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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.
Step 2
Why this answer is correct
The correct answer is A. \(^{n}C_m2^{n-m}\). 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.
Step 3
Exam Tip
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।
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.
Step 2
Why this answer is correct
The correct answer is B. \(^{n}C_3 2^{n-3}\). 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.
Step 3
Exam Tip
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।
Differentiate ((1+x)^n) and multiply by (x) to obtain this sum. In exams think of the derivative method when a factor (r) appears.
Step 2
Why this answer is correct
The correct answer is A. (nx(1+x)^{n-1}). Differentiate ((1+x)^n) and multiply by (x) to obtain this sum. In exams think of the derivative method when a factor (r) appears.
Step 3
Exam Tip
((1+x)^n) को differentiate करके (x) से multiply करने पर यह sum मिलता है। परीक्षा में (r) factor दिखे तो derivative method सोचें।
Two differentiations produce the factor (r(r-1)), and \(x^2\) restores the power. In exams use the second derivative for (r(r-1)).
Step 2
Why this answer is correct
The correct answer is A. (n(n-1)x-2(1+x)^{n-2}). Two differentiations produce the factor (r(r-1)), and \(x^2\) restores the power. In exams use the second derivative for (r(r-1)).
Step 3
Exam Tip
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।
Breaking powers into falling factorials allows standard binomial sums. In exams write \(r^3\) in falling form instead of expanding directly.
Step 2
Why this answer is correct
The correct answer is A. (r-3=r(r-1)(r-2)+3r(r-1)+r). Breaking powers into falling factorials allows standard binomial sums. In exams write \(r^3\) in falling form instead of expanding directly.
Step 3
Exam Tip
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।
A. पहले बड़ा समूह चुनना या पहले marked (s)-समूह चुनना एक ही कार्य है/Choosing the large group first or choosing the marked (s)-group first is the same task
Step 1
Concept
Both sides count the same selection with (s) special members inside an (r)-group. In exams change the order of nested selection.
Step 2
Why this answer is correct
The correct answer is A. पहले बड़ा समूह चुनना या पहले marked (s)-समूह चुनना एक ही कार्य है / Choosing the large group first or choosing the marked (s)-group first is the same task. Both sides count the same selection with (s) special members inside an (r)-group. In exams change the order of nested selection.
Step 3
Exam Tip
दोनों तरफ (r)-समूह के भीतर (s) विशेष सदस्यों वाला same selection गिना जाता है। परीक्षा में nested selection को order बदलकर देखें।
In sequential selection the internal order of each selected block is removed. In exams form a multinomial denominator when many labelled groups appear.
Step 2
Why this answer is correct
The correct answer is A. (\frac{n!}{a!b!c!(n-a-b-c)!}). In sequential selection the internal order of each selected block is removed. In exams form a multinomial denominator when many labelled groups appear.
Step 3
Exam Tip
Sequential selection में हर selected block का internal order हटता है। परीक्षा में कई labelled groups दिखें तो multinomial denominator बनाएं।
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.
Step 2
Why this answer is correct
The correct answer is A. \(2!\cdot2!\). 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.
Step 3
Exam Tip
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।
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.
Step 2
Why this answer is correct
The correct answer is B. (\frac{15!}{(5!)3 3!}). 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.
Step 3
Exam Tip
तीनों groups same size और unlabelled हैं इसलिए group order (3!) भी हटता है। परीक्षा में equal unlabelled groups में extra (3!) याद रखें।
A. (^{r}C_k\sum_{i=0}^{r-k}(-1)^i{}^{r-k}C_i(r-k-i)^n)
Step 1
Concept
First choose the empty boxes, then distribute onto the remaining boxes. In exams use selection plus inclusion-exclusion for exactly empty boxes.
Step 2
Why this answer is correct
The correct answer is A. (^{r}C_k\sum_{i=0}^{r-k}(-1)^i{}^{r-k}C_i(r-k-i)^n). First choose the empty boxes, then distribute onto the remaining boxes. In exams use selection plus inclusion-exclusion for exactly empty boxes.
Step 3
Exam Tip
पहले empty boxes चुनें, फिर बाकी boxes में onto distribution करें। परीक्षा में exactly empty boxes में selection plus inclusion-exclusion लगाएं।
Empty boxes are subtracted and added by inclusion-exclusion. In exams interpret onto as every labelled box being non-empty.
Step 2
Why this answer is correct
The correct answer is A. \(4^n-4\cdot3^n+6\cdot2^n-4\). Empty boxes are subtracted and added by inclusion-exclusion. In exams interpret onto as every labelled box being non-empty.
Step 3
Exam Tip
Empty boxes को inclusion-exclusion से घटाया और जोड़ा जाता है। परीक्षा में onto का मतलब हर labelled box non-empty समझें।
First count (3) labelled non-empty groups, then remove the (3!) label permutations. In exams divide the labelled count for unlabelled groups.
Step 2
Why this answer is correct
The correct answer is A. (\frac{1}{3!}\left\(3^n-3\cdot2^n+3\right\)). First count (3) labelled non-empty groups, then remove the (3!) label permutations. In exams divide the labelled count for unlabelled groups.
Step 3
Exam Tip
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।
After removing the minimum sum (14), (16) remains, so \({}^{16+4-1}C_{3}\) is obtained. In exams subtract unequal lower bounds first.
Step 2
Why this answer is correct
The correct answer is A. \(^{19}C_3\). After removing the minimum sum (14), (16) remains, so \({}^{16+4-1}C_{3}\) is obtained. In exams subtract unequal lower bounds first.
Step 3
Exam Tip
Minimum sum (14) हटाने पर (16) बचता है, इसलिए \({}^{16+4-1}C_{3}\) मिलता है। परीक्षा में unequal lower bounds पहले subtract करें।
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.
Step 2
Why this answer is correct
The correct answer is B. (1). 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.
Step 3
Exam Tip
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।
A violation starts at \(x_i\geq6\), so inclusion-exclusion applies. In exams use shift (6) for upper bound (5).
Step 2
Why this answer is correct
The correct answer is A. \(^{20}C_3-4{}^{14}C_3+6{}^{8}C_3-4{}^{2}C_3\). A violation starts at \(x_i\geq6\), so inclusion-exclusion applies. In exams use shift (6) for upper bound (5).
Step 3
Exam Tip
Violation \(x_i\geq6\) से शुरू होती है और inclusion-exclusion लागू होता है। परीक्षा में upper bound (5) हो तो shift (6) लें।
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.
Step 2
Why this answer is correct
The correct answer is A. \(^{5}C_3{}^{11}C_2\). 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.
Step 3
Exam Tip
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।
First choose the empty boxes, then distribute positively into the remaining (3) boxes. In exams split exactly empty into box selection and positive distribution.
Step 2
Why this answer is correct
The correct answer is A. \(^{5}C_2{}^{19}C_2\). First choose the empty boxes, then distribute positively into the remaining (3) boxes. In exams split exactly empty into box selection and positive distribution.
Step 3
Exam Tip
पहले empty boxes चुनें, फिर बाकी (3) boxes में positive distribution करें। परीक्षा में exactly empty को boxes selection और positive distribution में तोड़ें।
Comparing the alternating factorial expression for derangements gives this recurrence. In exams remember the inclusion-exclusion form for derangement recurrence.
Step 2
Why this answer is correct
The correct answer is A. Derangement inclusion-exclusion formula. Comparing the alternating factorial expression for derangements gives this recurrence. In exams remember the inclusion-exclusion form for derangement recurrence.
Step 3
Exam Tip
Derangement के alternating factorial expression को compare करने से यह recurrence मिलता है। परीक्षा में derangement recurrence के लिए inclusion-exclusion form याद रखें।
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.
Step 2
Why this answer is correct
The correct answer is A. \(^{7}C_2D_5\). 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.
Step 3
Exam Tip
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।
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.
Step 2
Why this answer is correct
The correct answer is A. (2(n-k-1)(n-2)!). 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.
Step 3
Exam Tip
(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।
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.
Step 2
Why this answer is correct
The correct answer is A. \(2\cdot6\cdot8!\). 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.
Step 3
Exam Tip
Positions के (10-3-1=6) choices और (A,B) order के (2) choices हैं। परीक्षा में between condition में position pairs गिनें।
Only (1) of the (3!) relative orders of those (3) books is allowed. In exams divide total by (k!) for fixed relative order.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{9!}{3!}\). Only (1) of the (3!) relative orders of those (3) books is allowed. In exams divide total by (k!) for fixed relative order.
Step 3
Exam Tip
उन (3) books के (3!) relative orders में केवल (1) allowed है। परीक्षा में fixed relative order में total को (k!) 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.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{12!}{8}\). Three independent before-after restrictions divide the count by \(2^3\). In exams divide by a power of (2) for independent pairs.
Step 3
Exam Tip
तीन स्वतंत्र before-after restrictions count को \(2^3\) से divide करती हैं। परीक्षा में independent pairs पर (2) की power से divide करें।
A. (A) को fix करके (B) की two possible circular positions देखें/Fix (A) and check the two possible circular positions of (B)
Step 1
Concept
After removing circular rotation, positions with fixed distance are counted. In exams fix one object first for circular distance.
Step 2
Why this answer is correct
The correct answer is A. (A) को fix करके (B) की two possible circular positions देखें / Fix (A) and check the two possible circular positions of (B). After removing circular rotation, positions with fixed distance are counted. In exams fix one object first for circular distance.
Step 3
Exam Tip
Circular rotation हटाने के बाद fixed distance वाली positions गिनी जाती हैं। परीक्षा में circular distance में पहले one object fix करें।
Total circular arrangements are (7!), and the adjacent block occurs in \(2\cdot6!\) ways. In exams handle circular not-adjacent by complement.
Step 2
Why this answer is correct
The correct answer is A. \(7!-2\cdot6!\). Total circular arrangements are (7!), and the adjacent block occurs in \(2\cdot6!\) ways. In exams handle circular not-adjacent by complement.
Step 3
Exam Tip
Total circular arrangements (7!) हैं और adjacent block \(2\cdot6!\) ways में आता है। परीक्षा में circular not adjacent को complement से करें।
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.
Step 2
Why this answer is correct
The correct answer is A. \(6!\cdot2^7\). 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.
Step 3
Exam Tip
(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।
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.
Step 2
Why this answer is correct
The correct answer is A. \(5!\cdot6!\). 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.
Step 3
Exam Tip
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।
In a bracelet, rotations and reflections are considered the same. In exams use (\frac{(n-1)!}{2}) for bracelet count.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{8!}{2}\). In a bracelet, rotations and reflections are considered the same. In exams use (\frac{(n-1)!}{2}) for bracelet count.
Step 3
Exam Tip
Bracelet में rotations और reflections same माने जाते हैं। परीक्षा में bracelet count के लिए (\frac{(n-1)!}{2}) लगाएं।
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.
Step 2
Why this answer is correct
The correct answer is A. \(^{8}P_4\). 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.
Step 3
Exam Tip
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।
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.
Step 2
Why this answer is correct
The correct answer is A. \(4\cdot7\cdot{}^{7}P_3\). 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.
Step 3
Exam Tip
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।
Choose exactly (2) odd positions, then multiply choices of odd and even digits. In exams choose positions first in exactly-type digit questions.
Step 2
Why this answer is correct
The correct answer is A. \(^{5}C_2\cdot3^2\cdot3^3\). Choose exactly (2) odd positions, then multiply choices of odd and even digits. In exams choose positions first in exactly-type digit questions.
Step 3
Exam Tip
Exactly (2) odd positions चुनें, फिर odd और even digits के choices multiply करें। परीक्षा में exactly type digit questions में positions first चुनें।
Every symbol appearing is an onto condition, so missing symbols are removed by inclusion-exclusion. In exams treat at least once as onto mapping.
Step 2
Why this answer is correct
The correct answer is A. (\sum_{i=0}^{4}(-1)^i{}^{4}C_i(4-i)6). Every symbol appearing is an onto condition, so missing symbols are removed by inclusion-exclusion. In exams treat at least once as onto mapping.
Step 3
Exam Tip
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।
First choose (3) symbols, then form onto strings on (7) positions. In exams use choose set plus onto count for exactly distinct symbols.
Step 2
Why this answer is correct
The correct answer is A. (^{5}C_3\left\(3^7-3\cdot2^7+3\right\)). First choose (3) symbols, then form onto strings on (7) positions. In exams use choose set plus onto count for exactly distinct symbols.
Step 3
Exam Tip
पहले (3) symbols चुनें, फिर (7) positions पर onto strings बनाएं। परीक्षा में exactly distinct symbols के लिए choose set plus onto count 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.
Step 2
Why this answer is correct
The correct answer is A. \(n\cdot{}^{n-1}C_{r-2}\cdot\frac{r!}{2!}\). 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.
Step 3
Exam Tip
Repeated symbol चुनें, बाकी (r-2) distinct symbols चुनें, फिर multiset arrange करें। परीक्षा में exactly one repeat में repeated item पहले fix करें।
This is the multinomial count of dividing (n) brackets into sizes (p,q,r,s). In exams treat exponents as group sizes.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{n!}{p!q!r!s!}\). This is the multinomial count of dividing (n) brackets into sizes (p,q,r,s). In exams treat exponents as group sizes.
Step 3
Exam Tip
यह (n) brackets को (p,q,r,s) sizes में बांटने का multinomial count है। परीक्षा में exponents को group sizes मानें।
The powers sum to (9), and the coefficient comes from the multinomial form. In exams use a repeated-arrangement style denominator in multinomial terms.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{9!}{4!3!2!}\). The powers sum to (9), and the coefficient comes from the multinomial form. In exams use a repeated-arrangement style denominator in multinomial terms.
Step 3
Exam Tip
Powers का sum (9) है और coefficient multinomial form से मिलता है। परीक्षा में multinomial term में repeated arrangements जैसा denominator रखें।
A. एक bracket से \(x^2\) या दो brackets से (x) चुनना/Choose \(x^2\) from one bracket or (x) from two brackets
Step 1
Concept
There are two disjoint cases to form \(x^2\). In exams make exponent-sum cases for polynomial expansion coefficients.
Step 2
Why this answer is correct
The correct answer is A. एक bracket से \(x^2\) या दो brackets से (x) चुनना / Choose \(x^2\) from one bracket or (x) from two brackets. There are two disjoint cases to form \(x^2\). In exams make exponent-sum cases for polynomial expansion coefficients.
Step 3
Exam Tip
\(x^2\) बनने के दो disjoint cases हैं। परीक्षा में polynomial expansion coefficients में exponent-sum cases बनाएं।
Case (1): choose one \(x^2\), case (2): choose two (x)'s. In exams add all cases that produce the same power.
Step 2
Why this answer is correct
The correct answer is A. \(^{8}C_1+{}^{8}C_2\). Case (1): choose one \(x^2\), case (2): choose two (x)'s. In exams add all cases that produce the same power.
Step 3
Exam Tip
Case (1): एक \(x^2\) चुनें, case (2): दो (x) चुनें। परीक्षा में same power पाने वाले सभी cases जोड़ें।
The cube roots of unity filter is used to separate modulo (3) classes. In exams distinguish (3)-step coefficient sums from even-odd sums.
Step 2
Why this answer is correct
The correct answer is A. Unity roots filter. The cube roots of unity filter is used to separate modulo (3) classes. In exams distinguish (3)-step coefficient sums from even-odd sums.
Step 3
Exam Tip
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।
A. \(\frac{{}^{n}C_{r+1}}{{}^{n}C_r}=\frac{n-r}{r+1}\)
Step 1
Concept
This ratio shows the direction of increase and decrease in consecutive binomial coefficients. In exams compare the ratio with (1) to locate the peak.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{{}^{n}C_{r+1}}{{}^{n}C_r}=\frac{n-r}{r+1}\). This ratio shows the direction of increase and decrease in consecutive binomial coefficients. In exams compare the ratio with (1) to locate the peak.
Step 3
Exam Tip
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।
Unequal equal-combination indices are complementary. In exams set the sum of lower indices equal to the upper index.
Step 2
Why this answer is correct
The correct answer is A. (2r+4=n). Unequal equal-combination indices are complementary. In exams set the sum of lower indices equal to the upper index.
Step 3
Exam Tip
Unequal equal-combination indices complementary होते हैं। परीक्षा में lower indices का sum upper index के बराबर करें।
Complementary indices give (2r-1+r+8=24), so (r=6). In exams check same-index and complement cases separately in equal combinations.
Step 2
Why this answer is correct
The correct answer is B. (6). Complementary indices give (2r-1+r+8=24), so (r=6). In exams check same-index and complement cases separately in equal combinations.
Step 3
Exam Tip
Complementary indices से (2r-1+r+8=24), इसलिए (r=6)। परीक्षा में equal combinations में same-index case और complement case अलग देखें।
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.
Step 2
Why this answer is correct
The correct answer is A. (2n-7r+2=0). 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.
Step 3
Exam Tip
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 complement of at least (2) special is (0) special or exactly (1) special. In exams subtract all unwanted cases in the complement.
Step 2
Why this answer is correct
The correct answer is A. \(^{n}C_r-{}^{n-s}C_r-s{}^{n-s}C_{r-1}\). The complement of at least (2) special is (0) special or exactly (1) special. In exams subtract all unwanted cases in the complement.
Step 3
Exam Tip
At least (2) special का complement (0) special या exactly (1) special है। परीक्षा में complement में सभी unwanted cases घटाएं।
Case (1): both are included, case (2): both are excluded. In exams split paired restrictions into two disjoint cases.
Step 2
Why this answer is correct
The correct answer is A. \(^{8}C_3+{}^{8}C_5\). Case (1): both are included, case (2): both are excluded. In exams split paired restrictions into two disjoint cases.
Step 3
Exam Tip
Case (1): दोनों शामिल हों, case (2): दोनों बाहर हों। परीक्षा में paired restriction को दो disjoint cases में तोड़ें।
Subtract selections containing both (A,B) from total selections. In exams complement is the shortest route for not-together selection.
Step 2
Why this answer is correct
The correct answer is A. \(^{9}C_4-{}^{7}C_2\). Subtract selections containing both (A,B) from total selections. In exams complement is the shortest route for not-together selection.
Step 3
Exam Tip
Total selections से (A,B) दोनों वाले selections घटते हैं। परीक्षा में not together selection में complement सबसे छोटा route है।
