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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?

Explanation opens after your attempt
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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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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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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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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