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।
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. 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।
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।
B. (r) चुनना और (n-r) छोड़ना समान निर्णय हैं/Choosing (r) and leaving (n-r) are equivalent decisions
Step 1
Concept
Choosing (r) objects is equivalent to not choosing (n-r) objects. In exams use complement selection to identify symmetry.
Step 2
Why this answer is correct
The correct answer is B. (r) चुनना और (n-r) छोड़ना समान निर्णय हैं / Choosing (r) and leaving (n-r) are equivalent decisions. Choosing (r) objects is equivalent to not choosing (n-r) objects. In exams use complement selection to identify symmetry.
Step 3
Exam Tip
(r) वस्तुएं चुनना उतना ही है जितना (n-r) वस्तुएं न चुनना। परीक्षा में complement selection से symmetry पहचानें।
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।
B. \(\frac{^{n}C_{r+1}}{^{n}C_r}=\frac{n-r}{r+1}\)
Step 1
Concept
Writing factorial forms and canceling common terms gives the ratio. In exams ratio method is fast for consecutive combinations.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{^{n}C_{r+1}}{^{n}C_r}=\frac{n-r}{r+1}\). Writing factorial forms and canceling common terms gives the ratio. In exams ratio method is fast for consecutive combinations.
Step 3
Exam Tip
Factorial form लिखकर common terms cancel करने से ratio मिलता है। परीक्षा में consecutive combinations में ratio method तेज होता है।
A permutation includes (r!) orders for each combination. In exams divide ordered count by (r!) to get unordered count.
Step 2
Why this answer is correct
The correct answer is B. \(^{n}C_r=\frac{^{n}P_r}{r!}\). A permutation includes (r!) orders for each combination. In exams divide ordered count by (r!) to get unordered count.
Step 3
Exam Tip
Permutation में हर combination के (r!) orders शामिल होते हैं। परीक्षा में unordered count पाने के लिए ordered count को (r!) से divide करें।
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. 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।
C. क्योंकि केवल समूह चाहिए/Because only a group is needed
Step 1
Concept
Only membership is counted in a committee, not order. In exams use combination for a group without posts.
Step 2
Why this answer is correct
The correct answer is C. क्योंकि केवल समूह चाहिए / Because only a group is needed. Only membership is counted in a committee, not order. In exams use combination for a group without posts.
Step 3
Exam Tip
Committee में केवल सदस्यता गिनी जाती है, क्रम नहीं। परीक्षा में group बिना पद के हो तो combination लगाएं।
B. पहले स्थान पर (n), फिर (n-1), फिर घटती choices मिलती हैं/Choices are (n), then (n-1), then decreasing
Step 1
Concept
After each chosen object available choices decrease. In exams write decreasing products for distinct arrangements.
Step 2
Why this answer is correct
The correct answer is B. पहले स्थान पर (n), फिर (n-1), फिर घटती choices मिलती हैं / Choices are (n), then (n-1), then decreasing. After each chosen object available choices decrease. In exams write decreasing products for distinct arrangements.
Step 3
Exam Tip
हर चुने गए object के बाद available choices कम होती जाती हैं। परीक्षा में distinct arrangement में decreasing product लिखें।
This is the symmetry of choosing (r) and leaving (n-r). In exams use this relation when opposite indices sum to (n).
Step 2
Why this answer is correct
The correct answer is C. \(^{n}C_r=^{n}C_{n-r}\). This is the symmetry of choosing (r) and leaving (n-r). In exams use this relation when opposite indices sum to (n).
Step 3
Exam Tip
यह (r) चुनने और (n-r) छोड़ने की symmetry है। परीक्षा में opposite indices का sum (n) हो तो यह relation लगाएं।
A. किसी भी वस्तु को न चुनने का एक तरीका है/There is one way to choose no object
Step 1
Concept
The empty selection is also a valid selection. In exams remember both (0!) and empty choice as (1).
Step 2
Why this answer is correct
The correct answer is A. किसी भी वस्तु को न चुनने का एक तरीका है / There is one way to choose no object. The empty selection is also a valid selection. In exams remember both (0!) and empty choice as (1).
Step 3
Exam Tip
Empty selection भी एक valid selection है। परीक्षा में (0!) और empty choice दोनों का मान (1) याद रखें।
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. सभी (n) वस्तुओं को क्रम में लगाने पर (n!) arrangements मिलती हैं/Arranging all (n) objects gives (n!) arrangements
Step 1
Concept
When all objects are arranged \(^{n}P_n=\frac{n!}{0!}=n!\). In exams use (0!=1).
Step 2
Why this answer is correct
The correct answer is A. सभी (n) वस्तुओं को क्रम में लगाने पर (n!) arrangements मिलती हैं / Arranging all (n) objects gives (n!) arrangements. When all objects are arranged \(^{n}P_n=\frac{n!}{0!}=n!\). In exams use (0!=1).
Step 3
Exam Tip
जब सभी objects arrange होते हैं तो \(^{n}P_n=\frac{n!}{0!}=n!\)। परीक्षा में (0!=1) जरूर लगाएं।
Changing order changes the code. In exams usually use permutation relation for codes or passwords.
Step 2
Why this answer is correct
The correct answer is B. \(^{6}P_4=^{6}C_4\cdot4!\). Changing order changes the code. In exams usually use permutation relation for codes or passwords.
Step 3
Exam Tip
Code में order बदलने से code बदल जाता है। परीक्षा में code या password में सामान्यतः permutation relation लगाएं।
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. क्योंकि (1) object का order अलग परिणाम नहीं बनाता/Because the order of (1) object creates no different result
Step 1
Concept
One object can be arranged in (1!) way. In exams permutation and combination give the same value for (r=1).
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (1) object का order अलग परिणाम नहीं बनाता / Because the order of (1) object creates no different result. One object can be arranged in (1!) way. In exams permutation and combination give the same value for (r=1).
Step 3
Exam Tip
एक वस्तु को arrange करने का (1!) तरीका है। परीक्षा में (r=1) पर permutation और combination समान मान दें।
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।
This is a shifted form of Pascal identity. In exams add adjacent lower-row combinations to form the upper-row combination.
Step 2
Why this answer is correct
The correct answer is B. Pascal identity. This is a shifted form of Pascal identity. In exams add adjacent lower-row combinations to form the upper-row combination.
Step 3
Exam Tip
यह Pascal identity का shifted रूप है। परीक्षा में adjacent lower-row combinations जोड़कर upper-row combination बनाएं।
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. 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।
When reflection is also same each circular arrangement is counted twice. In exams remember (\frac{(n-1)!}{2}) for necklace type.
Step 2
Why this answer is correct
The correct answer is B. (2). When reflection is also same each circular arrangement is counted twice. In exams remember (\frac{(n-1)!}{2}) for necklace type.
Step 3
Exam Tip
Reflection भी same होने पर हर circular arrangement दो बार गिनी जाती है। परीक्षा में necklace type में (\frac{(n-1)!}{2}) याद रखें।
A. जब repetition allowed हो और order important हो/When repetition is allowed and order is important
Step 1
Concept
Each of the (r) positions has (n) independent choices. In exams do not write a decreasing product when repetition is allowed.
Step 2
Why this answer is correct
The correct answer is A. जब repetition allowed हो और order important हो / When repetition is allowed and order is important. Each of the (r) positions has (n) independent choices. In exams do not write a decreasing product when repetition is allowed.
Step 3
Exam Tip
हर (r) स्थान पर (n) choices स्वतंत्र रूप से रहती हैं। परीक्षा में repetition allowed हो तो decreasing product नहीं लिखें।
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. क्योंकि factorial (n!) में चुनी न गई वस्तुओं की arrangements भी शामिल होती हैं/Because (n!) includes arrangements of unselected objects too
Step 1
Concept
In the formula (n!) loses both unwanted tail ((n-r)!) and order (r!). In exams understand both parts of the combination denominator.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि factorial (n!) में चुनी न गई वस्तुओं की arrangements भी शामिल होती हैं / Because (n!) includes arrangements of unselected objects too. In the formula (n!) loses both unwanted tail ((n-r)!) and order (r!). In exams understand both parts of the combination denominator.
Step 3
Exam Tip
Formula में (n!) से unwanted tail ((n-r)!) और order (r!) दोनों हटते हैं। परीक्षा में combination denominator के दोनों भाग समझें।
Near the maximum \(\frac{^{n}C_{r+1}}{^{n}C_r}\) is around (1). In exams use the ratio to identify transition from increasing to decreasing.
Step 2
Why this answer is correct
The correct answer is A. (1). Near the maximum \(\frac{^{n}C_{r+1}}{^{n}C_r}\) is around (1). In exams use the ratio to identify transition from increasing to decreasing.
Step 3
Exam Tip
Maximum के पास \(\frac{^{n}C_{r+1}}{^{n}C_r}\) (1) के आसपास होता है। परीक्षा में increasing से decreasing transition ratio से पहचानें।
A. (7+9=16) इसलिए complementary selection है/(7+9=16), so it is complementary selection
Step 1
Concept
When indices sum to (n), \(^{n}C_r=^{n}C_{n-r}\). In exams checking the sum is the fastest method.
Step 2
Why this answer is correct
The correct answer is A. (7+9=16) इसलिए complementary selection है / (7+9=16), so it is complementary selection. When indices sum to (n), \(^{n}C_r=^{n}C_{n-r}\). In exams checking the sum is the fastest method.
Step 3
Exam Tip
Indices का sum (n) होने पर \(^{n}C_r=^{n}C_{n-r}\)। परीक्षा में sum check सबसे तेज तरीका है।
In the falling product \(9\cdot8\cdot7\cdot6\), the remaining tail (5!) is removed. In exams write the denominator tail after the last chosen factor.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{9!}{5!}\). In the falling product \(9\cdot8\cdot7\cdot6\), the remaining tail (5!) is removed. In exams write the denominator tail after the last chosen factor.
Step 3
Exam Tip
Falling product \(9\cdot8\cdot7\cdot6\) में बाकी tail (5!) हटता है। परीक्षा में last chosen factor के बाद denominator tail लिखें।
\(^{10}P_4\) is an ordered count and \(^{10}C_4\) is unordered. In exams divide by (r!) to remove order.
Step 2
Why this answer is correct
The correct answer is B. (4!) से भाग / Divide by (4!). \(^{10}P_4\) is an ordered count and \(^{10}C_4\) is unordered. In exams divide by (r!) to remove order.
Step 3
Exam Tip
\(^{10}P_4\) ordered count है और \(^{10}C_4\) unordered count है। परीक्षा में order हटाने के लिए (r!) से divide करें।
To get unordered pairs from ordered pairs (2!) orders are removed. This connection is very useful in pair selection exams.
Step 2
Why this answer is correct
The correct answer is B. \(^{n}C_2=\frac{^{n}P_2}{2!}\). To get unordered pairs from ordered pairs (2!) orders are removed. This connection is very useful in pair selection exams.
Step 3
Exam Tip
Ordered pair से unordered pair पाने के लिए (2!) orders हटते हैं। परीक्षा में pair selection में यह connection बहुत उपयोगी है।
After choosing (p), the remaining (q) are fixed, so \(\frac{n!}{p!q!}\). In exams use factorial division for fixed group sizes.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{n!}{p!q!}\). After choosing (p), the remaining (q) are fixed, so \(\frac{n!}{p!q!}\). In exams use factorial division for fixed group sizes.
Step 3
Exam Tip
पहले (p) चुनने पर बाकी (q) तय हो जाते हैं, इसलिए \(\frac{n!}{p!q!}\)। परीक्षा में group sizes fixed हों तो factorial division 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।
It is obtained by taking the ratio of \(^{n}C_r\) and \(^{n}C_{r-1}\). In exams simplify adjacent combinations using factorial ratios.
Step 2
Why this answer is correct
The correct answer is A. Consecutive combination ratio derivation. It is obtained by taking the ratio of \(^{n}C_r\) and \(^{n}C_{r-1}\). In exams simplify adjacent combinations using factorial ratios.
Step 3
Exam Tip
यह \(^{n}C_r\) और \(^{n}C_{r-1}\) का ratio लेकर मिलता है। परीक्षा में adjacent combinations को factorial ratio से simplify करें।
The denominator in the pair formula is (2!), and (2!=2). In exams simplify pair formulas mentally.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (2!=2) / Because (2!=2). The denominator in the pair formula is (2!), and (2!=2). In exams simplify pair formulas mentally.
Step 3
Exam Tip
Pair formula में denominator (2!) ही होता है और (2!=2)। परीक्षा में pair simplification मानसिक रूप से करें।
A. \(\frac{20!}{3!17!}\) में (17!) cancel होता है/(17!) cancels in \(\frac{20!}{3!17!}\)
Step 1
Concept
Write (20!) as \(20\cdot19\cdot18\cdot17!\) and cancel (17!). In exams do not expand large factorials fully.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{20!}{3!17!}\) में (17!) cancel होता है / (17!) cancels in \(\frac{20!}{3!17!}\). Write (20!) as \(20\cdot19\cdot18\cdot17!\) and cancel (17!). In exams do not expand large factorials fully.
Step 3
Exam Tip
(20!) को \(20\cdot19\cdot18\cdot17!\) लिखकर (17!) cancel करें। परीक्षा में large factorials expand पूरा न करें।
A. Permutation में complement symmetry सामान्यतः नहीं होती/Complement symmetry generally does not hold in permutations
Step 1
Concept
Complement symmetry is a property of combinations, not permutations. In exams keep identities of \(^{n}C_r\) and \(^{n}P_r\) separate.
Step 2
Why this answer is correct
The correct answer is A. Permutation में complement symmetry सामान्यतः नहीं होती / Complement symmetry generally does not hold in permutations. Complement symmetry is a property of combinations, not permutations. In exams keep identities of \(^{n}C_r\) and \(^{n}P_r\) separate.
Step 3
Exam Tip
Complement symmetry combination की property है, permutation की नहीं। परीक्षा में \(^{n}C_r\) और \(^{n}P_r\) की identities अलग रखें।
A. Combination में chosen और not chosen sets complement होते हैं, permutation में ordered length बदल जाती है/In combinations chosen and not chosen sets are complements, while in permutations ordered length changes
Step 1
Concept
Combination counts only selection so complement works. In exams do not use complement symmetry when ordered slots appear.
Step 2
Why this answer is correct
The correct answer is A. Combination में chosen और not chosen sets complement होते हैं, permutation में ordered length बदल जाती है / In combinations chosen and not chosen sets are complements, while in permutations ordered length changes. Combination counts only selection so complement works. In exams do not use complement symmetry when ordered slots appear.
Step 3
Exam Tip
Combination सिर्फ selection गिनता है इसलिए complement works करता है। परीक्षा में ordered slots दिखें तो complement symmetry न लगाएं।
The complement of at least (1) girl is no girl. In exams total minus unwanted is fast for at least conditions.
Step 2
Why this answer is correct
The correct answer is A. \(^{9}C_3-^{5}C_3\). The complement of at least (1) girl is no girl. In exams total minus unwanted is fast for at least conditions.
Step 3
Exam Tip
At least (1) girl का complement no girl है। परीक्षा में at least condition में total minus unwanted तेज होता है।
With exactly (2) women the remaining (2) must be men. In exams use cases or direct product of combinations for exactly conditions.
Step 2
Why this answer is correct
The correct answer is A. \(^{5}C_2\cdot^{6}C_2\). With exactly (2) women the remaining (2) must be men. In exams use cases or direct product of combinations for exactly conditions.
Step 3
Exam Tip
Exactly (2) women के साथ बाकी (2) men चुनने होंगे। परीक्षा में exactly condition में cases या direct product of combinations लगाएं।
A. Block को एक object मानकर \(7!\cdot2!\)/Treat the block as one object and use \(7!\cdot2!\)
Step 1
Concept
Treating two together people as one block gives (7) objects to arrange and (2!) ways inside the block. In exams use block method for together conditions.
Step 2
Why this answer is correct
The correct answer is A. Block को एक object मानकर \(7!\cdot2!\) / Treat the block as one object and use \(7!\cdot2!\). Treating two together people as one block gives (7) objects to arrange and (2!) ways inside the block. In exams use block method for together conditions.
Step 3
Exam Tip
दो साथ लोगों को एक block मानने से (7) objects arrange होते हैं और block के अंदर (2!) ways हैं। परीक्षा में together condition में block method लगाएं।
A. क्योंकि कोई भी (3) points एक unique triangle बनाते हैं और order important नहीं है/Because any (3) points form one unique triangle and order is not important
Step 1
Concept
A triangle is determined by the set of vertices, not their order. In exams use combinations for geometric selection.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि कोई भी (3) points एक unique triangle बनाते हैं और order important नहीं है / Because any (3) points form one unique triangle and order is not important. A triangle is determined by the set of vertices, not their order. In exams use combinations for geometric selection.
Step 3
Exam Tip
Triangle केवल vertices के set से तय होता है, उनके order से नहीं। परीक्षा में geometric selection में combination लगाएं।
