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. क्योंकि प्रत्येक वस्तु के लिए चुनना या न चुनना दो विकल्प हैं और खाली चयन हटाया जाता है/Because each object has two choices, select or not select, and the empty selection is removed
Step 1
Concept
Each object has two independent choices, so total subsets are \(2^n\), and for at least (1) the empty set is removed. In exams use total minus unwanted for at least conditions.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि प्रत्येक वस्तु के लिए चुनना या न चुनना दो विकल्प हैं और खाली चयन हटाया जाता है / Because each object has two choices, select or not select, and the empty selection is removed. Each object has two independent choices, so total subsets are \(2^n\), and for at least (1) the empty set is removed. In exams use total minus unwanted for at least conditions.
Step 3
Exam Tip
हर वस्तु के लिए दो स्वतंत्र विकल्प होने से कुल subsets \(2^n\) होते हैं, और कम से कम (1) के लिए empty set हटता है। परीक्षा में at least condition में total minus unwanted सोचें।
A. Identical letters की internal permutations same result देती हैं/Internal permutations of identical letters give the same result
Step 1
Concept
Interchanging identical letters does not create a new arrangement. In exams divide by factorials of repeated counts.
Step 2
Why this answer is correct
The correct answer is A. Identical letters की internal permutations same result देती हैं / Internal permutations of identical letters give the same result. Interchanging identical letters does not create a new arrangement. In exams divide by factorials of repeated counts.
Step 3
Exam Tip
समान letters की आपसी अदला-बदली नई arrangement नहीं बनाती। परीक्षा में repeated counts के factorials से divide करें।
A. क्योंकि चुनी गई वस्तु दोबारा नहीं चुनी जाती/Because a chosen object is not selected again
Step 1
Concept
Without repetition the available count decreases after one choice is used. In exams connect no repetition with a falling product.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि चुनी गई वस्तु दोबारा नहीं चुनी जाती / Because a chosen object is not selected again. Without repetition the available count decreases after one choice is used. In exams connect no repetition with a falling product.
Step 3
Exam Tip
Without repetition में एक choice use होने के बाद available count कम होता है। परीक्षा में no repetition को falling product से जोड़ें।
A. एक object को fixed मानकर rotation duplicates हटाए जाते हैं/One object is fixed to remove rotational duplicates
Step 1
Concept
Rotations in a circle are considered the same so one position is fixed. In exams divide linear (n!) by (n) for circular permutation.
Step 2
Why this answer is correct
The correct answer is A. एक object को fixed मानकर rotation duplicates हटाए जाते हैं / One object is fixed to remove rotational duplicates. Rotations in a circle are considered the same so one position is fixed. In exams divide linear (n!) by (n) for circular permutation.
Step 3
Exam Tip
Circle में rotations same माने जाते हैं इसलिए एक position fixed करते हैं। परीक्षा में circular permutation में linear (n!) को (n) से divide करें।
To get combinations (3!) arrangements are removed. In exams divide ordered triples by (3!) to get unordered triples.
Step 2
Why this answer is correct
The correct answer is B. (\frac{n(n-1)(n-2)}{3!}). To get combinations (3!) arrangements are removed. In exams divide ordered triples by (3!) to get unordered triples.
Step 3
Exam Tip
Combination पाने के लिए (3!) arrangements हटाए जाते हैं। परीक्षा में ordered triple से unordered triple में divide by (3!) करें।
A. क्योंकि ordered pairs को unordered pairs में बदलने के लिए (2!) से भाग देते हैं/Because ordered pairs are divided by (2!) to become unordered pairs
Step 1
Concept
(AB) and (BA) are the same pair so repetition is removed. In exams (\frac{n(n-1)}{2}) is often useful for pairs.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि ordered pairs को unordered pairs में बदलने के लिए (2!) से भाग देते हैं / Because ordered pairs are divided by (2!) to become unordered pairs. (AB) and (BA) are the same pair so repetition is removed. In exams (\frac{n(n-1)}{2}) is often useful for pairs.
Step 3
Exam Tip
(AB) और (BA) same pair हैं इसलिए दोहराव हटता है। परीक्षा में pairs के लिए अक्सर (\frac{n(n-1)}{2}) उपयोगी है।
A. एक वस्तु चुनने के लिए (n) independent choices हैं/There are (n) independent choices for selecting one object
Step 1
Concept
In single selection order does not arise and choices are (n). In exams both \(^{n}C_1\) and \(^{n}P_1\) equal (n).
Step 2
Why this answer is correct
The correct answer is A. एक वस्तु चुनने के लिए (n) independent choices हैं / There are (n) independent choices for selecting one object. In single selection order does not arise and choices are (n). In exams both \(^{n}C_1\) and \(^{n}P_1\) equal (n).
Step 3
Exam Tip
Single selection में order का प्रश्न नहीं आता और choices (n) हैं। परीक्षा में \(^{n}C_1\) और \(^{n}P_1\) दोनों (n) होते हैं।
A. (n) वस्तुओं में से सभी (n) चुनने का केवल एक तरीका है/There is exactly one way to choose all (n) objects
Step 1
Concept
When all objects are chosen the selection is fixed. In exams quickly identify extreme cases \(^{n}C_0\) and \(^{n}C_n\).
Step 2
Why this answer is correct
The correct answer is A. (n) वस्तुओं में से सभी (n) चुनने का केवल एक तरीका है / There is exactly one way to choose all (n) objects. When all objects are chosen the selection is fixed. In exams quickly identify extreme cases \(^{n}C_0\) and \(^{n}C_n\).
Step 3
Exam Tip
सभी वस्तुएं चुनने पर selection fixed हो जाता है। परीक्षा में extreme cases \(^{n}C_0\) और \(^{n}C_n\) को जल्दी पहचानें।
A. क्योंकि दोनों पद अलग हैं/Because the two posts are different
Step 1
Concept
Switching captain and vice-captain changes the outcome. In exams treat different roles as order important.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि दोनों पद अलग हैं / Because the two posts are different. Switching captain and vice-captain changes the outcome. In exams treat different roles as order important.
Step 3
Exam Tip
Captain और vice-captain बदलने से outcome बदलता है। परीक्षा में roles अलग हों तो order important मानें।
B. क्योंकि (E) दो बार आया है/Because (E) appears twice
Step 1
Concept
Interchanging identical (E)'s does not give a new arrangement. In exams divide by factorials of repeated identical items.
Step 2
Why this answer is correct
The correct answer is B. क्योंकि (E) दो बार आया है / Because (E) appears twice. Interchanging identical (E)'s does not give a new arrangement. In exams divide by factorials of repeated identical items.
Step 3
Exam Tip
समान (E) की आपसी अदला-बदली नई arrangement नहीं देती। परीक्षा में repeated identical items के factorial से divide करें।
A. एक fixed वस्तु शामिल है या शामिल नहीं है/A fixed object is included or not included
Step 1
Concept
Make cases on one fixed object: include it or exclude it. In exams derive such identities by inclusion cases.
Step 2
Why this answer is correct
The correct answer is A. एक fixed वस्तु शामिल है या शामिल नहीं है / A fixed object is included or not included. Make cases on one fixed object: include it or exclude it. In exams derive such identities by inclusion cases.
Step 3
Exam Tip
किसी fixed वस्तु पर case बनाएं: उसे लें या न लें। परीक्षा में ऐसी identities को inclusion case से derive करें।
After (r) decreasing factors the remaining tail is ((n-r)!). In exams put the missing tail in the denominator.
Step 2
Why this answer is correct
The correct answer is B. (^{n}P_r=\frac{n!}{(n-r)!}). After (r) decreasing factors the remaining tail is ((n-r)!). In exams put the missing tail in the denominator.
Step 3
Exam Tip
घटते हुए (r) गुणकों के बाद बचे ((n-r)!) से factorial पूरा होता है। परीक्षा में missing tail को denominator बनाएं।
A. क्योंकि (r) वस्तुओं की व्यवस्था नहीं गिनी जाती/Because arrangements of (r) objects are not counted
Step 1
Concept
In combinations order is not important so (r!) duplicate arrangements are removed. In exams divide when order is ignored.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (r) वस्तुओं की व्यवस्था नहीं गिनी जाती / Because arrangements of (r) objects are not counted. In combinations order is not important so (r!) duplicate arrangements are removed. In exams divide when order is ignored.
Step 3
Exam Tip
Combination में क्रम का महत्व नहीं होता इसलिए (r!) duplicate arrangements हटाए जाते हैं। परीक्षा में order ignored हो तो division याद रखें।
Each selected group can be arranged in (r!) ways. In exams treat permutation as selection followed by arrangement.
Step 2
Why this answer is correct
The correct answer is B. \(^{n}P_r=^{n}C_r\times r!\). Each selected group can be arranged in (r!) ways. In exams treat permutation as selection followed by arrangement.
Step 3
Exam Tip
चयन के हर समूह को (r!) तरीकों से सजाया जा सकता है। परीक्षा में permutation को selection के बाद arrangement समझें।
The general ratio is \(\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}\). Putting (r=6) gives \(\frac{n-5}{6}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{n-5}{6}\). The general ratio is \(\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}\). Putting (r=6) gives \(\frac{n-5}{6}\).
Step 3
Exam Tip
सामान्य अनुपात \(\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}\) है। (r=6) रखने पर \(\frac{n-5}{6}\) मिलता है।
After cancelling ((n-4)!), four factors from (n) to (n-3) remain. In exams the number of numerator factors is (r).
Step 2
Why this answer is correct
The correct answer is A. (n(n-1)(n-2)(n-3)). After cancelling ((n-4)!), four factors from (n) to (n-3) remain. In exams the number of numerator factors is (r).
Step 3
Exam Tip
((n-4)!) कटने के बाद चार factors (n) से (n-3) तक बचते हैं। परीक्षा में numerator factors की संख्या (r) होती है।
The general ratio is \(\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}\). Putting (r=5) gives \(\frac{n-4}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{n-4}{5}\). The general ratio is \(\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}\). Putting (r=5) gives \(\frac{n-4}{5}\).
Step 3
Exam Tip
सामान्य अनुपात \(\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}\) है। यहाँ (r=5) रखने पर \(\frac{n-4}{5}\) मिलता है।
In (n!), internal orders of the unchosen ((n-r)) objects are extra, so they are removed. In exams identify the remaining part in the permutation formula.
Step 2
Why this answer is correct
The correct answer is A. न चुनी गई वस्तुएँ / Unchosen objects. In (n!), internal orders of the unchosen ((n-r)) objects are extra, so they are removed. In exams identify the remaining part in the permutation formula.
Step 3
Exam Tip
(n!) में न चुनी गई ((n-r)) वस्तुओं के अंदरूनी क्रम extra हैं इसलिए उन्हें हटाया जाता है। परीक्षा में permutation formula में remaining part पहचानें।
In (^{n}C_3=\frac{n!}{3!(n-3)!}), cancelling ((n-3)!) leaves (n(n-1)(n-2)). In exams write the cancellation step clearly.
Step 2
Why this answer is correct
The correct answer is B. (n(n-1)(n-2)). In (^{n}C_3=\frac{n!}{3!(n-3)!}), cancelling ((n-3)!) leaves (n(n-1)(n-2)). In exams write the cancellation step clearly.
Step 3
Exam Tip
(^{n}C_3=\frac{n!}{3!(n-3)!}) में ((n-3)!) कटने पर (n(n-1)(n-2)) बचता है। परीक्षा में cancellation step साफ लिखें।
A. (r)वें स्थान के विकल्प/Choices for the (r)th position
Step 1
Concept
After filling the first (r-1) positions (n-r+1) choices remain for the (r)th position. In exams connect recurrence with the next-place idea.
Step 2
Why this answer is correct
The correct answer is A. (r)वें स्थान के विकल्प / Choices for the (r)th position. After filling the first (r-1) positions (n-r+1) choices remain for the (r)th position. In exams connect recurrence with the next-place idea.
Step 3
Exam Tip
पहले (r-1) स्थान भरने के बाद (r)वें स्थान पर (n-r+1) विकल्प बचते हैं। परीक्षा में recurrence को अगले स्थान के विचार से जोड़ें।
A. क्योंकि शुरुआत में (n) वस्तुओं को क्रम में जमाकर गिना जा सकता है/Because initially (n) objects can be counted by arranging in order
Step 1
Concept
(n!) creates a larger ordered count which is corrected by (r!(n-r)!). In exams understand the overcounting correction.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि शुरुआत में (n) वस्तुओं को क्रम में जमाकर गिना जा सकता है / Because initially (n) objects can be counted by arranging in order. (n!) creates a larger ordered count which is corrected by (r!(n-r)!). In exams understand the overcounting correction.
Step 3
Exam Tip
(n!) से ordered arrangements की बड़ी गिनती बनती है जिसे (r!(n-r)!) से सही किया जाता है। परीक्षा में overcounting correction समझें।
A. क्योंकि एक व्यक्ति को fixed मानकर rotations हटाते हैं/Because one person is fixed to remove rotations
Step 1
Concept
A mere rotation in a circle does not make a new seating. In exams use the one fixed method in circular arrangement.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि एक व्यक्ति को fixed मानकर rotations हटाते हैं / Because one person is fixed to remove rotations. A mere rotation in a circle does not make a new seating. In exams use the one fixed method in circular arrangement.
Step 3
Exam Tip
वृत्त में केवल घूमना नई seating नहीं बनाता। परीक्षा में circular arrangement में one fixed method लगाएँ।
A. (r)वें स्थान के विकल्प/Choices for the (r)th position
Step 1
Concept
After filling the first (r-1) positions (n-r+1) choices remain for the (r)th position. In exams connect recurrence with the next-place idea.
Step 2
Why this answer is correct
The correct answer is A. (r)वें स्थान के विकल्प / Choices for the (r)th position. After filling the first (r-1) positions (n-r+1) choices remain for the (r)th position. In exams connect recurrence with the next-place idea.
Step 3
Exam Tip
पहले (r-1) स्थान भरने के बाद (r)वें स्थान पर (n-r+1) विकल्प बचते हैं। परीक्षा में recurrence को next-place idea से जोड़ें।
Simplifying the factorial form gives the factor \(\frac{n}{r}\). In exams check adjacent upper index identities by ratio.
Step 2
Why this answer is correct
The correct answer is B. \(^{n}C_r=\frac{n}{r},^{n-1}C_{r-1}\). Simplifying the factorial form gives the factor \(\frac{n}{r}\). In exams check adjacent upper index identities by ratio.
Step 3
Exam Tip
Factorial form से सरल करने पर \(\frac{n}{r}\) factor आता है। परीक्षा में adjacent upper index identities को ratio से जाँचें।
Adding the remaining factor ((n-r)!) makes the numerator (n!). In exams identify the missing tail while forming factorial form.
Step 2
Why this answer is correct
The correct answer is B. ((n-r)!). Adding the remaining factor ((n-r)!) makes the numerator (n!). In exams identify the missing tail while forming factorial form.
Step 3
Exam Tip
बाकी गुणनखंड ((n-r)!) जोड़ने से ऊपर (n!) बनता है। परीक्षा में factorial रूप बनाते समय missing tail पहचानें।
The choices are (n) for the first position then (n-1) down to (1). In exams connect complete arrangement with factorial.
Step 2
Why this answer is correct
The correct answer is B. (n!). The choices are (n) for the first position then (n-1) down to (1). In exams connect complete arrangement with factorial.
Step 3
Exam Tip
पहले स्थान के लिए (n) फिर (n-1) और अंत तक (1) विकल्प मिलते हैं। परीक्षा में complete arrangement को factorial से जोड़ें।
A. चुने गए और न चुने गए समूहों के अंदर क्रम महत्वहीन होता है/Order inside chosen and not chosen groups is irrelevant
Step 1
Concept
In (n!) internal orders inside chosen and unchosen groups are counted extra. In exams connect group division with factorial denominators.
Step 2
Why this answer is correct
The correct answer is A. चुने गए और न चुने गए समूहों के अंदर क्रम महत्वहीन होता है / Order inside chosen and not chosen groups is irrelevant. In (n!) internal orders inside chosen and unchosen groups are counted extra. In exams connect group division with factorial denominators.
Step 3
Exam Tip
(n!) में चुने और छोड़े दोनों समूहों के अंदरूनी क्रम extra गिने जाते हैं। परीक्षा में group division को factorial denominator से जोड़ें।
