A. दो अलग समूहों से कुल (r) वस्तुएं चुनना/Choosing total (r) objects from two separate groups
Step 1
Concept
Choose (k) from the first group and (r-k) from the second, then add all cases. In exams identify Vandermonde in two-group selection.
Step 2
Why this answer is correct
The correct answer is A. दो अलग समूहों से कुल (r) वस्तुएं चुनना / Choosing total (r) objects from two separate groups. Choose (k) from the first group and (r-k) from the second, then add all cases. In exams identify Vandermonde in two-group selection.
Step 3
Exam Tip
पहले समूह से (k) और दूसरे से (r-k) चुनकर सभी cases जुड़ते हैं। परीक्षा में दो समूहों वाली selection में वेंडरमोंड पहचानें।
B. दो (n)-आकार के समूहों से कुल (n) वस्तुएं चुनना/Choosing total (n) objects from two groups of size (n)
Step 1
Concept
Choosing (n-r) from the second group equals \({}^{n}C_{n-r}={}^{n}C_r\). In exams connect square sums with two equal groups.
Step 2
Why this answer is correct
The correct answer is B. दो (n)-आकार के समूहों से कुल (n) वस्तुएं चुनना / Choosing total (n) objects from two groups of size (n). Choosing (n-r) from the second group equals \({}^{n}C_{n-r}={}^{n}C_r\). In exams connect square sums with two equal groups.
Step 3
Exam Tip
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।
Choose (r) objects first, then arrange both chosen and unchosen lists. In exams factorial cancellation also gives this quickly.
Step 2
Why this answer is correct
The correct answer is B. (n!). Choose (r) objects first, then arrange both chosen and unchosen lists. In exams factorial cancellation also gives this quickly.
Step 3
Exam Tip
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।
A. एक चुने हुए (r)-समूह में एक सदस्य को चिह्नित करना/Marking one member in a chosen (r)-group
Step 1
Concept
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.
Step 2
Why this answer is correct
The correct answer is A. एक चुने हुए (r)-समूह में एक सदस्य को चिह्नित करना / Marking one member in a chosen (r)-group. 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.
Step 3
Exam Tip
बाईं ओर समूह चुनकर mark करते हैं, दाईं ओर marked member पहले चुनते हैं। परीक्षा में (r) factor को marked choice मानें।
A. चुने हुए समूह में ordered दो marked सदस्य/Ordered two marked members in the selected group
Step 1
Concept
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)).
Step 2
Why this answer is correct
The correct answer is A. चुने हुए समूह में ordered दो marked सदस्य / Ordered two marked members in the selected group. 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)).
Step 3
Exam Tip
(r)-समूह में पहला और दूसरा marked member क्रम सहित चुने जाते हैं। परीक्षा में (r(r-1)) देखकर ordered marking सोचें।
A. marked member चुनने के बाद बाकी (n-1) members स्वतंत्र रूप से चुने या छोड़े जाते हैं/After choosing the marked member, the remaining (n-1) members are freely chosen or left
Step 1
Concept
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.
Step 2
Why this answer is correct
The correct answer is A. marked member चुनने के बाद बाकी (n-1) members स्वतंत्र रूप से चुने या छोड़े जाते हैं / After choosing the marked member, the remaining (n-1) members are freely chosen or left. 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.
Step 3
Exam Tip
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(n-1)) ways to choose two ordered marked members and the remaining (n-2) are chosen freely. In exams two marks lead to \(2^{n-2}\).
Step 2
Why this answer is correct
The correct answer is B. (n(n-1)2^{n-2}). 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 \(2^{n-2}\).
Step 3
Exam Tip
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।
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.
Step 2
Why this answer is correct
The correct answer is A. \(^{n}C_k2^{n-k}\). 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.
Step 3
Exam Tip
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।
First choose the innermost (t) members, then add (s-t), then (r-s) members. In exams count nested choices backward by layers.
Step 2
Why this answer is correct
The correct answer is A. \(^{n}C_t{}^{n-t}C_{s-t}{}^{n-s}C_{r-s}\). First choose the innermost (t) members, then add (s-t), then (r-s) members. In exams count nested choices backward by layers.
Step 3
Exam Tip
पहले सबसे अंदर के (t) members चुनें, फिर (s-t), फिर (r-s) members जोड़ें। परीक्षा में nested choices को layers में उल्टा गिनें।
Labelled groups are fixed and order inside each group is not counted. In exams put group-size factorials in the multinomial denominator.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{n!}{a!b!c!}\). Labelled groups are fixed and order inside each group is not counted. In exams put group-size factorials in the multinomial denominator.
Step 3
Exam Tip
Labelled groups fixed हैं और हर group के अंदर order नहीं गिना जाता। परीक्षा में multinomial denominator में group-size factorials रखें।
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.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{12!}{4!4!4!3!}\). 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.
Step 3
Exam Tip
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।
Choose (a) first, then (b) from the remaining, and (c) is automatically fixed. In exams match sequential selection with factorial form.
Step 2
Why this answer is correct
The correct answer is A. (a+b+c=n). Choose (a) first, then (b) from the remaining, and (c) is automatically fixed. In exams match sequential selection with factorial form.
Step 3
Exam Tip
पहले (a) चुनें, फिर बचे में से (b), और (c) अपने आप तय होता है। परीक्षा में sequential selection को factorial form से मिलाएं।
Cases with empty boxes are removed from total functions by inclusion-exclusion. In exams read onto as every box being non-empty.
Step 2
Why this answer is correct
The correct answer is A. (\sum_{k=0}^{r}(-1)^k{}^{r}C_k(r-k)^n). Cases with empty boxes are removed from total functions by inclusion-exclusion. In exams read onto as every box being non-empty.
Step 3
Exam Tip
Total functions से empty boxes वाले cases inclusion-exclusion द्वारा हटते हैं। परीक्षा में onto का अर्थ हर box non-empty समझें।
Three empty-box choices are subtracted, then over-subtraction of two empty boxes is added. In exams apply inclusion-exclusion carefully for onto distribution.
Step 2
Why this answer is correct
The correct answer is A. \(3^n-3\cdot2^n+3\). Three empty-box choices are subtracted, then over-subtraction of two empty boxes is added. In exams apply inclusion-exclusion carefully for onto distribution.
Step 3
Exam Tip
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।
From total \(2^n\) assignments, the two all-in-one empty-box cases are invalid. In exams subtract empty cases separately for labelled boxes.
Step 2
Why this answer is correct
The correct answer is A. \(2^n-2\). From total \(2^n\) assignments, the two all-in-one empty-box cases are invalid. In exams subtract empty cases separately for labelled boxes.
Step 3
Exam Tip
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।
In the labelled count, interchanging the two groups counts each division twice. In exams divide by (2) for two unlabelled groups.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{2^n-2}{2}\). In the labelled count, interchanging the two groups counts each division twice. In exams divide by (2) for two unlabelled groups.
Step 3
Exam Tip
Labelled count में दोनों groups की अदला-बदली दो बार गिनती है। परीक्षा में unlabelled दो groups के लिए (2) से divide 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.
Step 2
Why this answer is correct
The correct answer is A. Stars and bars. This is multiset selection and the answer is \({}^{n+r-1}C_r\). In exams handle repetition with no order by the bars method.
Step 3
Exam Tip
यह multiset selection है और answer \({}^{n+r-1}C_r\) होता है। परीक्षा में repetition with no order को bars method से करें।
Give (2) first to the four variables, leaving (17). In exams subtract the lower bound and apply non-negative stars and bars.
Step 2
Why this answer is correct
The correct answer is B. \(^{20}C_3\). Give (2) first to the four variables, leaving (17). In exams subtract the lower bound and apply non-negative stars and bars.
Step 3
Exam Tip
चार variables को पहले (2) देने पर (17) बचता है। परीक्षा में lower bound घटाकर non-negative stars and bars 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.
Step 2
Why this answer is correct
The correct answer is A. \(^{4}C_2{}^{17}C_1\). 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.
Step 3
Exam Tip
पहले zero variables चुनें, फिर बाकी दो variables positive sum (18) बनाते हैं। परीक्षा में exactly zero cases में positive distribution लगाएं।
Cases with \(x_i\geq7\) are subtracted and double violations are added. In exams shift by (7) for bounded solutions.
Step 2
Why this answer is correct
The correct answer is A. \(^{17}C_2-3{}^{10}C_2+3{}^{3}C_2\). Cases with \(x_i\geq7\) are subtracted and double violations are added. In exams shift by (7) for bounded solutions.
Step 3
Exam Tip
Upper bound तोड़ने पर \(x_i\geq7\) के cases घटते और double violations जुड़ते हैं। परीक्षा में bounded solutions में (7) shift करें।
A violation begins with \(x_i\geq5\), so inclusion-exclusion applies. In exams use a subtract shift of (5) for upper limit (4).
Step 2
Why this answer is correct
The correct answer is A. \(^{13}C_3-4{}^{8}C_3+6{}^{3}C_3\). A violation begins with \(x_i\geq5\), so inclusion-exclusion applies. In exams use a subtract shift of (5) for upper limit (4).
Step 3
Exam Tip
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. पहले letter की गलत position चुनने के लिए/To choose the wrong position of the first letter
Step 1
Concept
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.
Step 2
Why this answer is correct
The correct answer is A. पहले letter की गलत position चुनने के लिए / To choose the wrong position of the first letter. 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.
Step 3
Exam Tip
पहला letter अपनी original position छोड़कर (n-1) positions में जा सकता है। परीक्षा में derangement recurrence में first object की गलत choice देखें।
Permutations with fixed points are subtracted and added alternately. In exams think of inclusion-exclusion when no object is in its correct place.
Step 2
Why this answer is correct
The correct answer is A. Inclusion-exclusion. Permutations with fixed points are subtracted and added alternately. In exams think of inclusion-exclusion when no object is in its correct place.
Step 3
Exam Tip
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।
This is the derangement of (5) objects and \(D_5=44\). In exams treat letters-envelope mismatch as a derangement pattern.
Step 2
Why this answer is correct
The correct answer is A. \(D_5=44\). This is the derangement of (5) objects and \(D_5=44\). In exams treat letters-envelope mismatch as a derangement pattern.
Step 3
Exam Tip
यह (5) objects का derangement है और \(D_5=44\) होता है। परीक्षा में letters-envelopes mismatch को derangement pattern मानें।
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.
Step 2
Why this answer is correct
The correct answer is A. \(^{n}C_kD_{n-k}\). 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.
Step 3
Exam Tip
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।
A. हर circular arrangement (n) rotations से linear arrangements में गिनी जाती है/Each circular arrangement is counted as (n) rotations in linear arrangements
Step 1
Concept
Rotations are duplicates in the linear count (n!). In exams divide rotational overcount by (n) in circular arrangements.
Step 2
Why this answer is correct
The correct answer is A. हर circular arrangement (n) rotations से linear arrangements में गिनी जाती है / Each circular arrangement is counted as (n) rotations in linear arrangements. Rotations are duplicates in the linear count (n!). In exams divide rotational overcount by (n) in circular arrangements.
Step 3
Exam Tip
Linear (n!) count में rotations duplicate होते हैं। परीक्षा में circular arrangement में rotational overcount को (n) से divide करें।
A. जब rotations और reflections दोनों same माने जाएं/When both rotations and reflections are considered the same
Step 1
Concept
In a bracelet, mirror images are considered the same. In exams divide the circular count by (2) when reflection is the same.
Step 2
Why this answer is correct
The correct answer is A. जब rotations और reflections दोनों same माने जाएं / When both rotations and reflections are considered the same. In a bracelet, mirror images are considered the same. In exams divide the circular count by (2) when reflection is the same.
Step 3
Exam Tip
Bracelet में mirror images same मानी जाती हैं। परीक्षा में reflection same होने पर circular count को (2) से divide करें।
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.
Step 2
Why this answer is correct
The correct answer is A. \(5!\cdot3!\). 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.
Step 3
Exam Tip
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।
Subtract adjacent block arrangements of (A,B) from total circular arrangements. In exams handle circular not-adjacent by complement.
Step 2
Why this answer is correct
The correct answer is A. ((n-1)!-2(n-2)!). Subtract adjacent block arrangements of (A,B) from total circular arrangements. In exams handle circular not-adjacent by complement.
Step 3
Exam Tip
Total circular arrangements से (A,B) adjacent block के arrangements घटते हैं। परीक्षा में circular not adjacent को complement से 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.
Step 2
Why this answer is correct
The correct answer is A. (n!(n-1)!). 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.
Step 3
Exam Tip
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।
Only (1) of the (k!) relative orders of the special objects is allowed. In exams divide total arrangements by (k!) for fixed relative order.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{n!}{k!}\). Only (1) of the (k!) relative orders of the special objects is allowed. In exams divide total arrangements by (k!) for fixed relative order.
Step 3
Exam Tip
Special objects के (k!) relative orders में केवल (1) allowed है। परीक्षा में fixed relative order में total arrangements को (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.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{10!}{4}\). Each of the two independent relative-order restrictions halves the count. In exams divide by \(2^k\) for independent before-after pairs.
Step 3
Exam Tip
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।
A. क्योंकि (0) last में valid है पर first में invalid है/Because (0) is valid at the last place but invalid at the first place
Step 1
Concept
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.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (0) last में valid है पर first में invalid है / Because (0) is valid at the last place but invalid at the first place. 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.
Step 3
Exam Tip
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।
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).
Step 2
Why this answer is correct
The correct answer is A. \(6\cdot5\cdot4+5\cdot5\cdot4\). 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).
Step 3
Exam Tip
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।
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.
Step 2
Why this answer is correct
The correct answer is A. \(^{4}C_2\cdot3^2\cdot4^2\). 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.
Step 3
Exam Tip
Even positions चुनें, उन पर (3) even choices और बाकी पर (4) odd choices हैं। परीक्षा में exactly condition में positions first चुनें।
Subtract all-distinct strings from total repetition-allowed strings. In exams solve at least repeat by the no-repeat complement.
Step 2
Why this answer is correct
The correct answer is A. \(n^r-{}^{n}P_r\). Subtract all-distinct strings from total repetition-allowed strings. In exams solve at least repeat by the no-repeat complement.
Step 3
Exam Tip
Total repeated-allowed strings से all distinct strings घटाएं। परीक्षा में at least repeat को complement no repeat से करें।
First choose (s) symbols, then map the (r) positions onto those (s) symbols. In exams use the onto idea for exactly distinct symbols.
Step 2
Why this answer is correct
The correct answer is A. (^{n}C_s s! S(r,s)). First choose (s) symbols, then map the (r) positions onto those (s) symbols. In exams use the onto idea for exactly distinct symbols.
Step 3
Exam Tip
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।
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.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{n!}{p!q!r!}\). 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.
Step 3
Exam Tip
(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।
The exponents sum to (8), and the coefficient comes from the multinomial form. In exams treat powers as group sizes.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{8!}{3!2!3!}\). The exponents sum to (8), and the coefficient comes from the multinomial form. In exams treat powers as group sizes.
Step 3
Exam Tip
Exponents का sum (8) है और coefficient multinomial form से मिलता है। परीक्षा में powers को group sizes मानें।
A. ((1+1)^n) और ((1-1)^n) को जोड़कर/By adding ((1+1)^n) and ((1-1)^n)
Step 1
Concept
On adding, odd-index terms cancel out. In exams use (x=1) and (x=-1) together for even-odd binomial sums.
Step 2
Why this answer is correct
The correct answer is A. ((1+1)^n) और ((1-1)^n) को जोड़कर / By adding ((1+1)^n) and ((1-1)^n). On adding, odd-index terms cancel out. In exams use (x=1) and (x=-1) together for even-odd binomial sums.
Step 3
Exam Tip
जोड़ने पर odd-index terms cancel हो जाते हैं। परीक्षा में even-odd binomial sums में (x=1) और (x=-1) साथ प्रयोग करें।
The roots of unity filter separates indices modulo (3). In exams recognize that such sums differ from the usual even-odd method.
Step 2
Why this answer is correct
The correct answer is A. Roots of unity filter. The roots of unity filter separates indices modulo (3). In exams recognize that such sums differ from the usual even-odd method.
Step 3
Exam Tip
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।
A. \(\frac{{}^{n}C_{r-1}}{{}^{n}C_{r-2}}=\frac{n-r+2}{r-1}\)
Step 1
Concept
Factorial cancellation in consecutive combinations gives this ratio. In exams use adjacent ratios instead of calculating long values.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{{}^{n}C_{r-1}}{{}^{n}C_{r-2}}=\frac{n-r+2}{r-1}\). Factorial cancellation in consecutive combinations gives this ratio. In exams use adjacent ratios instead of calculating long values.
Step 3
Exam Tip
Consecutive combinations में factorial cancellation से यह ratio मिलता है। परीक्षा में लंबी values निकालने के बजाय adjacent ratio लगाएं।
A. Ratio (1) से अधिक होने पर sequence बढ़ती है और (1) से कम होने पर घटती है/The sequence increases when the ratio is greater than (1) and decreases when it is less than (1)
Step 1
Concept
Near the maximum, the sequence transitions from increasing to decreasing. In exams locate the binomial coefficient peak by ratios.
Step 2
Why this answer is correct
The correct answer is A. Ratio (1) से अधिक होने पर sequence बढ़ती है और (1) से कम होने पर घटती है / The sequence increases when the ratio is greater than (1) and decreases when it is less than (1). Near the maximum, the sequence transitions from increasing to decreasing. In exams locate the binomial coefficient peak by ratios.
Step 3
Exam Tip
Maximum के पास increasing से decreasing transition होता है। परीक्षा में binomial coefficient peak को ratio से locate करें।
A. \({}^{n}C_{\frac{n-1}{2}}\) और \({}^{n}C_{\frac{n+1}{2}}\)/\({}^{n}C_{\frac{n-1}{2}}\) and \({}^{n}C_{\frac{n+1}{2}}\)
Step 1
Concept
For odd (n), the two central complementary indices give equal maxima. In exams remember two middle terms in the odd case.
Step 2
Why this answer is correct
The correct answer is A. \({}^{n}C_{\frac{n-1}{2}}\) और \({}^{n}C_{\frac{n+1}{2}}\) / \({}^{n}C_{\frac{n-1}{2}}\) and \({}^{n}C_{\frac{n+1}{2}}\). For odd (n), the two central complementary indices give equal maxima. In exams remember two middle terms in the odd case.
Step 3
Exam Tip
Odd (n) में दो central complementary indices बराबर maximum देते हैं। परीक्षा में odd case में दो middle terms याद रखें।
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.
Step 2
Why this answer is correct
The correct answer is B. (4). 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.
Step 3
Exam Tip
Unequal equal-combination indices complementary होते हैं, इसलिए (3r-1+r+5=20)। परीक्षा में lower indices का sum upper index के बराबर करें।
The ratio \(\frac{n-r}{r+1}=\frac{2}{3}\) gives (3n-3r=2r+2). In exams cross-multiply consecutive combination ratios.
Step 2
Why this answer is correct
The correct answer is A. (3n-5r=2). The ratio \(\frac{n-r}{r+1}=\frac{2}{3}\) gives (3n-3r=2r+2). In exams cross-multiply consecutive combination ratios.
Step 3
Exam Tip
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।
