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।
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।
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।
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।
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।
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 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।
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।
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।
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।
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।
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।
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।
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।
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।
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।
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।
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।
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।
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।
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।
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।
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।
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।
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।
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।
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।
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।
\(\frac{8!}{4!\cdot2!\cdot2!}=\frac{8\cdot7\cdot6\cdot5}{4}=420\). Cancel (4!) first and calculate with smaller factors.
Step 2
Why this answer is correct
The correct answer is A. (420). \(\frac{8!}{4!\cdot2!\cdot2!}=\frac{8\cdot7\cdot6\cdot5}{4}=420\). Cancel (4!) first and calculate with smaller factors.
Step 3
Exam Tip
\(\frac{8!}{4!\cdot2!\cdot2!}=\frac{8\cdot7\cdot6\cdot5}{4}=420\) होता है। पहले (4!) काटकर छोटे पदों में गणना करें।
