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A. एक चुने हुए सदस्य को चिह्नित करके गिनना/Counting by marking one selected member
Step 1
Concept
Choosing the marked member first gives (n) choices and each group is counted (r) times. In exams treat the denominator (r) as overcount.
Step 2
Why this answer is correct
The correct answer is A. एक चुने हुए सदस्य को चिह्नित करके गिनना / Counting by marking one selected member. Choosing the marked member first gives (n) choices and each group is counted (r) times. In exams treat the denominator (r) as overcount.
Step 3
Exam Tip
पहले चिह्नित सदस्य चुनने पर (n) विकल्प आते हैं और हर समूह (r) बार गिना जाता है। परीक्षा में (r) वाला denominator overcount समझें।
C. \(\frac{{}^{n}C_r}{{}^{n-1}C_{r-1}}=\frac{n}{r}\)
Step 1
Concept
Canceling factorial forms gives \(\frac{n}{r}\). In exams use the ratio method for adjacent upper and lower indices.
Step 2
Why this answer is correct
The correct answer is C. \(\frac{{}^{n}C_r}{{}^{n-1}C_{r-1}}=\frac{n}{r}\). Canceling factorial forms gives \(\frac{n}{r}\). In exams use the ratio method for adjacent upper and lower indices.
Step 3
Exam Tip
Factorial रूप cancel करने पर \(\frac{n}{r}\) मिलता है। परीक्षा में adjacent upper और lower indices में ratio method उपयोग करें।
Adjacent terms of the same row give the middle term in the next row. In exams apply Pascal identity when (r) and (r-1) appear together.
Step 2
Why this answer is correct
The correct answer is A. \(^{n+1}C_r\). Adjacent terms of the same row give the middle term in the next row. In exams apply Pascal identity when (r) and (r-1) appear together.
Step 3
Exam Tip
एक ही पंक्ति के adjacent terms अगली पंक्ति का बीच वाला term देते हैं। परीक्षा में (r) और (r-1) साथ दिखें तो Pascal identity लगाएं।
Unequal equal-combination indices are complementary, so (r+r+3=15). 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. (6). Unequal equal-combination indices are complementary, so (r+r+3=15). 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 (r+r+3=15)। परीक्षा में lower indices का sum upper index के बराबर रखें।
The complementary condition gives (2r+r+3=18), so (r=5). In exams consider two cases for equality: same index or complementary index.
Step 2
Why this answer is correct
The correct answer is C. (5). The complementary condition gives (2r+r+3=18), so (r=5). In exams consider two cases for equality: same index or complementary index.
Step 3
Exam Tip
Complementary condition से (2r+r+3=18), इसलिए (r=5)। परीक्षा में equality के दो cases सोचें: same index या complementary index।
A. चुनी गई (r) वस्तुओं के सभी क्रम/All orders of the selected (r) objects
Step 1
Concept
Combination gives only the group and (r!) adds arrangements of that group. In exams treat permutation as selection plus arrangement.
Step 2
Why this answer is correct
The correct answer is A. चुनी गई (r) वस्तुओं के सभी क्रम / All orders of the selected (r) objects. Combination gives only the group and (r!) adds arrangements of that group. In exams treat permutation as selection plus arrangement.
Step 3
Exam Tip
Combination केवल समूह देता है और (r!) उस समूह के arrangements जोड़ता है। परीक्षा में permutation को selection plus arrangement समझें।
A. Permutation में order counted होता है और combination में नहीं/Order is counted in permutation but not in combination
Step 1
Concept
To get unordered count from ordered count, (r!) orders are removed. In exams divide when order is ignored.
Step 2
Why this answer is correct
The correct answer is A. Permutation में order counted होता है और combination में नहीं / Order is counted in permutation but not in combination. To get unordered count from ordered count, (r!) orders are removed. In exams divide when order is ignored.
Step 3
Exam Tip
Ordered count से unordered count पाने के लिए (r!) orders हटाए जाते हैं। परीक्षा में order ignored हो तो divide करें।
Subtract selections of (0), (1), and (2) from all subsets. In exams handle at least by total minus small unwanted cases.
Step 2
Why this answer is correct
The correct answer is A. \(2^n-1-n-{}^{n}C_2\). Subtract selections of (0), (1), and (2) from all subsets. In exams handle at least by total minus small unwanted cases.
Step 3
Exam Tip
सभी subsets में से (0), (1), और (2) selections घटाते हैं। परीक्षा में at least को total minus small unwanted cases से करें।
At most (3) includes selections of (0), (1), (2), and (3). In exams add all allowed cases when there is an upper limit.
Step 2
Why this answer is correct
The correct answer is B. \(1+n+{}^{n}C_2+{}^{n}C_3\). At most (3) includes selections of (0), (1), (2), and (3). In exams add all allowed cases when there is an upper limit.
Step 3
Exam Tip
At most (3) में (0), (1), (2), और (3) selections शामिल होते हैं। परीक्षा में upper limit हो तो सभी allowed cases जोड़ें।
In stars and bars, (18) stars and (3) bars are arranged. In exams use \(^{n+r-1}C_{r-1}\) for non-negative solutions.
Step 2
Why this answer is correct
The correct answer is B. \(^{21}C_3\). In stars and bars, (18) stars and (3) bars are arranged. In exams use \(^{n+r-1}C_{r-1}\) for non-negative solutions.
Step 3
Exam Tip
Stars and bars में (18) stars और (3) bars arrange होते हैं। परीक्षा में अऋणात्मक हलों के लिए \(^{n+r-1}C_{r-1}\) लगाएं।
Giving (3) first to each variable leaves (12), then count non-negative solutions. In exams shift the minimum condition.
Step 2
Why this answer is correct
The correct answer is A. \(^{14}C_2\). Giving (3) first to each variable leaves (12), then count non-negative solutions. In exams shift the minimum condition.
Step 3
Exam Tip
हर variable को पहले (3) देने पर (12) बचता है, फिर अऋणात्मक हल गिनते हैं। परीक्षा में minimum condition को shift करें।
For identical pens and distinct students, use (9) stars and (3) bars. In exams remember stars and bars for zero-allowed distribution.
Step 2
Why this answer is correct
The correct answer is B. \(^{12}C_3\). For identical pens and distinct students, use (9) stars and (3) bars. In exams remember stars and bars for zero-allowed distribution.
Step 3
Exam Tip
Identical pens और distinct students के लिए (9) stars तथा (3) bars लगते हैं। परीक्षा में zero allowed distribution में stars and bars याद रखें।
First choose the empty person, then distribute positively among the remaining (3) persons. In exams use choose empty plus positive stars and bars for exactly empty conditions.
Step 2
Why this answer is correct
The correct answer is A. \(^{4}C_1\cdot{}^{13}C_2\). First choose the empty person, then distribute positively among the remaining (3) persons. In exams use choose empty plus positive stars and bars for exactly empty conditions.
Step 3
Exam Tip
पहले empty person चुनें, फिर बाकी (3) persons में positive distribution करें। परीक्षा में exactly empty condition में choose empty plus positive stars-bars लगाएं।
A. जब हर object independently किसी भी box में जा सकता है/When each object can independently go to any box
Step 1
Concept
Each distinct object has (r) independent choices. In exams use the power rule for distinct objects and unrestricted boxes.
Step 2
Why this answer is correct
The correct answer is A. जब हर object independently किसी भी box में जा सकता है / When each object can independently go to any box. Each distinct object has (r) independent choices. In exams use the power rule for distinct objects and unrestricted boxes.
Step 3
Exam Tip
हर distinct object के पास (r) independent choices होते हैं। परीक्षा में distinct objects और unrestricted boxes में power rule लगाएं।
Each prize has (10) independent choices. In exams use the power formula when distinct prizes and repeated recipients are allowed.
Step 2
Why this answer is correct
The correct answer is B. \(10^6\). Each prize has (10) independent choices. In exams use the power formula when distinct prizes and repeated recipients are allowed.
Step 3
Exam Tip
हर prize के लिए (10) independent choices हैं। परीक्षा में distinct prizes और repeated recipients allowed हों तो power formula लें।
Prizes are distinct and recipients cannot repeat, so it becomes an ordered assignment. In exams treat distinct prizes without repetition as permutation.
Step 2
Why this answer is correct
The correct answer is C. \(^{10}P_6\). Prizes are distinct and recipients cannot repeat, so it becomes an ordered assignment. In exams treat distinct prizes without repetition as permutation.
Step 3
Exam Tip
Prizes अलग हैं और recipients repeat नहीं हो सकते, इसलिए ordered assignment बनता है। परीक्षा में distinct prizes without repetition को permutation समझें।
In identical prize distribution, use (6) stars and (9) bars. In exams apply stars and bars for identical items and distinct receivers.
Step 2
Why this answer is correct
The correct answer is B. \(^{15}C_9\). In identical prize distribution, use (6) stars and (9) bars. In exams apply stars and bars for identical items and distinct receivers.
Step 3
Exam Tip
Identical prizes distribution में (6) stars और (9) bars लगते हैं। परीक्षा में identical items और distinct receivers में stars and bars लगाएं।
The three posts are different, so the order of selection is meaningful. In exams use permutation for different posts.
Step 2
Why this answer is correct
The correct answer is B. \(^{8}P_3\). The three posts are different, so the order of selection is meaningful. In exams use permutation for different posts.
Step 3
Exam Tip
तीनों पद अलग हैं, इसलिए order of selection meaningful है। परीक्षा में different posts हों तो permutation लगाएं।
A. क्योंकि committee में क्रम का महत्व नहीं है/Because order has no importance in a committee
Step 1
Concept
A committee without posts counts only the group. In exams use combination when there are no posts.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि committee में क्रम का महत्व नहीं है / Because order has no importance in a committee. A committee without posts counts only the group. In exams use combination when there are no posts.
Step 3
Exam Tip
बिना पद वाली committee में केवल group गिना जाता है। परीक्षा में पद न हों तो combination उपयोग करें।
Subtract collinear (3)-point selections from total (3)-point selections. In exams subtract invalid selections.
Step 2
Why this answer is correct
The correct answer is A. \(^{10}C_3-{}^{4}C_3\). Subtract collinear (3)-point selections from total (3)-point selections. In exams subtract invalid selections.
Step 3
Exam Tip
Total (3)-point selections से collinear (3)-point selections हटते हैं। परीक्षा में invalid selections घटाएं।
A selection of four vertices determines a quadrilateral. In exams use combinations of vertices when shapes are formed from polygons.
Step 2
Why this answer is correct
The correct answer is A. \(^{n}C_4\). A selection of four vertices determines a quadrilateral. In exams use combinations of vertices when shapes are formed from polygons.
Step 3
Exam Tip
चार vertices का selection एक quadrilateral तय करता है। परीक्षा में polygon से shapes बनें तो vertices का combination लें।
A. क्योंकि handshake में pair unordered होता है/Because a handshake pair is unordered
Step 1
Concept
A handshake of (A) with (B) and (B) with (A) is the same. In exams treat mutual relations as combinations.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि handshake में pair unordered होता है / Because a handshake pair is unordered. A handshake of (A) with (B) and (B) with (A) is the same. In exams treat mutual relations as combinations.
Step 3
Exam Tip
(A) का (B) से handshake और (B) का (A) से handshake same है। परीक्षा में mutual relation को combination मानें।
Treating the (3) special books as one block gives (5) objects to arrange. In exams also count (3!) arrangements inside the block.
Step 2
Why this answer is correct
The correct answer is A. \(5!\cdot3!\). Treating the (3) special books as one block gives (5) objects to arrange. In exams also count (3!) arrangements inside the block.
Step 3
Exam Tip
(3) special books को one block मानने पर (5) objects arrange होते हैं। परीक्षा में block के अंदर (3!) arrangements भी गिनें।
Subtract the arrangements where (A) and (B) are together as a block from total arrangements. In exams use complement for not together.
Step 2
Why this answer is correct
The correct answer is A. \(9!-8!\cdot2!\). Subtract the arrangements where (A) and (B) are together as a block from total arrangements. In exams use complement for not together.
Step 3
Exam Tip
Total arrangements से (A) और (B) together block arrangements घटते हैं। परीक्षा में not together को complement से करना आसान है।
After (6) boys, (7) gaps are formed, and (5) girls are arranged in them. In exams use the gap method for no two together.
Step 2
Why this answer is correct
The correct answer is B. \(^{7}C_5\cdot5!\). After (6) boys, (7) gaps are formed, and (5) girls are arranged in them. In exams use the gap method for no two together.
Step 3
Exam Tip
(6) boys के बाद (7) gaps बनते हैं, जिनमें (5) girls arrange होती हैं। परीक्षा में no two together में gap method लगाएं।
There are (2) patterns for the starting gender and both groups arrange in (5!) ways. In exams do not forget the (2) patterns for equal alternate groups.
Step 2
Why this answer is correct
The correct answer is B. \(2\cdot5!\cdot5!\). There are (2) patterns for the starting gender and both groups arrange in (5!) ways. In exams do not forget the (2) patterns for equal alternate groups.
Step 3
Exam Tip
Starting gender के (2) patterns होते हैं और दोनों groups अपने-अपने (5!) ways में arrange होते हैं। परीक्षा में equal alternate groups में (2) pattern न भूलें।
Removing circular rotations gives (7!), and divide by (2) when reflection is the same. In exams read the mirror-image condition in necklace problems.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{7!}{2}\). Removing circular rotations gives (7!), and divide by (2) when reflection is the same. In exams read the mirror-image condition in necklace problems.
Step 3
Exam Tip
Circular rotations हटाने से (7!) आता है और reflection same होने पर (2) से divide करते हैं। परीक्षा में necklace में mirror image condition पढ़ें।
A. एक व्यक्ति fix करके rotations हटाए जाते हैं/One person is fixed to remove rotations
Step 1
Concept
Rotations give the same arrangements at a round table. In exams write ((n-1)!) for circular seating.
Step 2
Why this answer is correct
The correct answer is A. एक व्यक्ति fix करके rotations हटाए जाते हैं / One person is fixed to remove rotations. Rotations give the same arrangements at a round table. In exams write ((n-1)!) for circular seating.
Step 3
Exam Tip
Round table में rotations same arrangements देते हैं। परीक्षा में circular seating में ((n-1)!) लिखें।
The circular arrangement of (5) couple-blocks is (4!) and each block has (2) internal orders. In exams use one less factorial for circular blocks.
Step 2
Why this answer is correct
The correct answer is B. \(4!\cdot2^5\). The circular arrangement of (5) couple-blocks is (4!) and each block has (2) internal orders. In exams use one less factorial for circular blocks.
Step 3
Exam Tip
(5) couple-blocks की circular arrangement (4!) है और हर block में (2) internal orders हैं। परीक्षा में circular blocks में one less factorial लें।
A. क्योंकि (A) और (R) दो-दो बार आते हैं/Because (A) and (R) appear twice each
Step 1
Concept
Interchanging identical letters does not create a new arrangement. In exams put factorials of repeated letters in the denominator.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (A) और (R) दो-दो बार आते हैं / Because (A) and (R) appear twice each. Interchanging identical letters does not create a new arrangement. In exams put factorials of repeated letters in the denominator.
Step 3
Exam Tip
समान letters की आपसी अदला-बदली नई arrangement नहीं देती। परीक्षा में repeated letters के factorials denominator में रखें।
(S) appears three times, (T) appears three times, and (I) appears twice. In exams count letter frequencies before word arrangement.
Step 2
Why this answer is correct
The correct answer is A. (3!3!2!). (S) appears three times, (T) appears three times, and (I) appears twice. In exams count letter frequencies before word arrangement.
Step 3
Exam Tip
(S) तीन बार, (T) तीन बार और (I) दो बार आता है। परीक्षा में word arrangement से पहले letter frequencies गिनें।
Only (1) of the (3!) possible 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. \(\frac{8!}{3!}\). Only (1) of the (3!) possible 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 की (3!) possible relative orders में केवल (1) allowed है। परीक्षा में fixed relative order में total को (k!) से divide करें।
Only one of the (3!) relative orders of these (3) people is valid. In exams handle fixed order restriction by symmetry division.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{10!}{3!}\). Only one of the (3!) relative orders of these (3) people is valid. In exams handle fixed order restriction by symmetry division.
Step 3
Exam Tip
इन (3) लोगों के (3!) relative orders में केवल एक valid है। परीक्षा में fixed order restriction को symmetry division से करें।
The first digit cannot be (0), and the remaining two places have (6) choices. In exams handle the leading-zero restriction separately.
Step 2
Why this answer is correct
The correct answer is B. \(5\cdot6\cdot6\). The first digit cannot be (0), and the remaining two places have (6) choices. In exams handle the leading-zero restriction separately.
Step 3
Exam Tip
First digit (0) नहीं हो सकती, बाकी दो places पर (6) choices हैं। परीक्षा में leading zero restriction अलग रखें।
A. हर स्थान पर (7) independent choices हैं/Each position has (7) independent choices
Step 1
Concept
When repetition is allowed, a selected symbol remains available again. In exams use the power rule for independent positions.
Step 2
Why this answer is correct
The correct answer is A. हर स्थान पर (7) independent choices हैं / Each position has (7) independent choices. When repetition is allowed, a selected symbol remains available again. In exams use the power rule for independent positions.
Step 3
Exam Tip
Repetition allowed होने पर selected symbol फिर उपलब्ध रहता है। परीक्षा में independent positions में power rule लगाएं।
Order matters in passwords and repetition is not allowed. In exams use permutation for ordered slots without repetition.
Step 2
Why this answer is correct
The correct answer is C. \(^{7}P_5\). Order matters in passwords and repetition is not allowed. In exams use permutation for ordered slots without repetition.
Step 3
Exam Tip
Password में order important है और repetition नहीं है। परीक्षा में without repetition ordered slots में permutation लगाएं।
A. क्योंकि (r) brackets से (b) चुनते हैं/Because (b) is chosen from (r) brackets
Step 1
Concept
To form \(b^r\), choose (r) brackets from (n) brackets. In exams connect binomial coefficients with bracket selection.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (r) brackets से (b) चुनते हैं / Because (b) is chosen from (r) brackets. To form \(b^r\), choose (r) brackets from (n) brackets. In exams connect binomial coefficients with bracket selection.
Step 3
Exam Tip
\(b^r\) बनाने के लिए (n) brackets में से (r) brackets चुनते हैं। परीक्षा में binomial coefficient को bracket selection से जोड़ें।
A. सभी possible subset sizes को जोड़ना/Adding all possible subset sizes
Step 1
Concept
The left side adds subsets of every size and the right side gives choose-or-not choices for each object. In exams count subset identities in two ways.
Step 2
Why this answer is correct
The correct answer is A. सभी possible subset sizes को जोड़ना / Adding all possible subset sizes. The left side adds subsets of every size and the right side gives choose-or-not choices for each object. In exams count subset identities in two ways.
Step 3
Exam Tip
Left side हर size के subsets को जोड़ता है और right side प्रत्येक object के choose or not choose choices देता है। परीक्षा में subset identity को दो तरीकों से गिनें।
A. एक selected सदस्य को mark करना/Marking one selected member
Step 1
Concept
Choose a subset and mark one member, or first choose the marked member and freely choose the rest. In exams treat the factor (r) as a marked choice.
Step 2
Why this answer is correct
The correct answer is A. एक selected सदस्य को mark करना / Marking one selected member. Choose a subset and mark one member, or first choose the marked member and freely choose the rest. In exams treat the factor (r) as a marked choice.
Step 3
Exam Tip
पहले subset चुनकर उसमें एक member mark करें या पहले marked member चुनकर बाकी freely चुनें। परीक्षा में (r) factor को marked choice समझें।
First choose the marked pair, then freely choose from the remaining (n-2) objects. In exams choose marked objects first in nested combination sums.
Step 2
Why this answer is correct
The correct answer is A. \(^{n}C_2 2^{n-2}\). First choose the marked pair, then freely choose from the remaining (n-2) objects. In exams choose marked objects first in nested combination sums.
Step 3
Exam Tip
पहले marked pair चुनें, फिर बाकी (n-2) objects freely choose करें। परीक्षा में nested combination sums में marked objects पहले चुनें।
When choosing (r) objects from two groups, (k) objects are taken from the first group. In exams identify Vandermonde in two-group selection.
Step 2
Why this answer is correct
The correct answer is B. Vandermonde identity. When choosing (r) objects from two groups, (k) objects are taken from the first group. In exams identify Vandermonde in two-group selection.
Step 3
Exam Tip
दो groups से कुल (r) objects चुनने में first group से (k) objects लिए जाते हैं। परीक्षा में two-group selection में Vandermonde पहचानें।
Choosing (a) objects first and then (b) from the remaining objects is like labelled grouping. In exams convert sequential selections into factorial division.
Step 2
Why this answer is correct
The correct answer is A. (\frac{n!}{a!b!(n-a-b)!}). Choosing (a) objects first and then (b) from the remaining objects is like labelled grouping. In exams convert sequential selections into factorial division.
Step 3
Exam Tip
पहले (a) objects और फिर बचे हुए से (b) objects चुनना labelled grouping जैसा है। परीक्षा में sequential selections को factorial division में बदलें।
