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Treat (A) and (B) as one block, so circular arrangements are (6!) and internal arrangements are (2!). Exam tip: use block method for together conditions.
Step 2
Why this answer is correct
The correct answer is A. (1440). Treat (A) and (B) as one block, so circular arrangements are (6!) and internal arrangements are (2!). Exam tip: use block method for together conditions.
Step 3
Exam Tip
(A) और (B) को एक खंड मानें, इसलिए (7) वस्तुओं की गोल व्यवस्था (6!) और अंदर (2!) तरीके हैं। परीक्षा में साथ वाली शर्त पर खंड विधि लगाएं।
Subtract arrangements where the (3) special books form one block, that is \(5!\times3!\), from total (7!). Exam tip: for not-together cases, subtract unfavorable cases.
Step 2
Why this answer is correct
The correct answer is B. (4320). Subtract arrangements where the (3) special books form one block, that is \(5!\times3!\), from total (7!). Exam tip: for not-together cases, subtract unfavorable cases.
Step 3
Exam Tip
कुल (7!) से उन व्यवस्थाओं को घटाएं जिनमें (3) विशेष पुस्तकें एक खंड बनती हैं, यानी \(5!\times3!\)। परीक्षा में निषेध शर्त के लिए कुल से प्रतिकूल घटाएं।
There are (7) letters and (R) repeats twice, so the count is \(\frac{7!}{2!}\). Exam tip: divide by factorials of repeated letters.
Step 2
Why this answer is correct
The correct answer is B. (1260). There are (7) letters and (R) repeats twice, so the count is \(\frac{7!}{2!}\). Exam tip: divide by factorials of repeated letters.
Step 3
Exam Tip
कुल (7) अक्षर हैं और (R) दो बार आता है, इसलिए संख्या \(\frac{7!}{2!}\) है। परीक्षा में समान अक्षरों के फैक्टोरियल से भाग दें।
Treat (TT) as one block, giving (8) objects with (M) and (E) repeated, so the count is \(\frac{8!}{2!,2!}\). Exam tip: form the block first, then handle repetitions.
Step 2
Why this answer is correct
The correct answer is A. (10080). Treat (TT) as one block, giving (8) objects with (M) and (E) repeated, so the count is \(\frac{8!}{2!,2!}\). Exam tip: form the block first, then handle repetitions.
Step 3
Exam Tip
(TT) को एक खंड मानने पर (8) वस्तुएं मिलती हैं जिनमें (M), (E) और (T)-खंड जैसी पुनरावृत्तियां संभलती हैं, संख्या \(\frac{8!}{2!,2!}\) है। परीक्षा में पहले खंड बनाकर फिर समान अक्षर देखें।
The last digit must be (0) or (5); counting both cases separately gives (300). Exam tip: never allow zero in the first place.
Step 2
Why this answer is correct
The correct answer is C. (300). The last digit must be (0) or (5); counting both cases separately gives (300). Exam tip: never allow zero in the first place.
Step 3
Exam Tip
अंतिम अंक (0) या (5) होगा; दोनों मामलों को अलग गिनने पर कुल (300) मिलता है। परीक्षा में शून्य को प्रथम स्थान पर न आने दें।
Each selection of (4) digits has exactly (1) increasing arrangement, so the count is \(\binom{8}{4}=70\). Exam tip: when order is fixed, selection alone is counted.
Step 2
Why this answer is correct
The correct answer is B. (70). Each selection of (4) digits has exactly (1) increasing arrangement, so the count is \(\binom{8}{4}=70\). Exam tip: when order is fixed, selection alone is counted.
Step 3
Exam Tip
किसी भी (4) अंकों के चयन का केवल (1) बढ़ता क्रम होता है, इसलिए संख्या \(\binom{8}{4}=70\) है। परीक्षा में क्रम तय हो तो चयन ही पर्याप्त होता है।
There are (5) position pairs for (P) and (Q), with (2!) orders, and the remaining (7) people can be arranged in (7!) ways. Exam tip: count distance conditions by position pairs.
Step 2
Why this answer is correct
The correct answer is C. (60480). There are (5) position pairs for (P) and (Q), with (2!) orders, and the remaining (7) people can be arranged in (7!) ways. Exam tip: count distance conditions by position pairs.
Step 3
Exam Tip
(P) और (Q) की जगहों के (5) जोड़े हैं और उनका क्रम (2!) है, बाकी (7) लोग (7!) तरीकों से बैठते हैं। परीक्षा में दूरी वाली शर्त को स्थान-जोड़ों से गिनें।
Arrange (6) men in (6!) ways, then place (4) women in \(^{7}P_{4}\) of the (7) gaps. Exam tip: use the gaps method for non-adjacent placement.
Step 2
Why this answer is correct
The correct answer is A. (604800). Arrange (6) men in (6!) ways, then place (4) women in \(^{7}P_{4}\) of the (7) gaps. Exam tip: use the gaps method for non-adjacent placement.
Step 3
Exam Tip
पहले (6) पुरुषों को (6!) तरीकों से बैठाएं, फिर (7) खाली स्थानों में (4) महिलाओं को \(^{7}P_{4}\) तरीकों से रखें। परीक्षा में अलग-अलग रखने के लिए gaps विधि लगाएं।
Treat (MM) as one block; now there are (10) objects with (A) and (T) repeated twice, so the count is \(\frac{10!}{2!,2!}\). Exam tip: after blocking, do not forget remaining repetitions.
Step 2
Why this answer is correct
The correct answer is B. (907200). Treat (MM) as one block; now there are (10) objects with (A) and (T) repeated twice, so the count is \(\frac{10!}{2!,2!}\). Exam tip: after blocking, do not forget remaining repetitions.
Step 3
Exam Tip
(MM) को एक खंड मानें; अब (10) वस्तुओं में (A) और (T) दो-दो बार हैं, इसलिए संख्या \(\frac{10!}{2!,2!}\) है। परीक्षा में खंड बनाने के बाद शेष पुनरावृत्ति न भूलें।
Subtract unfavorable cases where both special objects are selected, \(^{8}P_{2}\times{}^{4}P_{2}\), from total \(^{10}P_{4}\). Exam tip: at most one means not both.
Step 2
Why this answer is correct
The correct answer is B. (2688). Subtract unfavorable cases where both special objects are selected, \(^{8}P_{2}\times{}^{4}P_{2}\), from total \(^{10}P_{4}\). Exam tip: at most one means not both.
Step 3
Exam Tip
दोनों विशेष वस्तुओं के चुने जाने वाले \(^{8}P_{2}\times{}^{4}P_{2}\) प्रतिकूल मामलों को कुल \(^{10}P_{4}\) से घटाएं। परीक्षा में अधिकतम एक का अर्थ दोनों साथ नहीं है।
This is the derangement of (5) objects, so the number is (!5=44). Exam tip: identify derangement when no object goes to its own position.
Step 2
Why this answer is correct
The correct answer is A. (44). This is the derangement of (5) objects, so the number is (!5=44). Exam tip: identify derangement when no object goes to its own position.
Step 3
Exam Tip
यह (5) वस्तुओं का derangement है, जिसकी संख्या (!5=44) होती है। परीक्षा में किसी भी सही मिलान के न होने पर derangement पहचानें।
Total circular arrangements are ((6-1)!), and adjacent (A,B) arrangements are \(2!\times4!\). Exam tip: for not adjacent, subtract the adjacent case from total.
Step 2
Why this answer is correct
The correct answer is C. (72). Total circular arrangements are ((6-1)!), and adjacent (A,B) arrangements are \(2!\times4!\). Exam tip: for not adjacent, subtract the adjacent case from total.
Step 3
Exam Tip
कुल गोल व्यवस्थाएं ((6-1)!) हैं और (A,B) साथ होने पर \(2!\times4!\) व्यवस्थाएं हैं। परीक्षा में नहीं साथ के लिए कुल से साथ वाला मामला घटाएं।
Treat vowels (U,E) as one block; among (5) objects, (S) and (C) repeat twice, giving \(\frac{5!}{2!,2!}\times2!\). Exam tip: count the internal order of the vowel block separately.
Step 2
Why this answer is correct
The correct answer is D. (120). Treat vowels (U,E) as one block; among (5) objects, (S) and (C) repeat twice, giving \(\frac{5!}{2!,2!}\times2!\). Exam tip: count the internal order of the vowel block separately.
Step 3
Exam Tip
स्वर (U,E) को एक खंड मानें; (5) वस्तुओं में (S) दो और (C) दो हैं, इसलिए \(\frac{5!}{2!,2!}\times2!\) मिलता है। परीक्षा में स्वर-खंड के अंदरूनी क्रम को अलग गिनें।
The last place has (4) choices from (1,3,5,7), the first place then has (6) nonzero choices, followed by (6) and (5) choices. Exam tip: handle last digit and first digit restrictions first.
Step 2
Why this answer is correct
The correct answer is B. (720). The last place has (4) choices from (1,3,5,7), the first place then has (6) nonzero choices, followed by (6) and (5) choices. Exam tip: handle last digit and first digit restrictions first.
Step 3
Exam Tip
अंतिम स्थान (1,3,5,7) में से (4) तरीकों से और पहला स्थान शून्य छोड़कर (6) तरीकों से भरेगा, फिर (6) और (5) तरीके मिलेंगे। परीक्षा में अंतिम अंक और प्रथम अंक की शर्त पहले संभालें।
For a necklace, both rotation and reflection are identical, so the count is (\frac{(9-1)!}{2}). Exam tip: remember the difference between circular arrangements and necklaces.
Step 2
Why this answer is correct
The correct answer is A. (20160). For a necklace, both rotation and reflection are identical, so the count is (\frac{(9-1)!}{2}). Exam tip: remember the difference between circular arrangements and necklaces.
Step 3
Exam Tip
माला में घूर्णन और पलटना दोनों समान होते हैं, इसलिए संख्या (\frac{(9-1)!}{2}) है। परीक्षा में circular और necklace में अंतर याद रखें।
Choose the (4) correct letters in \(\binom{6}{4}\) ways, and derange the remaining (2) letters in (!2=1) way. Exam tip: split exact conditions into fixed and deranged parts.
Step 2
Why this answer is correct
The correct answer is C. (15). Choose the (4) correct letters in \(\binom{6}{4}\) ways, and derange the remaining (2) letters in (!2=1) way. Exam tip: split exact conditions into fixed and deranged parts.
Step 3
Exam Tip
सही जाने वाले (4) पत्र \(\binom{6}{4}\) तरीकों से चुनें और बाकी (2) का derangement (!2=1) है। परीक्षा में ठीक वाली शर्त के लिए fixed और deranged भाग अलग करें।
Subtract the cases where all (5) vowels form one block, \(4!\times5!\), from total (8!). Exam tip: not all together means not in one single block.
Step 2
Why this answer is correct
The correct answer is A. (38880). Subtract the cases where all (5) vowels form one block, \(4!\times5!\), from total (8!). Exam tip: not all together means not in one single block.
Step 3
Exam Tip
कुल (8!) से सभी (5) स्वरों को एक खंड मानकर \(4!\times5!\) घटाएं। परीक्षा में not all together का अर्थ सभी एक ही खंड में नहीं है।
Arrange the boys in (7!) ways and place the (3) girls in \(^{8}P_{3}\) of the (8) gaps. Exam tip: permute separated objects into gaps.
Step 2
Why this answer is correct
The correct answer is B. (1693440). Arrange the boys in (7!) ways and place the (3) girls in \(^{8}P_{3}\) of the (8) gaps. Exam tip: permute separated objects into gaps.
Step 3
Exam Tip
पहले लड़कों को (7!) तरीकों से बैठाएं और (8) gaps में (3) लड़कियों को \(^{8}P_{3}\) तरीकों से रखें। परीक्षा में separated objects को gaps में permute करें।
Choose the other (4) people from (11) in \(\binom{11}{4}\) ways and arrange the (5) people in (5!) ways. Exam tip: include the compulsory object first.
Step 2
Why this answer is correct
The correct answer is C. (55440). Choose the other (4) people from (11) in \(\binom{11}{4}\) ways and arrange the (5) people in (5!) ways. Exam tip: include the compulsory object first.
Step 3
Exam Tip
(A) के साथ बाकी (4) लोग \(\binom{11}{4}\) तरीकों से चुनें और (5!) तरीकों से व्यवस्थित करें। परीक्षा में compulsory object को पहले शामिल मानें।
There are (11) letters, with (B) and (I) repeated twice, so the count is \(\frac{11!}{2!,2!}\). Exam tip: divide by the factorial of each repeated letter.
Step 2
Why this answer is correct
The correct answer is A. (19958400). There are (11) letters, with (B) and (I) repeated twice, so the count is \(\frac{11!}{2!,2!}\). Exam tip: divide by the factorial of each repeated letter.
Step 3
Exam Tip
कुल (11) अक्षर हैं और (B,I) दो-दो बार आते हैं, इसलिए संख्या \(\frac{11!}{2!,2!}\) है। परीक्षा में हर repeated letter के factorial से अलग-अलग भाग दें।
Place the two special players in the (6) inner positions in \(^{6}P_{2}\) ways and arrange the remaining (6) players in (6!) ways. Exam tip: handle restricted positions first.
Step 2
Why this answer is correct
The correct answer is C. (30240). Place the two special players in the (6) inner positions in \(^{6}P_{2}\) ways and arrange the remaining (6) players in (6!) ways. Exam tip: handle restricted positions first.
Step 3
Exam Tip
दो विशेष खिलाड़ियों को (6) भीतरी स्थानों में \(^{6}P_{2}\) तरीकों से और बाकी (6) खिलाड़ियों को (6!) तरीकों से रखें। परीक्षा में प्रतिबंधित स्थानों को पहले अलग करें।
Treat the (4) blue balls as one block, giving (6!) outside arrangements and (4!) internal arrangements. Exam tip: count internal orders for distinct objects inside a block.
Step 2
Why this answer is correct
The correct answer is A. (17280). Treat the (4) blue balls as one block, giving (6!) outside arrangements and (4!) internal arrangements. Exam tip: count internal orders for distinct objects inside a block.
Step 3
Exam Tip
(4) नीली गेंदों को एक खंड मानें, तब (6!) बाहरी व्यवस्थाएं और (4!) अंदरूनी व्यवस्थाएं मिलती हैं। परीक्षा में distinct objects वाले खंड के अंदर क्रम जरूर गिनें।
Group digits by residues modulo (3), count (5)-digit selections whose sum is divisible by (3), then arrange each selection in (5!) ways. Exam tip: use digit sum for divisibility by (3).
Step 2
Why this answer is correct
The correct answer is C. (4560). Group digits by residues modulo (3), count (5)-digit selections whose sum is divisible by (3), then arrange each selection in (5!) ways. Exam tip: use digit sum for divisibility by (3).
Step 3
Exam Tip
अंकों को (3) शेष-वर्गों में बांटकर ऐसे (5)-अंकीय चयन गिनें जिनका योग (3) से विभाज्य हो, फिर प्रत्येक चयन को (5!) तरीकों से सजाएं। परीक्षा में divisibility by (3) के लिए digit sum प्रयोग करें।
Treat (ABC) as one block with fixed internal order, so the circular arrangement of (6) objects is ((6-1)!). Exam tip: if internal order is fixed, do not multiply by an internal factorial.
Step 2
Why this answer is correct
The correct answer is A. (720). Treat (ABC) as one block with fixed internal order, so the circular arrangement of (6) objects is ((6-1)!). Exam tip: if internal order is fixed, do not multiply by an internal factorial.
Step 3
Exam Tip
(ABC) को निश्चित अंदरूनी क्रम वाला एक खंड मानें, इसलिए कुल (6) वस्तुओं की गोल व्यवस्था ((6-1)!) है। परीक्षा में inside order fixed हो तो अंदर factorial न लगाएं।
There are (8) adjacent chair pairs and (2!) orders for the two special students, then the remaining (4) students occupy (7) chairs in \(^{7}P_{4}\) ways. Exam tip: with distinct chairs, account for empty chairs too.
Step 2
Why this answer is correct
The correct answer is B. (40320). There are (8) adjacent chair pairs and (2!) orders for the two special students, then the remaining (4) students occupy (7) chairs in \(^{7}P_{4}\) ways. Exam tip: with distinct chairs, account for empty chairs too.
Step 3
Exam Tip
दो विशेष छात्रों के लिए (8) adjacent chair-pairs और (2!) क्रम हैं, फिर बाकी (4) छात्र (7) बची कुर्सियों पर \(^{7}P_{4}\) तरीकों से बैठते हैं। परीक्षा में कुर्सियां distinct हों तो खाली कुर्सियों को भी ध्यान में रखें।
There are (12) letters; (N) and (E) occur four times each and (D) occurs twice, so the count is \(\frac{12!}{4!,4!,2!}\). Exam tip: correct repetition counts are crucial.
Step 2
Why this answer is correct
The correct answer is C. (6652800). There are (12) letters; (N) and (E) occur four times each and (D) occurs twice, so the count is \(\frac{12!}{4!,4!,2!}\). Exam tip: correct repetition counts are crucial.
Step 3
Exam Tip
कुल (12) अक्षर हैं; (N) और (E) चार-चार बार तथा (D) दो बार आते हैं, इसलिए संख्या \(\frac{12!}{4!,4!,2!}\) है। परीक्षा में repetition count सही लिखना सबसे जरूरी है।
Treat each couple as one block; the (4) blocks have ((4-1)!) circular arrangements and each block has (2!) internal orders. Exam tip: count the internal swap of every couple.
Step 2
Why this answer is correct
The correct answer is B. (384). Treat each couple as one block; the (4) blocks have ((4-1)!) circular arrangements and each block has (2!) internal orders. Exam tip: count the internal swap of every couple.
Step 3
Exam Tip
प्रत्येक जोड़े को एक खंड मानने पर (4) खंडों की गोल व्यवस्था ((4-1)!) और हर खंड के अंदर (2!) तरीके हैं। परीक्षा में हर couple के internal swap को अलग गिनें।
Choose the position of the special flag in (3) ways and fill the other (3) positions from (9) flags in \(^{9}P_{3}\) ways. Exam tip: count allowed positions of the compulsory object first.
Step 2
Why this answer is correct
The correct answer is B. (1512). Choose the position of the special flag in (3) ways and fill the other (3) positions from (9) flags in \(^{9}P_{3}\) ways. Exam tip: count allowed positions of the compulsory object first.
Step 3
Exam Tip
विशेष झंडे की स्थिति (3) तरीकों से चुने और बाकी (3) स्थान (9) झंडों से \(^{9}P_{3}\) तरीकों से भरें। परीक्षा में अनिवार्य वस्तु की allowed positions पहले गिनें।
Use inclusion-exclusion on total (8!), subtracting cases with (T) first and (E) last: (8!-7!-7!+6!). Exam tip: add back the overlap in two forbidden conditions.
Step 2
Why this answer is correct
The correct answer is B. (30960). Use inclusion-exclusion on total (8!), subtracting cases with (T) first and (E) last: (8!-7!-7!+6!). Exam tip: add back the overlap in two forbidden conditions.
Step 3
Exam Tip
कुल (8!) से (T) first और (E) last वाले मामलों को inclusion-exclusion से घटाएं, यानी (8!-7!-7!+6!)। परीक्षा में दो निषेध शर्तों में overlap जोड़ना न भूलें।
Arrange the (5) mathematics books in (5!) ways, then place the (4) physics books in \(^{6}P_{4}\) gaps. Exam tip: gaps are the safest method for no two together.
Step 2
Why this answer is correct
The correct answer is A. (86400). Arrange the (5) mathematics books in (5!) ways, then place the (4) physics books in \(^{6}P_{4}\) gaps. Exam tip: gaps are the safest method for no two together.
Step 3
Exam Tip
पहले (5) गणित पुस्तकें (5!) तरीकों से रखें, फिर (6) gaps में (4) भौतिकी पुस्तकें \(^{6}P_{4}\) तरीकों से रखें। परीक्षा में no two together के लिए gaps सबसे सुरक्षित तरीका है।
First choose ordered seat pairs for (A,B) with distance at least (3); there are (42) such pairs, then seat the remaining (4) people in \(^{6}P_{4}\) ways. Exam tip: position-pair counting is useful for numbered seats.
Step 2
Why this answer is correct
The correct answer is B. (16800). First choose ordered seat pairs for (A,B) with distance at least (3); there are (42) such pairs, then seat the remaining (4) people in \(^{6}P_{4}\) ways. Exam tip: position-pair counting is useful for numbered seats.
Step 3
Exam Tip
पहले (A,B) के सीट-जोड़ों में दूरी कम से कम (3) रखें; ऐसे ordered pairs (42) हैं, फिर बाकी (4) लोग \(^{6}P_{4}\) तरीकों से बैठते हैं। परीक्षा में numbered seats में position-pair गिनना उपयोगी है।
Total arrangements are \(\frac{7!}{2!}\), and among the (4) relative orders of vowels only (1) order is allowed, so divide by (4!). Exam tip: divide by the factorial when relative order is fixed.
Step 2
Why this answer is correct
The correct answer is D. (210). Total arrangements are \(\frac{7!}{2!}\), and among the (4) relative orders of vowels only (1) order is allowed, so divide by (4!). Exam tip: divide by the factorial when relative order is fixed.
Step 3
Exam Tip
कुल व्यवस्थाएं \(\frac{7!}{2!}\) हैं और (4) स्वरों के सापेक्ष क्रमों में केवल (1) क्रम स्वीकार्य है, इसलिए (4!) से भाग दें। परीक्षा में relative order fixed हो तो संबंधित factorial से divide करें।
In total (7!) arrangements, the two relative orders of (A) and (B) are equally likely, so half are valid. Exam tip: use symmetry for relative order conditions like left of.
Step 2
Why this answer is correct
The correct answer is A. (2520). In total (7!) arrangements, the two relative orders of (A) and (B) are equally likely, so half are valid. Exam tip: use symmetry for relative order conditions like left of.
Step 3
Exam Tip
कुल (7!) व्यवस्थाओं में (A) और (B) के दो सापेक्ष क्रम समान रूप से संभव हैं, इसलिए आधी व्यवस्थाएं मान्य हैं। परीक्षा में left of जैसी relative order शर्त के लिए symmetry प्रयोग करें।
There are (2) alternate patterns, and in each pattern Indians can be arranged in (5!) ways and foreigners in (5!) ways. Exam tip: equal groups in alternate arrangement give (2) patterns.
Step 2
Why this answer is correct
The correct answer is A. (28800). There are (2) alternate patterns, and in each pattern Indians can be arranged in (5!) ways and foreigners in (5!) ways. Exam tip: equal groups in alternate arrangement give (2) patterns.
Step 3
Exam Tip
दो alternate patterns संभव हैं और प्रत्येक में भारतीय (5!) तथा विदेशी (5!) तरीकों से बैठते हैं। परीक्षा में बराबर समूहों के alternate arrangement में (2) patterns होते हैं।
Choose the other (4) digits with (0) and (8) in \(\binom{7}{4}\) ways; for each selection subtract (5!) leading-zero cases from (6!). Exam tip: leading zero correction is mandatory.
Step 2
Why this answer is correct
The correct answer is B. (13440). Choose the other (4) digits with (0) and (8) in \(\binom{7}{4}\) ways; for each selection subtract (5!) leading-zero cases from (6!). Exam tip: leading zero correction is mandatory.
Step 3
Exam Tip
(0) और (8) के साथ बाकी (4) अंक \(\binom{7}{4}\) तरीकों से चुनें; प्रत्येक चयन में कुल (6!) से first zero वाले (5!) घटाएं। परीक्षा में leading zero का correction अनिवार्य है।
We assign (4) of the (6) rings to (4) distinct fingers with order, so \(^{6}P_{4}=360\). Exam tip: distinct fingers require permutations.
Step 2
Why this answer is correct
The correct answer is A. (360). We assign (4) of the (6) rings to (4) distinct fingers with order, so \(^{6}P_{4}=360\). Exam tip: distinct fingers require permutations.
Step 3
Exam Tip
(6) अंगूठियों में से (4) को (4) अलग-अलग उंगलियों पर क्रम सहित रखना है, इसलिए \(^{6}P_{4}=360\)। परीक्षा में distinct fingers होने पर permutation लगती है।
Fix (A); the opposite seat for (B) is fixed, and the remaining (8) people can be arranged in (8!) ways. Exam tip: fix one person to remove circular symmetry.
Step 2
Why this answer is correct
The correct answer is A. (80640). Fix (A); the opposite seat for (B) is fixed, and the remaining (8) people can be arranged in (8!) ways. Exam tip: fix one person to remove circular symmetry.
Step 3
Exam Tip
(A) को स्थिर करें, (B) की विपरीत सीट निश्चित है, और बाकी (8) लोग (8!) तरीकों से बैठते हैं। परीक्षा में circular symmetry हटाने के लिए एक व्यक्ति को स्थिर करें।
There are (11) letters; (I) occurs four times, (S) occurs four times, and (P) occurs twice, so the count is \(\frac{11!}{4!,4!,2!}\). Exam tip: recheck repeated-letter counts.
Step 2
Why this answer is correct
The correct answer is A. (34650). There are (11) letters; (I) occurs four times, (S) occurs four times, and (P) occurs twice, so the count is \(\frac{11!}{4!,4!,2!}\). Exam tip: recheck repeated-letter counts.
Step 3
Exam Tip
कुल (11) अक्षर हैं; (I) चार, (S) चार और (P) दो बार आते हैं, इसलिए \(\frac{11!}{4!,4!,2!}\) है। परीक्षा में समान अक्षरों की गिनती दोबारा जांचें।
Choose the (2) correctly assigned prizes in \(\binom{7}{2}\) ways and derange the remaining (5), giving (!5=44). Exam tip: use derangement for exactly correct cases.
Step 2
Why this answer is correct
The correct answer is B. (1854). Choose the (2) correctly assigned prizes in \(\binom{7}{2}\) ways and derange the remaining (5), giving (!5=44). Exam tip: use derangement for exactly correct cases.
Step 3
Exam Tip
सही जाने वाले (2) पुरस्कार \(\binom{7}{2}\) तरीकों से चुनें और बाकी (5) का derangement (!5=44) है। परीक्षा में exactly correct cases के लिए derangement का प्रयोग करें।
Choose the other (3) letters from (6), then subtract adjacent special-pair arrangements \(2!\times4!\) from total (5!). Exam tip: handle inclusion and adjacency restrictions separately.
Step 2
Why this answer is correct
The correct answer is C. (2160). Choose the other (3) letters from (6), then subtract adjacent special-pair arrangements \(2!\times4!\) from total (5!). Exam tip: handle inclusion and adjacency restrictions separately.
Step 3
Exam Tip
बाकी (3) अक्षर (6) में से चुनें, फिर (5!) कुल व्यवस्थाओं से special pair adjacent वाली \(2!\times4!\) व्यवस्थाएं घटाएं। परीक्षा में शामिल और adjacent restriction को अलग-अलग संभालें।
The thousands digit must be from (5,6,7,8,9), and the units digit must be even; case-wise counting gives (400). Exam tip: count overlapping digit choices case by case.
Step 2
Why this answer is correct
The correct answer is A. (400). The thousands digit must be from (5,6,7,8,9), and the units digit must be even; case-wise counting gives (400). Exam tip: count overlapping digit choices case by case.
Step 3
Exam Tip
हजारों का अंक (5,6,7,8,9) में से और इकाई अंक सम होना चाहिए; leading और ending cases अलग गिनने पर (400) मिलता है। परीक्षा में overlapping digit choices को case-wise गिनें।
Form an (AB) block among (8) objects, then subtract cases where (C) is adjacent to the block on either outside side; total is (2!\times\(8!-2\times7!\)). Exam tip: compound restrictions may need both block and subtraction.
Step 2
Why this answer is correct
The correct answer is D. (10080). Form an (AB) block among (8) objects, then subtract cases where (C) is adjacent to the block on either outside side; total is (2!\times\(8!-2\times7!\)). Exam tip: compound restrictions may need both block and subtraction.
Step 3
Exam Tip
(AB) खंड बनाकर (8) वस्तुओं की व्यवस्थाएं लें, फिर (C) के खंड से adjacent होने वाले (2) बाहरी स्थानों को घटाएं; कुल (2!\times\(8!-2\times7!\)) है। परीक्षा में compound restriction में block और subtraction दोनों लग सकते हैं।
There are (4) position pairs at distance (2) and (2!) orders for questions (1) and (2), and the remaining (4) questions arrange in (4!) ways. Exam tip: exactly one between means position difference (2).
Step 2
Why this answer is correct
The correct answer is B. (192). There are (4) position pairs at distance (2) and (2!) orders for questions (1) and (2), and the remaining (4) questions arrange in (4!) ways. Exam tip: exactly one between means position difference (2).
Step 3
Exam Tip
प्रश्न (1) और (2) के लिए दूरी (2) वाले (4) position-pairs और (2!) क्रम हैं, बाकी (4) प्रश्न (4!) तरीकों से लगेंगे। परीक्षा में exactly one between का अर्थ position difference (2) है।
Choose the end for (P) and (E) in (2) ways; the remaining (5) letters include (E) repeated twice, so there are \(\frac{5!}{2!}\) arrangements. Exam tip: divide for a repeated end-letter left inside.
Step 2
Why this answer is correct
The correct answer is B. (240). Choose the end for (P) and (E) in (2) ways; the remaining (5) letters include (E) repeated twice, so there are \(\frac{5!}{2!}\) arrangements. Exam tip: divide for a repeated end-letter left inside.
Step 3
Exam Tip
एक सिरा (P) के लिए और दूसरा (E) के लिए (2) तरीकों से चुने; बाकी (5) अक्षरों में (E) दो बार है, इसलिए \(\frac{5!}{2!}\) व्यवस्थाएं हैं। परीक्षा में repeated end-letter बच जाने पर उसका factorial divide करें।
Subtract adjacent-special-car cases \(9\times2!\times{}^{8}P_{6}\) from total \(^{10}P_{8}\). Exam tip: when parking spots are distinct, start from total arrangements.
Step 2
Why this answer is correct
The correct answer is B. (1451520). Subtract adjacent-special-car cases \(9\times2!\times{}^{8}P_{6}\) from total \(^{10}P_{8}\). Exam tip: when parking spots are distinct, start from total arrangements.
Step 3
Exam Tip
कुल \(^{10}P_{8}\) से special cars adjacent वाली \(9\times2!\times{}^{8}P_{6}\) व्यवस्थाएं घटाएं। परीक्षा में parking spots distinct हों तो पहले कुल arrangements लें।
First arrange the (5) white balls around a circle in ((5-1)!) ways, then place (3) black balls in \(^{5}P_{3}\) circular gaps. Exam tip: in circular gaps, arrange the base objects first.
Step 2
Why this answer is correct
The correct answer is A. (2880). First arrange the (5) white balls around a circle in ((5-1)!) ways, then place (3) black balls in \(^{5}P_{3}\) circular gaps. Exam tip: in circular gaps, arrange the base objects first.
Step 3
Exam Tip
पहले (5) सफेद गेंदों की गोल व्यवस्था ((5-1)!) है, फिर (5) gaps में (3) काली गेंदें \(^{5}P_{3}\) तरीकों से रखी जाती हैं। परीक्षा में circular gaps में सफेद वस्तुओं को पहले fix करें।
Among total (7!) arrangements, the (3!) relative orders of (A,B,C) are equally likely, so \(\frac{7!}{3!}\) are valid. Exam tip: divide by factorial for fixed relative order.
Step 2
Why this answer is correct
The correct answer is A. (840). Among total (7!) arrangements, the (3!) relative orders of (A,B,C) are equally likely, so \(\frac{7!}{3!}\) are valid. Exam tip: divide by factorial for fixed relative order.
Step 3
Exam Tip
कुल (7!) व्यवस्थाओं में (A,B,C) के (3!) सापेक्ष क्रम समान हैं, इसलिए \(\frac{7!}{3!}\) मान्य हैं। परीक्षा में fixed relative order के लिए factorial से divide करें।
Treat each couple as one block with fixed internal order, so only the (4!) blocks are arranged. Exam tip: do not multiply by (2!) when internal order is fixed.
Step 2
Why this answer is correct
The correct answer is A. (24). Treat each couple as one block with fixed internal order, so only the (4!) blocks are arranged. Exam tip: do not multiply by (2!) when internal order is fixed.
Step 3
Exam Tip
हर दंपति को निश्चित अंदरूनी क्रम वाला एक खंड मानें, इसलिए केवल (4!) खंडों को सजाना है। परीक्षा में internal order fixed हो तो (2!) न लगाएं।
Zero can occupy only the hundreds or tens place, giving (2) choices; the other (3) places are filled from (6) nonzero digits in \(^{6}P_{3}\) ways. Exam tip: decide the allowed positions of (0) first.
Step 2
Why this answer is correct
The correct answer is B. (300). Zero can occupy only the hundreds or tens place, giving (2) choices; the other (3) places are filled from (6) nonzero digits in \(^{6}P_{3}\) ways. Exam tip: decide the allowed positions of (0) first.
Step 3
Exam Tip
(0) को केवल सैकड़ों या दहाइयों के स्थान पर रख सकते हैं, यानी (2) तरीके; फिर बाकी (3) स्थान (6) nonzero अंकों से \(^{6}P_{3}\) तरीकों से भरते हैं। परीक्षा में (0) की allowed positions पहले तय करें।
Treat (AB) and (CD) as two blocks; (6) objects can be arranged in (6!) ways and the blocks internally in \(2!\times2!\) ways. Exam tip: with multiple blocks, count each block's internal order.
Step 2
Why this answer is correct
The correct answer is A. (2880). Treat (AB) and (CD) as two blocks; (6) objects can be arranged in (6!) ways and the blocks internally in \(2!\times2!\) ways. Exam tip: with multiple blocks, count each block's internal order.
Step 3
Exam Tip
(AB) और (CD) को दो खंड मानें; कुल (6) वस्तुएं (6!) तरीकों से और दोनों खंडों के अंदर \(2!\times2!\) तरीके हैं। परीक्षा में कई blocks हों तो हर block का internal order गिनें।
