर alternate seating की count कौन-सी है?

There are (n) men and (n) women among (2n) people. What is the count of alternate seating around a round table?

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Correct Answer

A. (n!(n-1)!)

Explanation

Simple Explanation

पहले men को circle में ((n-1)!) ways से बैठाएं, फिर gaps में women को (n!) ways से रखें। परीक्षा में circular alternate में पहले एक group fix करें। / First seat men in a circle in ((n-1)!) ways, then place women in gaps in (n!) ways. In exams fix one group first for circular alternation.

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(n) distinct objects को row में रखना है और (k) special objects का relative order fixed है। Count क्या होगा?

(n) distinct objects are arranged in a row and the relative order of (k) special objects is fixed. What is the count?

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Correct Answer

A. \(\frac{n!}{k!}\)

Explanation

Simple Explanation

Special objects के (k!) relative orders में केवल (1) allowed है। परीक्षा में fixed relative order में total arrangements को (k!) से divide करें। / Only (1) of the (k!) relative orders of the special objects is allowed. In exams divide total arrangements by (k!) for fixed relative order.

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(10) people की line में (A) (B) से पहले और (C) (D) से पहले आए। Count क्या होगा?

In a line of (10) people, (A) must come before (B) and (C) before (D). What is the count?

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Correct Answer

B. \(\frac{10!}{4}\)

Explanation

Simple Explanation

दो स्वतंत्र relative order restrictions में हर एक count को आधा करता है। परीक्षा में independent before-after pairs पर \(2^k\) से divide करें। / Each of the two independent relative-order restrictions halves the count. In exams divide by \(2^k\) for independent before-after pairs.

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(0,1,2,3,4,5,6,7) से repetition बिना (5)-digit even numbers बनाने में (0) को last digit case अलग क्यों लिया जाता है?

Why is the case with (0) as the last digit treated separately when forming (5)-digit even numbers without repetition from (0,1,2,3,4,5,6,7)?

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Correct Answer

A. क्योंकि (0) last में valid है पर first में invalid हैBecause (0) is valid at the last place but invalid at the first place

Explanation

Simple Explanation

Even number में (0) unit place पर आ सकता है लेकिन leading digit नहीं बन सकता। परीक्षा में digit problems में zero cases अलग रखें। / In an even number, (0) may be the unit digit but cannot be the leading digit. In exams keep zero cases separate in digit problems.

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Digits (0,1,2,3,4,5,6) से repetition बिना (4)-digit numbers बनाने हैं जो (5) से divisible हों। सही count expression कौन-सा है?

Using digits (0,1,2,3,4,5,6) without repetition, (4)-digit numbers divisible by (5) are formed. Which count expression is correct?

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Correct Answer

A. \(6\cdot5\cdot4+5\cdot5\cdot4\)

Explanation

Simple Explanation

Last digit (0) या (5) हो सकता है, और (0) first digit पर restriction बदलता है। परीक्षा में divisibility by (5) में unit digit cases बनाएं। / The last digit can be (0) or (5), and (0) changes the first-digit restriction. In exams make unit-digit cases for divisibility by (5).

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Digits (1,2,3,4,5,6,7) से repetition allowed (4)-digit numbers में exactly (2) positions पर even digits हों, तो count कौन-सी है?

Using digits (1,2,3,4,5,6,7) with repetition allowed, if exactly (2) positions contain even digits in a (4)-digit number, what is the count?

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Correct Answer

A. \(^{4}C_2\cdot3^2\cdot4^2\)

Explanation

Simple Explanation

Even positions चुनें, उन पर (3) even choices और बाकी पर (4) odd choices हैं। परीक्षा में exactly condition में positions first चुनें। / Choose the even positions, then each has (3) even choices and the rest have (4) odd choices. In exams choose positions first in exactly conditions.

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(n) distinct symbols से length (r) strings बनती हैं जिनमें repetition allowed है लेकिन कम से कम एक symbol repeat हो। Count क्या है?

Length (r) strings are formed from (n) distinct symbols with repetition allowed, but at least one symbol repeats. What is the count?

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Correct Answer

A. \(n^r-{}^{n}P_r\)

Explanation

Simple Explanation

Total repeated-allowed strings से all distinct strings घटाएं। परीक्षा में at least repeat को complement no repeat से करें। / Subtract all-distinct strings from total repetition-allowed strings. In exams solve at least repeat by the no-repeat complement.

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(n) distinct symbols से length (r) strings बनती हैं और exactly (s) distinct symbols use हों। सही count कौन-सा है?

Length (r) strings are formed from (n) distinct symbols and exactly (s) distinct symbols are used. Which count is correct?

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Correct Answer

A. (^{n}C_s s! S(r,s))

Explanation

Simple Explanation

पहले (s) symbols चुनें, फिर (r) positions को उन (s) symbols पर onto map करें। परीक्षा में exactly distinct symbols में onto idea उपयोग करें। / First choose (s) symbols, then map the (r) positions onto those (s) symbols. In exams use the onto idea for exactly distinct symbols.

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((a+b+c)^n) में \(a^p b^q c^r\) का coefficient क्या है, यदि (p+q+r=n)?

What is the coefficient of \(a^p b^q c^r\) in ((a+b+c)^n), if (p+q+r=n)?

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Correct Answer

A. \(\frac{n!}{p!q!r!}\)

Explanation

Simple Explanation

(p) brackets से (a), (q) से (b), और (r) से (c) चुनने का multinomial count है। परीक्षा में multinomial coefficient को repeated arrangement जैसा समझें। / It is the multinomial count of choosing (a) from (p) brackets, (b) from (q), and (c) from (r). In exams treat multinomial coefficients like repeated arrangements.

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((x+y+z)8) में \(x^3y^2z^3\) का coefficient कौन-सा है?

What is the coefficient of \(x^3y^2z^3\) in ((x+y+z)8)?

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Correct Answer

A. \(\frac{8!}{3!2!3!}\)

Explanation

Simple Explanation

Exponents का sum (8) है और coefficient multinomial form से मिलता है। परीक्षा में powers को group sizes मानें। / The exponents sum to (8), and the coefficient comes from the multinomial form. In exams treat powers as group sizes.

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((1+x)^n) के even-index coefficients का sum कैसे derive होता है?

How is the sum of even-index coefficients of ((1+x)^n) derived?

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Correct Answer

A. ((1+1)^n) और ((1-1)^n) को जोड़करBy adding ((1+1)^n) and ((1-1)^n)

Explanation

Simple Explanation

जोड़ने पर odd-index terms cancel हो जाते हैं। परीक्षा में even-odd binomial sums में (x=1) और (x=-1) साथ प्रयोग करें। / On adding, odd-index terms cancel out. In exams use (x=1) and (x=-1) together for even-odd binomial sums.

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\({}^{n}C_0+{}^{n}C_3+{}^{n}C_6+\cdots\) जैसे sums को अलग करने के लिए कौन-सा advanced method उपयोगी है?

Which advanced method is useful for separating sums like \({}^{n}C_0+{}^{n}C_3+{}^{n}C_6+\cdots\)?

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Correct Answer

A. Roots of unity filter

Explanation

Simple Explanation

Indices modulo (3) अलग करने के लिए unity roots filter प्रयोग किया जाता है। परीक्षा में ऐसी sums को सामान्य even-odd method से अलग पहचानें। / The roots of unity filter separates indices modulo (3). In exams recognize that such sums differ from the usual even-odd method.

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यदि \(^{n}C_{r-2}:{}^{n}C_{r-1}:{}^{n}C_r\) दिए हों, तो formula derivation में कौन-सा ratio सबसे पहले उपयोगी होगा?

If \(^{n}C_{r-2}:{}^{n}C_{r-1}:{}^{n}C_r\) is given, which ratio is most useful first in the derivation?

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Correct Answer

A. \(\frac{{}^{n}C_{r-1}}{{}^{n}C_{r-2}}=\frac{n-r+2}{r-1}\)

Explanation

Simple Explanation

Consecutive combinations में factorial cancellation से यह ratio मिलता है। परीक्षा में लंबी values निकालने के बजाय adjacent ratio लगाएं। / Factorial cancellation in consecutive combinations gives this ratio. In exams use adjacent ratios instead of calculating long values.

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यदि \(^{n}C_r\) maximum term है, तो \(\frac{{}^{n}C_{r+1}}{{}^{n}C_r}\) के लिए कौन-सी condition उपयोगी है?

If \(^{n}C_r\) is a maximum term, which condition is useful for \(\frac{{}^{n}C_{r+1}}{{}^{n}C_r}\)?

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Correct Answer

A. Ratio (1) से अधिक होने पर sequence बढ़ती है और (1) से कम होने पर घटती हैThe sequence increases when the ratio is greater than (1) and decreases when it is less than (1)

Explanation

Simple Explanation

Maximum के पास increasing से decreasing transition होता है। परीक्षा में binomial coefficient peak को ratio से locate करें। / Near the maximum, the sequence transitions from increasing to decreasing. In exams locate the binomial coefficient peak by ratios.

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जब (n) even हो, तो \({}^{n}C_r\) का unique maximum किस (r) पर होता है?

When (n) is even, at which (r) does \({}^{n}C_r\) have a unique maximum?

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Correct Answer

A. \(r=\frac{n}{2}\)

Explanation

Simple Explanation

Even (n) में middle index single होता है। परीक्षा में symmetry और ratio से central term पहचानें। / For even (n), the middle index is single. In exams identify the central term by symmetry and ratio.

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जब (n) odd हो, तो \({}^{n}C_r\) के maximum terms कौन-से होते हैं?

When (n) is odd, which terms are maximum for \({}^{n}C_r\)?

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Correct Answer

A. \({}^{n}C_{\frac{n-1}{2}}\) और \({}^{n}C_{\frac{n+1}{2}}\)\({}^{n}C_{\frac{n-1}{2}}\) and \({}^{n}C_{\frac{n+1}{2}}\)

Explanation

Simple Explanation

Odd (n) में दो central complementary indices बराबर maximum देते हैं। परीक्षा में odd case में दो middle terms याद रखें। / For odd (n), the two central complementary indices give equal maxima. In exams remember two middle terms in the odd case.

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यदि \({}^{20}C_{3r-1}={}^{20}C_{r+5}\) और indices equal नहीं हैं, तो (r) क्या होगा?

If \({}^{20}C_{3r-1}={}^{20}C_{r+5}\) and the indices are not equal, what is (r)?

Explanation opens after your attempt
Correct Answer

B. (4)

Explanation

Simple Explanation

Unequal equal-combination indices complementary होते हैं, इसलिए (3r-1+r+5=20)। परीक्षा में lower indices का sum upper index के बराबर करें। / Unequal equal-combination indices are complementary, so (3r-1+r+5=20). In exams set the sum of lower indices equal to the upper index.

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यदि \({}^{n}P_3=10{}^{n}P_2\), तो (n) क्या होगा?

If \({}^{n}P_3=10{}^{n}P_2\), what is (n)?

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Correct Answer

C. (12)

Explanation

Simple Explanation

({}^{n}P_3=(n-2){}^{n}P_2), इसलिए (n-2=10)। परीक्षा में consecutive permutation relation से जल्दी solve करें। / ({}^{n}P_3=(n-2){}^{n}P_2), so (n-2=10). In exams solve quickly using consecutive permutation relations.

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यदि \(\frac{{}^{n}C_{r+1}}{{}^{n}C_r}=\frac{2}{3}\), तो (n), (r) के बीच कौन-सा relation बनेगा?

If \(\frac{{}^{n}C_{r+1}}{{}^{n}C_r}=\frac{2}{3}\), what relation is formed between (n) and (r)?

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Correct Answer

A. (3n-5r=2)

Explanation

Simple Explanation

Ratio \(\frac{n-r}{r+1}=\frac{2}{3}\) से (3n-3r=2r+2) मिलता है। परीक्षा में consecutive combination ratios को cross multiply करें। / The ratio \(\frac{n-r}{r+1}=\frac{2}{3}\) gives (3n-3r=2r+2). In exams cross-multiply consecutive combination ratios.

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\(\sum_{r=0}^{n}{}^{n}C_r{}^{n-r}C_m\) का सही सरल रूप कौन-सा है?

What is the correct simplified form of \(\sum_{r=0}^{n}{}^{n}C_r{}^{n-r}C_m\)?

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Correct Answer

A. \(^{n}C_m2^{n-m}\)

Explanation

Simple Explanation

पहले (m) विशेष सदस्यों को दूसरे भाग में रखिए और बाकी (n-m) सदस्यों को स्वतंत्र विकल्प दीजिए। परीक्षा में ऐसे sums में fixed marked set पहले चुनें। / First place the (m) special members in the second part and give free choices to the remaining (n-m) members. In exams choose the fixed marked set first in such sums.

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\(\sum_{r=0}^{n}{}^{n}C_r{}^{r}C_3\) को किस रूप में लिखा जाएगा?

In which form is \(\sum_{r=0}^{n}{}^{n}C_r{}^{r}C_3\) written?

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Correct Answer

B. \(^{n}C_3 2^{n-3}\)

Explanation

Simple Explanation

पहले (3) चिह्नित सदस्य चुनें और बाकी (n-3) सदस्य subset में आएं या न आएं। परीक्षा में अंदर वाले चयन को पहले गिनना तेज होता है। / Choose the (3) marked members first and let each of the remaining (n-3) members enter the subset or not. In exams count the inner selection first.

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\(\sum_{r=0}^{n}r{}^{n}C_r x^r\) किस derivative identity से जुड़ा है?

Which derivative identity is connected with \(\sum_{r=0}^{n}r{}^{n}C_r x^r\)?

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Correct Answer

A. (nx(1+x)^{n-1})

Explanation

Simple Explanation

((1+x)^n) को differentiate करके (x) से multiply करने पर यह sum मिलता है। परीक्षा में (r) factor दिखे तो derivative method सोचें। / Differentiate ((1+x)^n) and multiply by (x) to obtain this sum. In exams think of the derivative method when a factor (r) appears.

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(\sum_{r=0}^{n}r(r-1){}^{n}C_r x^r) का सही रूप कौन-सा है?

What is the correct form of (\sum_{r=0}^{n}r(r-1){}^{n}C_r x^r)?

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Correct Answer

A. (n(n-1)x-2(1+x)^{n-2})

Explanation

Simple Explanation

दो बार differentiation करने पर (r(r-1)) factor आता है और \(x^2\) से power restore होती है। परीक्षा में (r(r-1)) के लिए second derivative लगाएं। / Two differentiations produce the factor (r(r-1)), and \(x^2\) restores the power. In exams use the second derivative for (r(r-1)).

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\(\sum_{r=0}^{n}r^3{}^{n}C_r\) निकालने में कौन-सा decomposition सबसे उपयोगी है?

Which decomposition is most useful for evaluating \(\sum_{r=0}^{n}r^3{}^{n}C_r\)?

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Correct Answer

A. (r-3=r(r-1)(r-2)+3r(r-1)+r)

Explanation

Simple Explanation

Powers को falling factorials में तोड़ने से standard binomial sums लगते हैं। परीक्षा में \(r^3\) को सीधे expand करने के बजाय falling form लिखें। / Breaking powers into falling factorials allows standard binomial sums. In exams write \(r^3\) in falling form instead of expanding directly.

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\({}^{n}C_r{}^{r}C_s={}^{n}C_s{}^{n-s}C_{r-s}\) का मुख्य कारण क्या है?

What is the main reason for \({}^{n}C_r{}^{r}C_s={}^{n}C_s{}^{n-s}C_{r-s}\)?

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A. पहले बड़ा समूह चुनना या पहले marked (s)-समूह चुनना एक ही कार्य हैChoosing the large group first or choosing the marked (s)-group first is the same task

Explanation

Simple Explanation

दोनों तरफ (r)-समूह के भीतर (s) विशेष सदस्यों वाला same selection गिना जाता है। परीक्षा में nested selection को order बदलकर देखें। / Both sides count the same selection with (s) special members inside an (r)-group. In exams change the order of nested selection.

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\({}^{n}C_a{}^{n-a}C_b{}^{n-a-b}C_c\) किस factorial form में बदलेगा?

Into which factorial form does \({}^{n}C_a{}^{n-a}C_b{}^{n-a-b}C_c\) convert?

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Correct Answer

A. (\frac{n!}{a!b!c!(n-a-b-c)!})

Explanation

Simple Explanation

Sequential selection में हर selected block का internal order हटता है। परीक्षा में कई labelled groups दिखें तो multinomial denominator बनाएं। / In sequential selection the internal order of each selected block is removed. In exams form a multinomial denominator when many labelled groups appear.

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यदि (12) distinct objects को (2,2,4,4) आकार के unlabelled groups में बांटना हो, तो extra division कौन-सी होगी?

If (12) distinct objects are divided into unlabelled groups of sizes (2,2,4,4), what is the extra division?

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Correct Answer

A. \(2!\cdot2!\)

Explanation

Simple Explanation

दो size (2) groups और दो size (4) groups की अदला-बदली duplicate देती है। परीक्षा में equal-size unlabelled groups के factorials से extra divide करें। / Interchanging the two size (2) groups and the two size (4) groups gives duplicates. In exams divide extra by factorials of equal-size unlabelled groups.

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(15) distinct objects को (5,5,5) के unlabelled groups में बांटने का formula कौन-सा है?

What is the formula for dividing (15) distinct objects into unlabelled groups of (5,5,5)?

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Correct Answer

B. (\frac{15!}{(5!)3 3!})

Explanation

Simple Explanation

तीनों groups same size और unlabelled हैं इसलिए group order (3!) भी हटता है। परीक्षा में equal unlabelled groups में extra (3!) याद रखें। / All three groups have the same size and are unlabelled, so group order (3!) is also removed. In exams remember the extra (3!) for equal unlabelled groups.

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कौन-सा formula (n) distinct objects को (r) labelled boxes में exactly (k) empty boxes के साथ distribute करता है?

Which formula distributes (n) distinct objects into (r) labelled boxes with exactly (k) empty boxes?

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Correct Answer

A. (^{r}C_k\sum_{i=0}^{r-k}(-1)^i{}^{r-k}C_i(r-k-i)^n)

Explanation

Simple Explanation

पहले empty boxes चुनें, फिर बाकी boxes में onto distribution करें। परीक्षा में exactly empty boxes में selection plus inclusion-exclusion लगाएं। / First choose the empty boxes, then distribute onto the remaining boxes. In exams use selection plus inclusion-exclusion for exactly empty boxes.

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(n) distinct objects को (4) labelled boxes में onto भेजने की संख्या कौन-सी है?

What is the number of onto distributions of (n) distinct objects into (4) labelled boxes?

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Correct Answer

A. \(4^n-4\cdot3^n+6\cdot2^n-4\)

Explanation

Simple Explanation

Empty boxes को inclusion-exclusion से घटाया और जोड़ा जाता है। परीक्षा में onto का मतलब हर labelled box non-empty समझें। / Empty boxes are subtracted and added by inclusion-exclusion. In exams interpret onto as every labelled box being non-empty.

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(n) distinct objects को (3) non-empty unlabelled groups में बांटने की संख्या किससे जुड़ती है?

The number of ways to divide (n) distinct objects into (3) non-empty unlabelled groups is connected with which form?

Explanation opens after your attempt
Correct Answer

A. (\frac{1}{3!}\left\(3^n-3\cdot2^n+3\right\))

Explanation

Simple Explanation

पहले (3) labelled non-empty groups गिनते हैं, फिर labels की (3!) अदला-बदली हटाते हैं। परीक्षा में unlabelled groups के लिए labelled count divide करें। / First count (3) labelled non-empty groups, then remove the (3!) label permutations. In exams divide the labelled count for unlabelled groups.

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\(x_1+x_2+x_3+x_4=30\) में \(x_1\geq2\), \(x_2\geq3\), \(x_3\geq4\), \(x_4\geq5\) हो, तो count क्या है?

In \(x_1+x_2+x_3+x_4=30\), if \(x_1\geq2\), \(x_2\geq3\), \(x_3\geq4\), \(x_4\geq5\), what is the count?

Explanation opens after your attempt
Correct Answer

A. \(^{19}C_3\)

Explanation

Simple Explanation

Minimum sum (14) हटाने पर (16) बचता है, इसलिए \({}^{16+4-1}C_{3}\) मिलता है। परीक्षा में unequal lower bounds पहले subtract करें। / After removing the minimum sum (14), (16) remains, so \({}^{16+4-1}C_{3}\) is obtained. In exams subtract unequal lower bounds first.

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\(x_1+x_2+x_3=24\) में \(0\leq x_i\leq8\) हो, तो valid count किस expression से मिलेगा?

If \(x_1+x_2+x_3=24\) and \(0\leq x_i\leq8\), which expression gives the valid count?

Explanation opens after your attempt
Correct Answer

B. (1)

Explanation

Simple Explanation

कुल (24) और तीन variables की maximum (8) होने से केवल ((8,8,8)) संभव है। परीक्षा में inclusion-exclusion से पहले extreme feasibility देखें। / Since the total is (24) and the maximum of each of the three variables is (8), only ((8,8,8)) is possible. In exams check extreme feasibility before inclusion-exclusion.

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\(x_1+x_2+x_3+x_4=17\) में \(0\leq x_i\leq5\) हो, तो valid count कौन-सा है?

If \(x_1+x_2+x_3+x_4=17\) and \(0\leq x_i\leq5\), what is the valid count?

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Correct Answer

A. \(^{20}C_3-4{}^{14}C_3+6{}^{8}C_3-4{}^{2}C_3\)

Explanation

Simple Explanation

Violation \(x_i\geq6\) से शुरू होती है और inclusion-exclusion लागू होता है। परीक्षा में upper bound (5) हो तो shift (6) लें। / A violation starts at \(x_i\geq6\), so inclusion-exclusion applies. In exams use shift (6) for upper bound (5).

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\(x_1+x_2+x_3+x_4+x_5=12\) में exactly (3) variables positive हों, तो count क्या होगा?

In \(x_1+x_2+x_3+x_4+x_5=12\), if exactly (3) variables are positive, what is the count?

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Correct Answer

A. \(^{5}C_3{}^{11}C_2\)

Explanation

Simple Explanation

पहले (3) positive variables चुनें, फिर (12) को (3) positive parts में बांटें। परीक्षा में exactly positive variables में choose variables plus positive stars-bars करें। / Choose the (3) positive variables first, then split (12) into (3) positive parts. In exams use choose variables plus positive stars and bars for exactly positive variables.

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(20) identical balls को (5) boxes में बांटना है और exactly (2) boxes empty हों। सही count कौन-सी है?

(20) identical balls are distributed into (5) boxes and exactly (2) boxes are empty. Which count is correct?

Explanation opens after your attempt
Correct Answer

A. \(^{5}C_2{}^{19}C_2\)

Explanation

Simple Explanation

पहले empty boxes चुनें, फिर बाकी (3) boxes में positive distribution करें। परीक्षा में exactly empty को boxes selection और positive distribution में तोड़ें। / First choose the empty boxes, then distribute positively into the remaining (3) boxes. In exams split exactly empty into box selection and positive distribution.

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\(D_n\) के लिए (D_n=nD_{n-1}+(-1)^n) किस formula से निकलता है?

From which formula does (D_n=nD_{n-1}+(-1)^n) arise for \(D_n\)?

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Correct Answer

A. Derangement inclusion-exclusion formula

Explanation

Simple Explanation

Derangement के alternating factorial expression को compare करने से यह recurrence मिलता है। परीक्षा में derangement recurrence के लिए inclusion-exclusion form याद रखें। / Comparing the alternating factorial expression for derangements gives this recurrence. In exams remember the inclusion-exclusion form for derangement recurrence.

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Exactly (2) fixed points वाले permutations of (7) objects की संख्या क्या है?

What is the number of permutations of (7) objects with exactly (2) fixed points?

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Correct Answer

A. \(^{7}C_2D_5\)

Explanation

Simple Explanation

पहले सही रहने वाले (2) objects चुनें और बाकी (5) objects derange करें। परीक्षा में exactly fixed points के लिए choose fixed plus derange rest लगाएं। / First choose the (2) objects that stay fixed and derange the remaining (5). In exams use choose fixed plus derange rest for exactly fixed points.

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(6) letters और (6) envelopes में कोई letter सही envelope में न जाए, तो count कौन-सी है?

For (6) letters and (6) envelopes, if no letter goes into its correct envelope, which count is correct?

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Correct Answer

A. \(D_6=265\)

Explanation

Simple Explanation

यह (6) objects का derangement है और \(D_6=265\) होता है। परीक्षा में letters-envelope mismatch को derangement समझें। / This is a derangement of (6) objects and \(D_6=265\). In exams treat letter-envelope mismatch as derangement.

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(n) distinct people को row में arrange करना है और (A) तथा (B) के बीच exactly (k) people हों। Count कौन-सी है?

(n) distinct people are arranged in a row and exactly (k) people are between (A) and (B). Which count is correct?

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Correct Answer

A. (2(n-k-1)(n-2)!)

Explanation

Simple Explanation

(A,B) की positions distance (k+1) पर होती हैं और order के (2) choices हैं। परीक्षा में fixed gap problems में positions first count करें। / The positions of (A,B) are at distance (k+1), and there are (2) choices for order. In exams count positions first in fixed-gap problems.

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(10) people की line में (A) और (B) के बीच exactly (3) people हों, तो count क्या है?

In a line of (10) people, if exactly (3) people are between (A) and (B), what is the count?

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Correct Answer

A. \(2\cdot6\cdot8!\)

Explanation

Simple Explanation

Positions के (10-3-1=6) choices और (A,B) order के (2) choices हैं। परीक्षा में between condition में position pairs गिनें। / There are (10-3-1=6) position choices and (2) choices for the order of (A,B). In exams count position pairs for between conditions.

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(9) distinct books को shelf पर रखना है और (3) specified books का relative order fixed हो। Count क्या है?

(9) distinct books are arranged on a shelf and the relative order of (3) specified books is fixed. What is the count?

Explanation opens after your attempt
Correct Answer

A. \(\frac{9!}{3!}\)

Explanation

Simple Explanation

उन (3) books के (3!) relative orders में केवल (1) allowed है। परीक्षा में fixed relative order में total को (k!) से divide करें। / Only (1) of the (3!) relative orders of those (3) books is allowed. In exams divide total by (k!) for fixed relative order.

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(12) people की line में (A) before (B), (C) before (D), और (E) before (F) हो, तो count क्या होगा?

In a line of (12) people, if (A) is before (B), (C) is before (D), and (E) is before (F), what is the count?

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Correct Answer

A. \(\frac{12!}{8}\)

Explanation

Simple Explanation

तीन स्वतंत्र before-after restrictions count को \(2^3\) से divide करती हैं। परीक्षा में independent pairs पर (2) की power से divide करें। / Three independent before-after restrictions divide the count by \(2^3\). In exams divide by a power of (2) for independent pairs.

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(n) people की circular seating में (A) और (B) के बीच exactly (k) people one direction में हों, तो core counting idea क्या है?

In circular seating of (n) people, if exactly (k) people lie between (A) and (B) in one direction, what is the core counting idea?

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Correct Answer

A. (A) को fix करके (B) की two possible circular positions देखेंFix (A) and check the two possible circular positions of (B)

Explanation

Simple Explanation

Circular rotation हटाने के बाद fixed distance वाली positions गिनी जाती हैं। परीक्षा में circular distance में पहले one object fix करें। / After removing circular rotation, positions with fixed distance are counted. In exams fix one object first for circular distance.

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(8) people को round table पर बैठाना है और (A) तथा (B) adjacent न हों। Count कौन-सी है?

(8) people are seated around a round table and (A) and (B) are not adjacent. Which count is correct?

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Correct Answer

A. \(7!-2\cdot6!\)

Explanation

Simple Explanation

Total circular arrangements (7!) हैं और adjacent block \(2\cdot6!\) ways में आता है। परीक्षा में circular not adjacent को complement से करें। / Total circular arrangements are (7!), and the adjacent block occurs in \(2\cdot6!\) ways. In exams handle circular not-adjacent by complement.

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(7) couples को round table पर बैठाना है और हर couple साथ रहे। Count क्या होगा?

(7) couples are seated around a round table and every couple stays together. What is the count?

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Correct Answer

A. \(6!\cdot2^7\)

Explanation

Simple Explanation

(7) couple-blocks की circular arrangement (6!) है और हर block में (2) internal orders हैं। परीक्षा में circular block count में blocks minus one factorial लें। / The circular arrangement of (7) couple-blocks is (6!), and each block has (2) internal orders. In exams use blocks minus one factorial for circular block count.

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(6) men और (6) women को round table पर alternate बैठाने की count क्या है?

What is the count for seating (6) men and (6) women alternately around a round table?

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Correct Answer

A. \(5!\cdot6!\)

Explanation

Simple Explanation

पहले men को circle में (5!) ways से बैठाएं, फिर (6) gaps में women को (6!) ways से रखें। परीक्षा में circular alternate में starting factor (2) न लगाएं। / Seat the men in a circle in (5!) ways, then place the women in the (6) gaps in (6!) ways. In exams do not add a starting factor (2) in circular alternation.

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(9) distinct beads को bracelet में arrange करने पर count क्या होगा?

What is the count for arranging (9) distinct beads in a bracelet?

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Correct Answer

A. \(\frac{8!}{2}\)

Explanation

Simple Explanation

Bracelet में rotations और reflections same माने जाते हैं। परीक्षा में bracelet count के लिए (\frac{(n-1)!}{2}) लगाएं। / In a bracelet, rotations and reflections are considered the same. In exams use (\frac{(n-1)!}{2}) for bracelet count.

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(0,1,2,3,4,5,6,7,8) से repetition बिना (5)-digit even numbers बनाते समय (0) last digit case में count क्या होगा?

Using (0,1,2,3,4,5,6,7,8) without repetition, what is the count for (5)-digit even numbers when (0) is the last digit?

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Correct Answer

A. \(^{8}P_4\)

Explanation

Simple Explanation

Last digit (0) fix होने पर बाकी (4) places में (8) non-zero digits से ordered selection होता है। परीक्षा में zero-last case leading restriction हटाता है। / When the last digit is fixed as (0), the remaining (4) places use ordered selection from (8) non-zero digits. In exams the zero-last case removes the leading restriction.

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Digits (0,1,2,3,4,5,6,7,8) से repetition बिना (5)-digit even numbers में non-zero even last digit case का count क्या है?

Using digits (0,1,2,3,4,5,6,7,8) without repetition, what is the count for (5)-digit even numbers with a non-zero even last digit?

Explanation opens after your attempt
Correct Answer

A. \(4\cdot7\cdot{}^{7}P_3\)

Explanation

Simple Explanation

Last digit के (4) choices हैं, first digit के (7) non-zero choices बचते हैं, फिर (3) places fill होती हैं। परीक्षा में zero और non-zero even cases अलग करें। / There are (4) choices for the last digit, (7) remaining non-zero choices for the first digit, and then (3) places are filled. In exams separate zero and non-zero even cases.

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Digits (1,2,3,4,5,6) से repetition allowed (5)-digit numbers में exactly (2) odd digits हों, तो count कौन-सी है?

Using digits (1,2,3,4,5,6) with repetition allowed, if exactly (2) odd digits occur in (5)-digit numbers, what is the count?

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Correct Answer

A. \(^{5}C_2\cdot3^2\cdot3^3\)

Explanation

Simple Explanation

Exactly (2) odd positions चुनें, फिर odd और even digits के choices multiply करें। परीक्षा में exactly type digit questions में positions first चुनें। / Choose exactly (2) odd positions, then multiply choices of odd and even digits. In exams choose positions first in exactly-type digit questions.

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Length (6) strings (4) symbols से बनती हैं और हर symbol कम से कम एक बार आए। Count कौन-सी है?

Length (6) strings are formed from (4) symbols and every symbol appears at least once. Which count is correct?

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Correct Answer

A. (\sum_{i=0}^{4}(-1)^i{}^{4}C_i(4-i)6)

Explanation

Simple Explanation

हर symbol का आना onto condition है, इसलिए missing symbols को inclusion-exclusion से हटाते हैं। परीक्षा में at least once को onto mapping समझें। / Every symbol appearing is an onto condition, so missing symbols are removed by inclusion-exclusion. In exams treat at least once as onto mapping.

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Length (7) strings (5) symbols से बनती हैं और exactly (3) distinct symbols use हों। सही form कौन-सी है?

Length (7) strings are formed from (5) symbols and exactly (3) distinct symbols are used. Which form is correct?

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Correct Answer

A. (^{5}C_3\left\(3^7-3\cdot2^7+3\right\))

Explanation

Simple Explanation

पहले (3) symbols चुनें, फिर (7) positions पर onto strings बनाएं। परीक्षा में exactly distinct symbols के लिए choose set plus onto count करें। / First choose (3) symbols, then form onto strings on (7) positions. In exams use choose set plus onto count for exactly distinct symbols.

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(n) distinct symbols से length (r) strings में exactly one repeated symbol और बाकी all distinct हों, तो count का core expression क्या होगा?

For length (r) strings from (n) distinct symbols with exactly one repeated symbol and all others distinct, what is the core expression?

Explanation opens after your attempt
Correct Answer

A. \(n\cdot{}^{n-1}C_{r-2}\cdot\frac{r!}{2!}\)

Explanation

Simple Explanation

Repeated symbol चुनें, बाकी (r-2) distinct symbols चुनें, फिर multiset arrange करें। परीक्षा में exactly one repeat में repeated item पहले fix करें। / Choose the repeated symbol, choose the remaining (r-2) distinct symbols, then arrange the multiset. In exams fix the repeated item first for exactly one repeat.

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((a+b+c+d)^n) में \(a^p b^q c^r d^s\) का coefficient कौन-सा है, यदि (p+q+r+s=n)?

What is the coefficient of \(a^p b^q c^r d^s\) in ((a+b+c+d)^n), if (p+q+r+s=n)?

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Correct Answer

A. \(\frac{n!}{p!q!r!s!}\)

Explanation

Simple Explanation

यह (n) brackets को (p,q,r,s) sizes में बांटने का multinomial count है। परीक्षा में exponents को group sizes मानें। / This is the multinomial count of dividing (n) brackets into sizes (p,q,r,s). In exams treat exponents as group sizes.

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((x+y+z)9) में \(x^4y^3z^2\) का coefficient क्या होगा?

What is the coefficient of \(x^4y^3z^2\) in ((x+y+z)9)?

Explanation opens after your attempt
Correct Answer

A. \(\frac{9!}{4!3!2!}\)

Explanation

Simple Explanation

Powers का sum (9) है और coefficient multinomial form से मिलता है। परीक्षा में multinomial term में repeated arrangements जैसा denominator रखें। / The powers sum to (9), and the coefficient comes from the multinomial form. In exams use a repeated-arrangement style denominator in multinomial terms.

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(\(1+x+x^2\)^n) में \(x^2\) के coefficient को किस count से derive करेंगे?

How will the coefficient of \(x^2\) in (\(1+x+x^2\)^n) be derived?

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Correct Answer

A. एक bracket से \(x^2\) या दो brackets से (x) चुननाChoose \(x^2\) from one bracket or (x) from two brackets

Explanation

Simple Explanation

\(x^2\) बनने के दो disjoint cases हैं। परीक्षा में polynomial expansion coefficients में exponent-sum cases बनाएं। / There are two disjoint cases to form \(x^2\). In exams make exponent-sum cases for polynomial expansion coefficients.

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(\(1+x+x^2\)8) में \(x^2\) का coefficient कौन-सा है?

What is the coefficient of \(x^2\) in (\(1+x+x^2\)8)?

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Correct Answer

A. \(^{8}C_1+{}^{8}C_2\)

Explanation

Simple Explanation

Case (1): एक \(x^2\) चुनें, case (2): दो (x) चुनें। परीक्षा में same power पाने वाले सभी cases जोड़ें। / Case (1): choose one \(x^2\), case (2): choose two (x)'s. In exams add all cases that produce the same power.

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((1+x)^n) में ऐसे coefficients का sum जिनके indices (3) से विभाज्य हैं, अलग करने की advanced technique क्या है?

What advanced technique separates the sum of coefficients in ((1+x)^n) whose indices are divisible by (3)?

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Correct Answer

A. Unity roots filter

Explanation

Simple Explanation

Modulo (3) classes अलग करने के लिए cube roots of unity का filter उपयोग होता है। परीक्षा में (3)-step coefficient sums को even-odd से अलग पहचानें। / The cube roots of unity filter is used to separate modulo (3) classes. In exams distinguish (3)-step coefficient sums from even-odd sums.

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\({}^{n}C_r\) के maximum term के लिए ratio test में कौन-सा ratio उपयोग होता है?

Which ratio is used in the ratio test for the maximum term of \({}^{n}C_r\)?

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Correct Answer

A. \(\frac{{}^{n}C_{r+1}}{{}^{n}C_r}=\frac{n-r}{r+1}\)

Explanation

Simple Explanation

Consecutive binomial coefficients में यह ratio बढ़ने और घटने की दिशा बताता है। परीक्षा में peak खोजने के लिए ratio को (1) से compare करें। / This ratio shows the direction of increase and decrease in consecutive binomial coefficients. In exams compare the ratio with (1) to locate the peak.

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यदि \({}^{n}C_{r+1}>{}^{n}C_r\), तो कौन-सी inequality सही है?

If \({}^{n}C_{r+1}>{}^{n}C_r\), which inequality is correct?

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Correct Answer

A. (n-r>r+1)

Explanation

Simple Explanation

Ratio \(\frac{n-r}{r+1}>1\) होना चाहिए। परीक्षा में monotonicity के लिए consecutive ratio को (1) से compare करें। / The ratio \(\frac{n-r}{r+1}>1\) must hold. In exams compare consecutive ratios with (1) for monotonicity.

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यदि \({}^{n}C_{r}= {}^{n}C_{r+4}\) और indices बराबर नहीं हैं, तो (r) और (n) का relation क्या है?

If \({}^{n}C_{r}= {}^{n}C_{r+4}\) and the indices are not equal, what is the relation between (r) and (n)?

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Correct Answer

A. (2r+4=n)

Explanation

Simple Explanation

Unequal equal-combination indices complementary होते हैं। परीक्षा में lower indices का sum upper index के बराबर करें। / Unequal equal-combination indices are complementary. In exams set the sum of lower indices equal to the upper index.

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यदि \({}^{24}C_{2r-1}={}^{24}C_{r+8}\) और lower indices unequal हैं, तो (r) क्या होगा?

If \({}^{24}C_{2r-1}={}^{24}C_{r+8}\) and the lower indices are unequal, what is (r)?

Explanation opens after your attempt
Correct Answer

B. (6)

Explanation

Simple Explanation

Complementary indices से (2r-1+r+8=24), इसलिए (r=6)। परीक्षा में equal combinations में same-index case और complement case अलग देखें। / Complementary indices give (2r-1+r+8=24), so (r=6). In exams check same-index and complement cases separately in equal combinations.

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यदि \({}^{n}P_4=12{}^{n}P_3\), तो (n) का मान क्या है?

If \({}^{n}P_4=12{}^{n}P_3\), what is the value of (n)?

Explanation opens after your attempt
Correct Answer

B. (15)

Explanation

Simple Explanation

({}^{n}P_4=(n-3){}^{n}P_3), इसलिए (n-3=12)। परीक्षा में consecutive permutation relation सीधे लगाएं। / ({}^{n}P_4=(n-3){}^{n}P_3), so (n-3=12). In exams apply the consecutive permutation relation directly.

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यदि \(\frac{{}^{n}C_{r+1}}{{}^{n}C_r}=\frac{3}{4}\), तो कौन-सा relation सही है?

If \(\frac{{}^{n}C_{r+1}}{{}^{n}C_r}=\frac{3}{4}\), which relation is correct?

Explanation opens after your attempt
Correct Answer

A. (4n-7r=3)

Explanation

Simple Explanation

\(\frac{n-r}{r+1}=\frac{3}{4}\) से (4n-4r=3r+3) मिलता है। परीक्षा में ratio equations को cross multiply करें। / From \(\frac{n-r}{r+1}=\frac{3}{4}\), we get (4n-4r=3r+3). In exams cross-multiply ratio equations.

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यदि \(\frac{{}^{n}C_{r}}{{}^{n}C_{r-1}}=\frac{5}{2}\), तो relation कौन-सा बनेगा?

If \(\frac{{}^{n}C_{r}}{{}^{n}C_{r-1}}=\frac{5}{2}\), which relation is formed?

Explanation opens after your attempt
Correct Answer

A. (2n-7r+2=0)

Explanation

Simple Explanation

Ratio \(\frac{{}^{n}C_r}{{}^{n}C_{r-1}}=\frac{n-r+1}{r}\) है। परीक्षा में consecutive combination ratio का सही direction रखें। / The ratio is \(\frac{{}^{n}C_r}{{}^{n}C_{r-1}}=\frac{n-r+1}{r}\). In exams keep the direction of consecutive combination ratios correct.

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(n) distinct objects में से (r) चुनने पर कम से कम (2) special objects चाहिए, और special objects (s) हैं। Complement expression कौन-सा है?

When choosing (r) objects from (n) distinct objects, at least (2) special objects are required and there are (s) special objects. Which complement expression is correct?

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Correct Answer

A. \(^{n}C_r-{}^{n-s}C_r-s{}^{n-s}C_{r-1}\)

Explanation

Simple Explanation

At least (2) special का complement (0) special या exactly (1) special है। परीक्षा में complement में सभी unwanted cases घटाएं। / The complement of at least (2) special is (0) special or exactly (1) special. In exams subtract all unwanted cases in the complement.

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(10) students में से (5) की committee बनानी है, (A) और (B) दोनों साथ या दोनों बाहर हों। Count क्या है?

A committee of (5) is formed from (10) students, and (A) and (B) are either both included or both excluded. What is the count?

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Correct Answer

A. \(^{8}C_3+{}^{8}C_5\)

Explanation

Simple Explanation

Case (1): दोनों शामिल हों, case (2): दोनों बाहर हों। परीक्षा में paired restriction को दो disjoint cases में तोड़ें। / Case (1): both are included, case (2): both are excluded. In exams split paired restrictions into two disjoint cases.

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(9) players में से (4) चुनने हैं, (A) और (B) साथ चयनित न हों। Count कौन-सी है?

(4) players are chosen from (9) players, and (A) and (B) are not selected together. Which count is correct?

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Correct Answer

A. \(^{9}C_4-{}^{7}C_2\)

Explanation

Simple Explanation

Total selections से (A,B) दोनों वाले selections घटते हैं। परीक्षा में not together selection में complement सबसे छोटा route है। / Subtract selections containing both (A,B) from total selections. In exams complement is the shortest route for not-together selection.

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\(\sum_{k=r}^{n}{}^{k}C_r={}^{n+1}C_{r+1}\) किस counting idea से सिद्ध होती है?

The identity \(\sum_{k=r}^{n}{}^{k}C_r={}^{n+1}C_{r+1}\) is proved by which counting idea?

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Correct Answer

A. सबसे बड़े चुने हुए element को fix करनाFixing the largest chosen element

Explanation

Simple Explanation

यह hockey-stick identity है और सबसे बड़ा selected element cases बनाता है। परीक्षा में ऐसी staircase sums में last element method लगाएं। / This is the hockey-stick identity and the largest selected element creates cases. In exams use the last-element method for such staircase sums.

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\(\sum_{k=0}^{r}{}^{m+k}C_k={}^{m+r+1}C_r\) को पहचानने का सबसे अच्छा तरीका क्या है?

What is the best way to recognize \(\sum_{k=0}^{r}{}^{m+k}C_k={}^{m+r+1}C_r\)?

Explanation opens after your attempt
Correct Answer

B. इसे hockey-stick identity का shifted रूप माननाTreat it as a shifted form of the hockey-stick identity

Explanation

Simple Explanation

Upper और lower indices साथ बढ़ रहे हैं इसलिए यह diagonal sum है। परीक्षा में diagonal combination sum दिखे तो hockey-stick सोचें। / The upper and lower indices increase together so it is a diagonal sum. In exams think of hockey-stick when a diagonal combination sum appears.

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\(\sum_{r=0}^{n}{}^{n}C_r{}^{r}C_2\) का combinatorial अर्थ क्या है?

What is the combinatorial meaning of \(\sum_{r=0}^{n}{}^{n}C_r{}^{r}C_2\)?

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Correct Answer

B. एक subset चुनकर उसमें unordered marked pair चुननाChoosing a subset and selecting an unordered marked pair inside it

Explanation

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पहले marked pair चुनें और बाकी elements freely choose करें। परीक्षा में \({}^{r}C_2\) दिखे तो pair marking सोचें। / Choose the marked pair first and choose the remaining elements freely. In exams think of pair marking when \({}^{r}C_2\) appears.

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\(\sum_{r=0}^{n}{}^{n}C_r{}^{r}C_3\) का simplified form कौन-सा है?

What is the simplified form of \(\sum_{r=0}^{n}{}^{n}C_r{}^{r}C_3\)?

Explanation opens after your attempt
Correct Answer

C. \(^{n}C_3 2^{n-3}\)

Explanation

Simple Explanation

पहले (3) marked members चुनते हैं और बाकी (n-3) members freely चुने जाते हैं। परीक्षा में inner combination को पहले count करें। / Choose the (3) marked members first and freely choose the remaining (n-3) members. In exams count the inner combination first.

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\(\sum_{r=0}^{n}r{}^{n}C_r{}^{r}C_2\) में \(r{}^{r}C_2\) को किस रूप में बदलना सबसे उपयोगी है?

In \(\sum_{r=0}^{n}r{}^{n}C_r{}^{r}C_2\), what is the most useful form of \(r{}^{r}C_2\)?

Explanation opens after your attempt
Correct Answer

D. \(2{}^{r}C_2+3{}^{r}C_3\)

Explanation

Simple Explanation

पहले pair mark हो और फिर एक member mark हो तो cases pair के अंदर या बाहर बनते हैं। परीक्षा में product of marks को cases में तोड़ें। / If a pair is marked and then one member is marked, cases occur inside or outside the pair. In exams split products of marks into cases.

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\({}^{n}C_a{}^{a}C_b{}^{b}C_c\) को सही order बदलकर कैसे लिखा जा सकता है?

How can \({}^{n}C_a{}^{a}C_b{}^{b}C_c\) be written by changing the order correctly?

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Correct Answer

A. \(^{n}C_c{}^{n-c}C_{b-c}{}^{n-b}C_{a-b}\)

Explanation

Simple Explanation

सबसे अंदर के (c) elements पहले चुनें और फिर layers जोड़ें। परीक्षा में nested choices को अंदर से बाहर count करें। / Choose the innermost (c) elements first and then add layers. In exams count nested choices from inside to outside.

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\({}^{n}C_r{}^{r}C_s\) को \({}^{n}C_s{}^{n-s}C_{r-s}\) लिखने में कौन-सा शर्त जरूरी है?

What condition is necessary to write \({}^{n}C_r{}^{r}C_s\) as \({}^{n}C_s{}^{n-s}C_{r-s}\)?

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Correct Answer

A. \(s\leq r\leq n\)

Explanation

Simple Explanation

छोटा selected set बड़े selected set के अंदर होना चाहिए। परीक्षा में nested combination में valid index order पहले check करें। / The smaller selected set must lie inside the larger selected set. In exams first check valid index order in nested combinations.

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(\sum_{r=0}^{n}(-1)^r r{}^{n}C_r) का मान (n>1) के लिए क्या है?

What is the value of (\sum_{r=0}^{n}(-1)^r r{}^{n}C_r) for (n>1)?

Explanation opens after your attempt
Correct Answer

B. (0)

Explanation

Simple Explanation

((1+x)^n) को differentiate करके (x=-1) रखने पर यह zero मिलता है। परीक्षा में alternating weighted sums में derivative plus (x=-1) लगाएं। / Differentiate ((1+x)^n) and put (x=-1) to get zero. In exams use derivative plus (x=-1) for alternating weighted sums.

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(\sum_{r=0}^{n}(-1)^r r(r-1){}^{n}C_r) का मान (n>2) के लिए क्या होगा?

What is (\sum_{r=0}^{n}(-1)^r r(r-1){}^{n}C_r) for (n>2)?

Explanation opens after your attempt
Correct Answer

C. (0)

Explanation

Simple Explanation

Second derivative के बाद (x=-1) रखने पर ((1-1)^{n-2}) आता है। परीक्षा में degree से कम falling factor हो तो zero check करें। / After the second derivative and putting (x=-1), ((1-1)^{n-2}) appears. In exams check zero when the falling factor is below the degree.

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\(\sum_{r=0}^{n}r^2{}^{n}C_r\) को derive करने में \(r^2\) का कौन-सा split सही है?

Which split of \(r^2\) is correct for deriving \(\sum_{r=0}^{n}r^2{}^{n}C_r\)?

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Correct Answer

A. (r-2=r(r-1)+r)

Explanation

Simple Explanation

यह split standard first और second weighted sums जोड़ देता है। परीक्षा में powers को falling factorials में बदलें। / This split connects first and second weighted sums. In exams convert powers into falling factorials.

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\(\sum_{r=0}^{n}r^2{}^{n}C_r\) का simplified result क्या है?

What is the simplified result of \(\sum_{r=0}^{n}r^2{}^{n}C_r\)?

Explanation opens after your attempt
Correct Answer

C. (n(n+1)2^{n-2})

Explanation

Simple Explanation

(r-2=r(r-1)+r) लगाने पर दोनों standard sums जुड़ते हैं। परीक्षा में final form को (n(n+1)2^{n-2}) तक simplify करें। / Using (r-2=r(r-1)+r) adds two standard sums. In exams simplify the final form to (n(n+1)2^{n-2}).

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\(\sum_{r=0}^{n}r^3{}^{n}C_r\) के लिए कौन-सा final form सही है?

Which final form is correct for \(\sum_{r=0}^{n}r^3{}^{n}C_r\)?

Explanation opens after your attempt
Correct Answer

A. (n-2(n+3)2^{n-3})

Explanation

Simple Explanation

(r-3=r(r-1)(r-2)+3r(r-1)+r) से यह form मिलता है। परीक्षा में cubic sums में falling factorial decomposition लगाएं। / Using (r-3=r(r-1)(r-2)+3r(r-1)+r) gives this form. In exams use falling factorial decomposition for cubic sums.

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यदि (n) distinct objects को (a,a,b) आकार के unlabelled groups में बांटा जाए और (2a+b=n), तो extra division किससे होगा?

If (n) distinct objects are divided into unlabelled groups of sizes (a,a,b) and (2a+b=n), what is the extra division?

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Correct Answer

C. (2!)

Explanation

Simple Explanation

दो groups का size (a) समान है इसलिए उनकी अदला-बदली duplicate देती है। परीक्षा में equal-size unlabelled groups पर extra factorial divide करें। / Two groups have the same size (a), so interchanging them gives duplicates. In exams divide extra by factorial for equal-size unlabelled groups.

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(14) distinct objects को (3,3,4,4) के unlabelled groups में बांटने का denominator कौन-सा होगा?

What is the denominator for dividing (14) distinct objects into unlabelled groups of sizes (3,3,4,4)?

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Correct Answer

A. (3!3!4!4!2!2!)

Explanation

Simple Explanation

Internal orders और equal-size group swaps दोनों हटते हैं। परीक्षा में denominator में group sizes और equal-group factorials दोनों लिखें। / Both internal orders and equal-size group swaps are removed. In exams write both group sizes and equal-group factorials in the denominator.

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(n) distinct objects को (r) labelled boxes में exactly (r-1) non-empty boxes में भेजने की count कौन-सी है?

What is the count for sending (n) distinct objects into (r) labelled boxes with exactly (r-1) non-empty boxes?

Explanation opens after your attempt
Correct Answer

A. (^{r}C_1\left((r-1)^n-{}^{r-1}C_1(r-2)^n+\cdots\right))

Explanation

Simple Explanation

पहले empty box चुनें और बाकी (r-1) boxes में onto distribution करें। परीक्षा में exactly non-empty boxes में choose empty plus onto count करें। / Choose the empty box first and then distribute onto the remaining (r-1) boxes. In exams use choose empty plus onto count for exactly non-empty boxes.

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(n) distinct objects को (4) labelled boxes में exactly (2) boxes empty रखकर distribute करने की count क्या है?

What is the count for distributing (n) distinct objects into (4) labelled boxes with exactly (2) boxes empty?

Explanation opens after your attempt
Correct Answer

A. (^{4}C_2\(2^n-2\))

Explanation

Simple Explanation

पहले empty boxes चुनें फिर बाकी दो boxes दोनों non-empty होने चाहिए। परीक्षा में exactly empty में remaining boxes पर onto लगाएं। / Choose the empty boxes first and then the remaining two boxes must both be non-empty. In exams apply onto to the remaining boxes in exactly-empty cases.

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(n) distinct objects को (3) labelled boxes में सभी boxes non-empty रखने की count क्या है?

What is the count for placing (n) distinct objects into (3) labelled boxes with all boxes non-empty?

Explanation opens after your attempt
Correct Answer

A. \(3^n-3\cdot2^n+3\)

Explanation

Simple Explanation

Inclusion-exclusion से empty box cases हटते हैं। परीक्षा में non-empty labelled boxes को onto functions की तरह गिनें। / Empty-box cases are removed by inclusion-exclusion. In exams count non-empty labelled boxes like onto functions.

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\(x_1+x_2+x_3+x_4=22\) में \(x_1\geq1\), \(x_2\geq2\), \(x_3\geq3\), \(x_4\geq4\) हो, तो count क्या है?

In \(x_1+x_2+x_3+x_4=22\), if \(x_1\geq1\), \(x_2\geq2\), \(x_3\geq3\), \(x_4\geq4\), what is the count?

Explanation opens after your attempt
Correct Answer

B. \(^{15}C_3\)

Explanation

Simple Explanation

Minimum sum (10) हटाने पर (12) बचता है और (4) variables में distribute होता है। परीक्षा में lower bounds subtract करके stars and bars लगाएं। / After removing the minimum sum (10), (12) remains and is distributed among (4) variables. In exams subtract lower bounds and use stars and bars.

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\(x_1+x_2+x_3=19\) में \(0\leq x_i\leq7\) हो, तो valid count किस expression से मिलेगा?

If \(x_1+x_2+x_3=19\) and \(0\leq x_i\leq7\), which expression gives the valid count?

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Correct Answer

A. \(^{21}C_2-3{}^{13}C_2+3{}^{5}C_2\)

Explanation

Simple Explanation

Upper violation \(x_i\geq8\) है और inclusion-exclusion applied होता है। परीक्षा में upper bound (7) के लिए shift (8) लें। / The upper violation is \(x_i\geq8\) and inclusion-exclusion is applied. In exams use shift (8) for upper bound (7).

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\(x_1+x_2+x_3+x_4=16\) में exactly (1) variable zero हो और बाकी positive हों, तो count क्या है?

In \(x_1+x_2+x_3+x_4=16\), if exactly (1) variable is zero and the rest are positive, what is the count?

Explanation opens after your attempt
Correct Answer

A. \(^{4}C_1{}^{15}C_2\)

Explanation

Simple Explanation

Zero variable चुनें और बाकी (3) variables में positive sum (16) बांटें। परीक्षा में exactly zero cases को positive distribution में बदलें। / Choose the zero variable and split positive sum (16) among the remaining (3) variables. In exams convert exactly-zero cases into positive distribution.

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(18) identical balls को (6) boxes में रखना है और exactly (3) boxes non-empty हों, तो count कौन-सी है?

(18) identical balls are placed into (6) boxes and exactly (3) boxes are non-empty. Which count is correct?

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Correct Answer

A. \(^{6}C_3{}^{17}C_2\)

Explanation

Simple Explanation

पहले (3) non-empty boxes चुनें फिर (18) balls को (3) positive parts में बांटें। परीक्षा में exactly non-empty को choose boxes plus positive stars-bars करें। / First choose the (3) non-empty boxes and then split (18) balls into (3) positive parts. In exams use choose boxes plus positive stars and bars for exactly non-empty.

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(D_n=(n-1)\(D_{n-1}+D_{n-2}\)) में \(D_{n-2}\) case किस स्थिति से आता है?

In (D_n=(n-1)\(D_{n-1}+D_{n-2}\)), from which situation does the \(D_{n-2}\) case arise?

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Correct Answer

B. जब पहला object और उसका target object एक दूसरे की जगह बदल लेंWhen the first object and its target object swap places

Explanation

Simple Explanation

यदि दो objects आपस में swap करते हैं तो बाकी (n-2) objects derange होते हैं। परीक्षा में derangement recurrence में swap और non-swap cases अलग करें। / If two objects swap with each other, the remaining (n-2) objects are deranged. In exams separate swap and non-swap cases in derangement recurrence.

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\(D_7\) का मान कौन-सा है?

What is the value of \(D_7\)?

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Correct Answer

A. (1854)

Explanation

Simple Explanation

Derangement recurrence से (D_7=6\(D_6+D_5\)=1854) मिलता है। परीक्षा में छोटे \(D_n\) values recurrence से याद रखें। / Using the derangement recurrence, (D_7=6\(D_6+D_5\)=1854). In exams remember small \(D_n\) values through recurrence.

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Exactly (3) fixed points वाले permutations of (8) objects की संख्या क्या है?

What is the number of permutations of (8) objects with exactly (3) fixed points?

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Correct Answer

A. \(^{8}C_3D_5\)

Explanation

Simple Explanation

पहले fixed (3) objects चुनें और बाकी (5) objects derange करें। परीक्षा में exactly fixed points के लिए choose fixed plus derange rest लगाएं। / First choose the (3) fixed objects and derange the remaining (5) objects. In exams use choose fixed plus derange rest for exactly fixed points.

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(7) letters और envelopes में exactly (2) letters सही envelope में जाएं, तो count क्या होगा?

With (7) letters and envelopes, if exactly (2) letters go into correct envelopes, what is the count?

Explanation opens after your attempt
Correct Answer

A. \(^{7}C_2D_5\)

Explanation

Simple Explanation

सही letters चुनें और बाकी letters को wrong envelopes में derange करें। परीक्षा में exactly correct letters को fixed-point formula से करें। / Choose the correct letters and derange the remaining letters into wrong envelopes. In exams solve exactly correct letters using the fixed-point formula.

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(n) people की line में (A) और (B) के बीच exactly (k) people हों, तो count क्या है?

In a line of (n) people, if exactly (k) people are between (A) and (B), what is the count?

Explanation opens after your attempt
Correct Answer

A. (2(n-k-1)(n-2)!)

Explanation

Simple Explanation

(A,B) के position pairs (n-k-1) हैं और order के (2) choices हैं। परीक्षा में fixed gap में positions पहले गिनें। / There are (n-k-1) position pairs for (A,B) and (2) choices for their order. In exams count positions first in fixed-gap problems.

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(11) people की line में (A) और (B) के बीच exactly (4) people हों, तो count क्या है?

In a line of (11) people, if exactly (4) people are between (A) and (B), what is the count?

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Correct Answer

A. \(2\cdot6\cdot9!\)

Explanation

Simple Explanation

Position pairs (11-4-1=6) हैं और (A,B) का order (2) तरीकों से हो सकता है। परीक्षा में बीच वाले people को अलग चुनने की जरूरत नहीं होती। / There are (11-4-1=6) position pairs and (A,B) can be ordered in (2) ways. In exams there is no need to separately choose the people between them.

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(10) people को round table पर बैठाना है और (A) तथा (B) के बीच clockwise exactly (3) people हों। Count क्या है?

(10) people are seated around a round table and exactly (3) people lie clockwise between (A) and (B). What is the count?

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Correct Answer

A. (8!)

Explanation

Simple Explanation

(A) को fix करने पर (B) की position fixed हो जाती है और बाकी (8) people arrange होते हैं। परीक्षा में one-direction circular gap में one person fix करें। / After fixing (A), the position of (B) is fixed and the remaining (8) people are arranged. In exams fix one person in one-direction circular gap problems.

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(9) people को round table पर बैठाना है और (A) तथा (B) adjacent न हों। Count क्या है?

(9) people are seated around a round table and (A) and (B) are not adjacent. What is the count?

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Correct Answer

A. \(8!-2\cdot7!\)

Explanation

Simple Explanation

Total circular arrangements (8!) हैं और adjacent block \(2\cdot7!\) ways देता है। परीक्षा में circular not-adjacent को complement से करें। / Total circular arrangements are (8!) and the adjacent block gives \(2\cdot7!\) ways. In exams handle circular not-adjacent by complement.

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(8) men और (8) women को round table पर alternate बैठाने की count कौन-सी है?

What is the count for seating (8) men and (8) women alternately around a round table?

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Correct Answer

A. \(7!\cdot8!\)

Explanation

Simple Explanation

पहले men को circularly (7!) ways में बैठाएं और gaps में women को (8!) ways में रखें। परीक्षा में circular alternate में extra (2) factor न लगाएं। / Seat the men circularly in (7!) ways and place the women in the gaps in (8!) ways. In exams do not add an extra factor (2) in circular alternation.

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(10) distinct beads की necklace arrangements में rotations same लेकिन reflections different हों, तो count क्या है?

For necklace arrangements of (10) distinct beads where rotations are the same but reflections are different, what is the count?

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Correct Answer

A. (9!)

Explanation

Simple Explanation

केवल rotations duplicate हैं इसलिए circular count ((10-1)!) है। परीक्षा में reflection condition पढ़कर ही (2) से divide करें। / Only rotations are duplicates, so the circular count is ((10-1)!). In exams divide by (2) only after reading the reflection condition.

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(10) distinct beads की bracelet arrangements में count क्या होगा?

What is the count for bracelet arrangements of (10) distinct beads?

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A. \(\frac{9!}{2}\)

Explanation

Simple Explanation

Bracelet में rotations और reflections दोनों same मानी जाती हैं। परीक्षा में bracelet के लिए (\frac{(n-1)!}{2}) use करें। / In a bracelet, both rotations and reflections are considered the same. In exams use (\frac{(n-1)!}{2}) for bracelets.

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Digits (0,1,2,3,4,5,6,7,8,9) से repetition बिना (6)-digit numbers बनते हैं और number even हो। (0) last digit case का count क्या है?

Using digits (0,1,2,3,4,5,6,7,8,9) without repetition, (6)-digit even numbers are formed. What is the count when (0) is the last digit?

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A. \(^{9}P_5\)

Explanation

Simple Explanation

Last digit (0) fix होने पर first place पर zero issue नहीं रहता और (9) non-zero digits से (5) places भरते हैं। परीक्षा में zero-last case अलग करें। / When the last digit is fixed as (0), there is no leading-zero issue and (5) places are filled from (9) non-zero digits. In exams separate the zero-last case.

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Digits (0) से (9) तक repetition बिना (6)-digit even numbers में non-zero even last digit case का count क्या होगा?

Using digits (0) to (9) without repetition, what is the count for (6)-digit even numbers with a non-zero even last digit?

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Correct Answer

A. \(4\cdot8\cdot{}^{8}P_4\)

Explanation

Simple Explanation

Last digit के (4) non-zero even choices हैं और first digit के (8) non-zero choices बचते हैं। परीक्षा में first और last restrictions को साथ संभालें। / There are (4) non-zero even choices for the last digit and (8) remaining non-zero choices for the first digit. In exams handle first and last restrictions together.

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Digits (1,2,3,4,5,6,7,8) से repetition allowed (6)-digit numbers में exactly (3) even digits हों, तो count क्या है?

Using digits (1,2,3,4,5,6,7,8) with repetition allowed, if exactly (3) even digits occur in (6)-digit numbers, what is the count?

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Correct Answer

A. \(^{6}C_3\cdot4^3\cdot4^3\)

Explanation

Simple Explanation

Even positions चुनें और फिर even तथा odd choices independently multiply करें। परीक्षा में exactly digit type में positions first choose करें। / Choose the even positions and then multiply even and odd choices independently. In exams choose positions first in exactly digit-type problems.

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Length (8) strings (5) symbols से बनती हैं और हर symbol कम से कम एक बार आए। Count का inclusion-exclusion form कौन-सा है?

Length (8) strings are formed from (5) symbols and every symbol appears at least once. Which inclusion-exclusion form gives the count?

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A. (\sum_{i=0}^{5}(-1)^i{}^{5}C_i(5-i)8)

Explanation

Simple Explanation

हर symbol का आना onto condition है और missing symbols हटते हैं। परीक्षा में at least once को inclusion-exclusion से करें। / Every symbol appearing is an onto condition and missing symbols are removed. In exams solve at least once by inclusion-exclusion.

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Length (9) strings (6) symbols से बनती हैं और exactly (4) distinct symbols use हों। सही count कौन-सी है?

Length (9) strings are formed from (6) symbols and exactly (4) distinct symbols are used. Which count is correct?

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Correct Answer

A. (^{6}C_4\sum_{i=0}^{4}(-1)^i{}^{4}C_i(4-i)9)

Explanation

Simple Explanation

पहले (4) symbols चुनें और फिर उन पर onto strings बनाएं। परीक्षा में exactly distinct symbols में choose set plus onto count करें। / First choose (4) symbols and then form onto strings on them. In exams use choose set plus onto count for exactly distinct symbols.

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Length (r) strings में exactly (s) distinct symbols use हों तो (s!,S(r,s)) किसे count करता है?

In length (r) strings with exactly (s) distinct symbols used, what does (s!,S(r,s)) count?

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A. (r) positions से selected (s) symbols पर onto assignmentsOnto assignments from (r) positions to selected (s) symbols

Explanation

Simple Explanation

Stirling part positions को non-empty groups में बांटता है और (s!) groups को symbols assign करता है। परीक्षा में exactly used symbols को onto mapping समझें। / The Stirling part partitions positions into non-empty groups and (s!) assigns groups to symbols. In exams treat exactly used symbols as onto mapping.

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((a+b+c+d)^{10}) में \(a^2b^3c^1d^4\) का coefficient क्या है?

What is the coefficient of \(a^2b^3c^1d^4\) in ((a+b+c+d)^{10})?

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Correct Answer

A. \(\frac{10!}{2!3!1!4!}\)

Explanation

Simple Explanation

Exponents का sum (10) है और coefficient multinomial form से मिलता है। परीक्षा में powers को group sizes मानें। / The exponents sum to (10) and the coefficient comes from the multinomial form. In exams treat powers as group sizes.

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(\(1+x+x^2\)^{10}) में \(x^3\) का coefficient किस expression से मिलेगा?

Which expression gives the coefficient of \(x^3\) in (\(1+x+x^2\)^{10})?

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Correct Answer

A. \(^{10}C_3+10\cdot9\)

Explanation

Simple Explanation

Cases हैं: तीन (x) चुनें या एक \(x^2\) और एक (x) चुनें। परीक्षा में same power बनाने वाले all cases जोड़ें। / The cases are: choose three (x)'s or choose one \(x^2\) and one (x). In exams add all cases that form the same power.

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(\(1+x+x^2\)^n) में \(x^4\) coefficient के cases में कौन-सा option सही है?

Which option correctly lists cases for the coefficient of \(x^4\) in (\(1+x+x^2\)^n)?

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Correct Answer

A. चार (x), या दो (x) और एक \(x^2\), या दो \(x^2\)Four (x)'s, or two (x)'s and one \(x^2\), or two \(x^2\)'s

Explanation

Simple Explanation

Exponent (4) बनाने वाले सभी disjoint choices जोड़ने पड़ते हैं। परीक्षा में polynomial coefficient में exponent partitions बनाएं। / All disjoint choices that form exponent (4) must be added. In exams make exponent partitions for polynomial coefficients.

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((1+x)^n) में indices \(0,3,6,\ldots\) वाले coefficients अलग करने के लिए कौन-सा method use होता है?

Which method is used to separate coefficients with indices \(0,3,6,\ldots\) in ((1+x)^n)?

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Correct Answer

A. Roots of unity filter

Explanation

Simple Explanation

Modulo (3) classes अलग करने के लिए cube roots of unity filter उपयोगी है। परीक्षा में three-step coefficient sums को advanced filter से पहचानें। / The cube roots of unity filter is useful for separating modulo (3) classes. In exams identify three-step coefficient sums with an advanced filter.

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यदि \({}^{n}C_{r+1}>{}^{n}C_r\), तो सही condition कौन-सी है?

If \({}^{n}C_{r+1}>{}^{n}C_r\), which condition is correct?

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Correct Answer

A. (n-r>r+1)

Explanation

Simple Explanation

Ratio \(\frac{n-r}{r+1}\) को (1) से बड़ा होना चाहिए। परीक्षा में increasing region ratio से identify करें। / The ratio \(\frac{n-r}{r+1}\) must be greater than (1). In exams identify the increasing region by ratio.

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यदि \({}^{n}C_{r+1}<{}^{n}C_r\), तो कौन-सी inequality सही है?

If \({}^{n}C_{r+1}<{}^{n}C_r\), which inequality is correct?

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Correct Answer

A. (n-r<r+1)

Explanation

Simple Explanation

Consecutive ratio (1) से कम होने पर coefficients घटने लगते हैं। परीक्षा में peak के बाद inequality उलटी हो जाती है। / When the consecutive ratio is less than (1), the coefficients start decreasing. In exams the inequality reverses after the peak.

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यदि \({}^{n}C_{r}= {}^{n}C_{r+6}\) और indices unequal हैं, तो relation कौन-सा है?

If \({}^{n}C_{r}= {}^{n}C_{r+6}\) and the indices are unequal, which relation is correct?

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Correct Answer

A. (2r+6=n)

Explanation

Simple Explanation

Unequal equal-combination indices complementary होते हैं। परीक्षा में lower indices का sum upper index के बराबर करें। / Unequal equal-combination indices are complementary. In exams set the sum of lower indices equal to the upper index.

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यदि \({}^{30}C_{3r-2}={}^{30}C_{r+8}\) और lower indices unequal हैं, तो (r) क्या है?

If \({}^{30}C_{3r-2}={}^{30}C_{r+8}\) and the lower indices are unequal, what is (r)?

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Correct Answer

A. (6)

Explanation

Simple Explanation

Complementary condition से (3r-2+r+8=30), इसलिए (r=6)। परीक्षा में same-index और complement cases अलग solve करें। / The complementary condition gives (3r-2+r+8=30), so (r=6). In exams solve same-index and complement cases separately.

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यदि \({}^{n}P_5=15{}^{n}P_4\), तो (n) का मान क्या होगा?

If \({}^{n}P_5=15{}^{n}P_4\), what is the value of (n)?

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Correct Answer

A. (19)

Explanation

Simple Explanation

({}^{n}P_5=(n-4){}^{n}P_4), इसलिए (n-4=15)। परीक्षा में consecutive permutation relation सीधे लगाएं। / ({}^{n}P_5=(n-4){}^{n}P_4), so (n-4=15). In exams apply the consecutive permutation relation directly.

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यदि \(\frac{{}^{n}C_{r+1}}{{}^{n}C_r}=\frac{4}{5}\), तो कौन-सा relation सही है?

If \(\frac{{}^{n}C_{r+1}}{{}^{n}C_r}=\frac{4}{5}\), which relation is correct?

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Correct Answer

A. (5n-9r=4)

Explanation

Simple Explanation

Ratio \(\frac{n-r}{r+1}=\frac{4}{5}\) से (5n-5r=4r+4) मिलता है। परीक्षा में combination ratio को cross multiply करें। / The ratio \(\frac{n-r}{r+1}=\frac{4}{5}\) gives (5n-5r=4r+4). In exams cross-multiply combination ratios.

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यदि \(\frac{{}^{n}C_{r}}{{}^{n}C_{r-1}}=\frac{7}{3}\), तो कौन-सा relation बनेगा?

If \(\frac{{}^{n}C_{r}}{{}^{n}C_{r-1}}=\frac{7}{3}\), which relation is formed?

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Correct Answer

A. (3n-10r+3=0)

Explanation

Simple Explanation

\(\frac{{}^{n}C_r}{{}^{n}C_{r-1}}=\frac{n-r+1}{r}\) होता है और cross multiplication से relation मिलता है। परीक्षा में ratio direction सही रखें। / \(\frac{{}^{n}C_r}{{}^{n}C_{r-1}}=\frac{n-r+1}{r}\), and cross multiplication gives the relation. In exams keep the ratio direction correct.

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(12) candidates में से (5) की team बनानी है और (A,B,C) में से कम से कम (2) selected हों। Count कौन-सी है?

A team of (5) is formed from (12) candidates and at least (2) of (A,B,C) are selected. Which count is correct?

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A. \(^{3}C_2{}^{9}C_3+{}^{3}C_3{}^{9}C_2\)

Explanation

Simple Explanation

At least (2) special के cases exactly (2) और exactly (3) हैं। परीक्षा में small special group में direct cases साफ रहते हैं। / The cases for at least (2) special are exactly (2) and exactly (3). In exams direct cases are clear for a small special group.

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