Treat the vowels as one block, giving (5!) arrangements and (3!/2!) internal arrangements. In exams, always check repeated letters.
Step 2
Why this answer is correct
The correct answer is B. (360). Treat the vowels as one block, giving (5!) arrangements and (3!/2!) internal arrangements. In exams, always check repeated letters.
Step 3
Exam Tip
स्वरों को एक खंड मानने पर (5!) व्यवस्थाएं और भीतर (3!/2!) व्यवस्थाएं मिलती हैं। परीक्षा में समान अक्षरों की पुनरावृत्ति जरूर जांचें।
There are (6) letters with (S) repeated four times, so the number is (6!/4!). In exams, do not forget to divide by repeated letters.
Step 2
Why this answer is correct
The correct answer is B. (30). There are (6) letters with (S) repeated four times, so the number is (6!/4!). In exams, do not forget to divide by repeated letters.
Step 3
Exam Tip
कुल (6) अक्षरों में (S) चार बार है, इसलिए संख्या (6!/4!) है। परीक्षा में समान अक्षरों से भाग देना न भूलें।
The last digit must be (0) or (5), and the leading zero case must be counted separately. For such questions, start with the unit digit.
Step 2
Why this answer is correct
The correct answer is A. (108). The last digit must be (0) or (5), and the leading zero case must be counted separately. For such questions, start with the unit digit.
Step 3
Exam Tip
अंतिम अंक (0) या (5) होगा और पहले अंक के शून्य होने की स्थिति अलग गिननी होगी। ऐसे प्रश्नों में अंतिम अंक से केस बनाएं।
Subtract the together cases \(2\cdot 6!\) from the total (7!). For restriction questions, subtracting bad cases is often easier.
Step 2
Why this answer is correct
The correct answer is A. (3600). Subtract the together cases \(2\cdot 6!\) from the total (7!). For restriction questions, subtracting bad cases is often easier.
Step 3
Exam Tip
कुल (7!) में से साथ बैठने वाली \(2\cdot 6!\) व्यवस्थाएं घटाएं। निषेध वाले प्रश्नों में कुल से खराब व्यवस्थाएं घटाना आसान होता है।
Treat the two people as one block, so (7) units have circular arrangements (6!), with (2!) internal ways. In circular arrangements, rotations are identical.
Step 2
Why this answer is correct
The correct answer is B. (1440). Treat the two people as one block, so (7) units have circular arrangements (6!), with (2!) internal ways. In circular arrangements, rotations are identical.
Step 3
Exam Tip
दो लोगों को एक खंड मानकर (7) इकाइयों की गोल व्यवस्था (6!) और भीतर (2!) तरीके हैं। गोल व्यवस्था में घूर्णन को स्थिर मानें।
There are (4) possible position pairs for (T) and (E), with (2) orders, and the remaining (6!) arrangements. In position-based questions, choose fixed positions first.
Step 2
Why this answer is correct
The correct answer is C. (1920). There are (4) possible position pairs for (T) and (E), with (2) orders, and the remaining (6!) arrangements. In position-based questions, choose fixed positions first.
Step 3
Exam Tip
(T) और (E) की जगहों के (4) संभावित जोड़े हैं और क्रम (2) तरीके से होगा, बाकी (6!) हैं। स्थान-आधारित प्रश्नों में पहले निश्चित स्थान चुनें।
The last digit can be (2,4,6) in (3) ways, and the remaining four places in \(^{6}P_{4}\) ways. For even numbers, fix the unit digit first.
Step 2
Why this answer is correct
The correct answer is B. (1080). The last digit can be (2,4,6) in (3) ways, and the remaining four places in \(^{6}P_{4}\) ways. For even numbers, fix the unit digit first.
Step 3
Exam Tip
अंतिम अंक (2,4,6) में से (3) तरीकों से होगा और शेष चार स्थान \(^{6}P_{4}\) तरीकों से भरेंगे। सम संख्या में इकाई अंक पहले तय करें।
There are (9) letters with (M,T,E) each repeated twice, so the count is (9!/(2!2!2!)). For difficult words, list letter counts first.
Step 2
Why this answer is correct
The correct answer is A. (45360). There are (9) letters with (M,T,E) each repeated twice, so the count is (9!/(2!2!2!)). For difficult words, list letter counts first.
Step 3
Exam Tip
कुल (9) अक्षरों में (M,T,E) दो-दो बार हैं, इसलिए संख्या (9!/(2!2!2!)) है। कठिन शब्दों में अक्षरों की गिनती पहले लिखें।
Arrange the boys in (5!) ways, then place the girls in \(^{6}P_{4}\) ways in (4) of the (6) gaps. The gap method is safest for such questions.
Step 2
Why this answer is correct
The correct answer is A. (86400). Arrange the boys in (5!) ways, then place the girls in \(^{6}P_{4}\) ways in (4) of the (6) gaps. The gap method is safest for such questions.
Step 3
Exam Tip
पहले लड़कों को (5!) तरीकों से बैठाएं और (6) खाली स्थानों में से (4) पर लड़कियां \(^{6}P_{4}\) तरीकों से बैठेंगी। अंतराल विधि ऐसे प्रश्नों में सबसे सुरक्षित है।
The two specified books occupy the ends in (2!) ways, and the remaining (4) books in (4!) ways. Fill the most restricted places first.
Step 2
Why this answer is correct
The correct answer is B. (48). The two specified books occupy the ends in (2!) ways, and the remaining (4) books in (4!) ways. Fill the most restricted places first.
Step 3
Exam Tip
दो निर्धारित पुस्तकें सिरों पर (2!) तरीकों से और शेष (4) पुस्तकें (4!) तरीकों से लगेंगी। पहले सबसे अधिक प्रतिबंधित स्थान भरें।
In the total (6!) arrangements, the two orders of (A) and (B) are equally likely, so half are valid. Symmetry saves time in order-condition questions.
Step 2
Why this answer is correct
The correct answer is C. (360). In the total (6!) arrangements, the two orders of (A) and (B) are equally likely, so half are valid. Symmetry saves time in order-condition questions.
Step 3
Exam Tip
कुल (6!) व्यवस्थाओं में (A) और (B) के दोनों क्रम बराबर हैं, इसलिए आधी व्यवस्थाएं सही हैं। क्रम-शर्त में सममिति से समय बचता है।
There are (4) odd positions among (8), and the (4) vowels occupy them in (4!) ways, while consonants also arrange in (4!) ways. When counts match, arrange groups separately.
Step 2
Why this answer is correct
The correct answer is A. (1440). There are (4) odd positions among (8), and the (4) vowels occupy them in (4!) ways, while consonants also arrange in (4!) ways. When counts match, arrange groups separately.
Step 3
Exam Tip
(8) स्थानों में (4) विषम स्थान हैं और (4) स्वर उन्हीं में (4!) तरीकों से आएंगे, व्यंजन भी (4!) तरीकों से। स्थान की समान संख्या दिखे तो सीधे अलग-अलग व्यवस्थित करें।
The digit sum must be divisible by (3); arrange valid (5)-digit sets and subtract leading-zero arrangements. Choosing digit sets first is useful.
Step 2
Why this answer is correct
The correct answer is C. (360). The digit sum must be divisible by (3); arrange valid (5)-digit sets and subtract leading-zero arrangements. Choosing digit sets first is useful.
Step 3
Exam Tip
अंकों का योग (3) से विभाज्य होना चाहिए; मान्य (5)-अंकीय समूहों की व्यवस्थाओं में अग्र शून्य घटाएं। पहले अंकों के समूह चुनना उपयोगी है।
For a necklace, the count is ((9-1)!/2) because both rotation and reflection are identical. Remember the difference between circular seating and necklaces.
Step 2
Why this answer is correct
The correct answer is A. (20160). For a necklace, the count is ((9-1)!/2) because both rotation and reflection are identical. Remember the difference between circular seating and necklaces.
Step 3
Exam Tip
माला में संख्या ((9-1)!/2) होती है क्योंकि घूर्णन और प्रतिबिंब दोनों समान माने जाते हैं। गोल मेज और माला के अंतर को याद रखें।
Add arrangements beginning with smaller available letters at each position, then add (1). In rank questions, write the letters in alphabetical order first.
Step 2
Why this answer is correct
The correct answer is B. (360). Add arrangements beginning with smaller available letters at each position, then add (1). In rank questions, write the letters in alphabetical order first.
Step 3
Exam Tip
हर स्थान पर उससे छोटे उपलब्ध अक्षरों से शुरू होने वाली व्यवस्थाएं जोड़कर अंत में (1) जोड़ते हैं। रैंक प्रश्नों में अक्षरों को पहले वर्णक्रम में लिखें।
This is the derangement of (4) objects, whose value is (9). For such questions, use the derangement formula or a small listing.
Step 2
Why this answer is correct
The correct answer is C. (9). This is the derangement of (4) objects, whose value is (9). For such questions, use the derangement formula or a small listing.
Step 3
Exam Tip
यह (4) वस्तुओं का पूर्ण विस्थापन है और मान (9) होता है। ऐसे प्रश्नों में विस्थापन सूत्र या छोटी सूची काम आती है।
There are (7) letters with (L) and (O) each repeated twice, so the count is (7!/(2!2!)). Repeated letters are not treated as distinct objects.
Step 2
Why this answer is correct
The correct answer is A. (1260). There are (7) letters with (L) and (O) each repeated twice, so the count is (7!/(2!2!)). Repeated letters are not treated as distinct objects.
Step 3
Exam Tip
(7) अक्षरों में (L) और (O) दो-दो बार हैं, इसलिए संख्या (7!/(2!2!)) है। समान अक्षर हर बार अलग वस्तु नहीं माने जाते।
Treat the two (L)'s as one block, giving (6) units with (O) repeated twice, so the count is (6!/2!). Use the block method for together conditions.
Step 2
Why this answer is correct
The correct answer is B. (360). Treat the two (L)'s as one block, giving (6) units with (O) repeated twice, so the count is (6!/2!). Use the block method for together conditions.
Step 3
Exam Tip
दोनों (L) को एक खंड मानने पर (6) इकाइयां हैं और (O) दो बार है, इसलिए (6!/2!) है। साथ वाली शर्त में खंड विधि लगाएं।
The thousands digit can be (4,5,6,7,8), and the remaining (3) places are filled in \(^{7}P_{3}\) ways. In comparison questions, fix the first digit first.
Step 2
Why this answer is correct
The correct answer is A. (840). The thousands digit can be (4,5,6,7,8), and the remaining (3) places are filled in \(^{7}P_{3}\) ways. In comparison questions, fix the first digit first.
Step 3
Exam Tip
हजार का अंक (4,5,6,7,8) में से होगा और बाकी (3) स्थान \(^{7}P_{3}\) तरीकों से भरेंगे। तुलना वाले प्रश्नों में पहले पहला अंक तय करें।
There are two possible patterns, and in each pattern men arrange in (6!) ways and women in (6!) ways. For equal alternate seating, remember the two patterns.
Step 2
Why this answer is correct
The correct answer is A. \(2\cdot 6!\cdot 6!\). There are two possible patterns, and in each pattern men arrange in (6!) ways and women in (6!) ways. For equal alternate seating, remember the two patterns.
Step 3
Exam Tip
बैठने की दो रूपरेखाएं संभव हैं और हर रूपरेखा में पुरुष (6!) तथा महिलाएं (6!) तरीकों से बैठते हैं। समान संख्या में वैकल्पिक बैठाने पर दो पैटर्न याद रखें।
The total distinct arrangements are (7!/2!), and among the possible vowel orders only one valid alphabetical order is allowed. Count order restrictions carefully when vowels repeat.
Step 2
Why this answer is correct
The correct answer is B. (840). The total distinct arrangements are (7!/2!), and among the possible vowel orders only one valid alphabetical order is allowed. Count order restrictions carefully when vowels repeat.
Step 3
Exam Tip
कुल (7!) व्यवस्थाओं में (E,E,U) के प्रभाव से भिन्न व्यवस्थाएं (7!/2!) हैं और स्वरों के क्रमों में केवल एक क्रम मान्य है। समान स्वरों सहित क्रम-शर्त सावधानी से गिनें।
Subtract the together cases (5!) from the total (6!/2!). For not-together conditions, subtract together arrangements from total.
Step 2
Why this answer is correct
The correct answer is A. (240). Subtract the together cases (5!) from the total (6!/2!). For not-together conditions, subtract together arrangements from total.
Step 3
Exam Tip
कुल (6!/2!) में से (O) साथ वाली (5!) व्यवस्थाएं घटाएं। न-साथ वाली शर्त में कुल से साथ वाली व्यवस्थाएं घटाएं।
Subtract the (6!) arrangements where the specified flag is at the top from the total (7!). For one forbidden position, subtraction is quick.
Step 2
Why this answer is correct
The correct answer is B. (4320). Subtract the (6!) arrangements where the specified flag is at the top from the total (7!). For one forbidden position, subtraction is quick.
Step 3
Exam Tip
कुल (7!) में से विशेष झंडे के ऊपर होने पर (6!) व्यवस्थाएं घटाएं। निषेध वाली एक जगह हो तो घटाने की विधि तेज होती है।
The last digit can be (1,3,5) in (3) ways; the first place then has (4) nonzero choices, followed by \(4\cdot 3\). For odd numbers, fix the unit digit first.
Step 2
Why this answer is correct
The correct answer is B. (144). The last digit can be (1,3,5) in (3) ways; the first place then has (4) nonzero choices, followed by \(4\cdot 3\). For odd numbers, fix the unit digit first.
Step 3
Exam Tip
अंतिम अंक (1,3,5) में से (3) तरीकों से होगा; पहले स्थान के लिए शून्य को छोड़कर (4) विकल्प और बाकी \(4\cdot 3\) हैं। विषम संख्या में इकाई अंक पहले तय करें।
There are (10) letters with (S) three times, (T) three times and (I) twice, so the count is (10!/(3!3!2!)). For long words, make a frequency table.
Step 2
Why this answer is correct
The correct answer is A. (50400). There are (10) letters with (S) three times, (T) three times and (I) twice, so the count is (10!/(3!3!2!)). For long words, make a frequency table.
Step 3
Exam Tip
(10) अक्षरों में (S) तीन बार, (T) तीन बार और (I) दो बार है, इसलिए संख्या (10!/(3!3!2!)) है। बड़े शब्दों में आवृत्ति तालिका बनाएं।
(N) and (R) can occupy the ends in (2!) ways, and the remaining (4) letters in (4!) ways. With end restrictions, fill the ends first.
Step 2
Why this answer is correct
The correct answer is B. (48). (N) and (R) can occupy the ends in (2!) ways, and the remaining (4) letters in (4!) ways. With end restrictions, fill the ends first.
Step 3
Exam Tip
(N) और (R) सिरों पर (2!) तरीकों से होंगे और शेष (4) अक्षर (4!) तरीकों से। सिरों पर शर्त हो तो पहले सिरों को भरें।
Treat (1) and (2) as a block, giving (5!) arrangements and (2!) internal orders. For adjacent digits, the block method is direct.
Step 2
Why this answer is correct
The correct answer is C. (240). Treat (1) and (2) as a block, giving (5!) arrangements and (2!) internal orders. For adjacent digits, the block method is direct.
Step 3
Exam Tip
(1) और (2) को खंड मानकर (5!) व्यवस्थाएं और खंड के भीतर (2!) क्रम हैं। साथ वाले अंकों में खंड विधि सीधी होती है।
This is an ordered selection of (4) books from (9), so the number is \(^{9}P_{4}\). Arrangement on a shelf requires order.
Step 2
Why this answer is correct
The correct answer is A. (3024). This is an ordered selection of (4) books from (9), so the number is \(^{9}P_{4}\). Arrangement on a shelf requires order.
Step 3
Exam Tip
यह (9) में से (4) पुस्तकों का क्रम सहित चयन है, इसलिए \(^{9}P_{4}\) होगा। शेल्फ पर लगाने में क्रम जरूरी है।
Adding arrangements starting with smaller available letters at each position and then adding (1) gives rank (84). In rank problems, count position by position.
Step 2
Why this answer is correct
The correct answer is C. (84). Adding arrangements starting with smaller available letters at each position and then adding (1) gives rank (84). In rank problems, count position by position.
Step 3
Exam Tip
हर स्थान पर छोटे उपलब्ध अक्षरों से शुरू होने वाली व्यवस्थाएं जोड़कर (1) जोड़ने पर रैंक (84) मिलती है। रैंक में शुरुआत से अंत तक स्थानवार गिनें।
Total circular arrangements are (7!), and adjacent cases are \(2\cdot 6!\), so subtraction gives (3600). In circular arrangements, start with ((n-1)!).
Step 2
Why this answer is correct
The correct answer is A. (3600). Total circular arrangements are (7!), and adjacent cases are \(2\cdot 6!\), so subtraction gives (3600). In circular arrangements, start with ((n-1)!).
Step 3
Exam Tip
कुल गोल व्यवस्थाएं (7!) हैं और साथ वाली \(2\cdot 6!\) हैं, इसलिए घटाने पर (3600) मिलता है। गोल व्यवस्था में कुल ((n-1)!) से शुरू करें।
There are (8) letters with (C,L,U) each repeated twice, so the count is (8!/(2!2!2!)). Identifying repetitions is the key step.
Step 2
Why this answer is correct
The correct answer is A. (10080). There are (8) letters with (C,L,U) each repeated twice, so the count is (8!/(2!2!2!)). Identifying repetitions is the key step.
Step 3
Exam Tip
(8) अक्षरों में (C,L,U) दो-दो बार हैं, इसलिए संख्या (8!/(2!2!2!)) है। आवृत्तियों को पहचानना मुख्य कदम है।
There are (4) position pairs for (A,B) and (2) orders, then (5!) ways for the rest. For a fixed gap, count position pairs.
Step 2
Why this answer is correct
The correct answer is B. (960). There are (4) position pairs for (A,B) and (2) orders, then (5!) ways for the rest. For a fixed gap, count position pairs.
Step 3
Exam Tip
(A,B) के स्थानों के (4) जोड़े और (2) क्रम हैं, फिर शेष (5!) तरीके हैं। निश्चित दूरी में स्थान-जोड़े गिनें।
If first is (6), count \(^{6}P_{4}\); if last is (6), subtract leading-zero cases from choices; then subtract the intersection. Use inclusion-exclusion for 'or'.
Step 2
Why this answer is correct
The correct answer is A. (600). If first is (6), count \(^{6}P_{4}\); if last is (6), subtract leading-zero cases from choices; then subtract the intersection. Use inclusion-exclusion for 'or'.
Step 3
Exam Tip
पहला (6) होने पर \(^{6}P_{4}\), अंतिम (6) होने पर अग्र शून्य घटाकर \(5\cdot ^{5}P_{3}\), और दोनों स्थिति का प्रतिच्छेद घटता है। 'या' में समावेशन-बहिष्करण लगाएं।
For (4) fingers, choose and assign rings in \(^{6}P_{4}\) ways. Placing objects in distinct positions is an ordered selection.
Step 2
Why this answer is correct
The correct answer is A. (360). For (4) fingers, choose and assign rings in \(^{6}P_{4}\) ways. Placing objects in distinct positions is an ordered selection.
Step 3
Exam Tip
(4) उंगलियों के लिए (6) अंगूठियों में से क्रम सहित चयन \(^{6}P_{4}\) है। अलग स्थानों पर वस्तु रखना क्रम सहित चयन होता है।
There are (11) letters with (I,N,A) each repeated twice, so the count is (11!/(2!2!2!)). In long words, note repeated letters separately.
Step 2
Why this answer is correct
The correct answer is A. (9979200). There are (11) letters with (I,N,A) each repeated twice, so the count is (11!/(2!2!2!)). In long words, note repeated letters separately.
Step 3
Exam Tip
(11) अक्षरों में (I,N,A) दो-दो बार हैं, इसलिए संख्या (11!/(2!2!2!)) है। बड़े शब्दों में दोहराए अक्षर अलग से नोट करें।
Treat the three women as one block, so (6) units have (5!) circular arrangements and the women have (3!) internal ways. In circular block problems, count units carefully.
Step 2
Why this answer is correct
The correct answer is A. (720). Treat the three women as one block, so (6) units have (5!) circular arrangements and the women have (3!) internal ways. In circular block problems, count units carefully.
Step 3
Exam Tip
तीन महिलाओं को एक खंड मानकर (6) इकाइयों की गोल व्यवस्था (5!) है और महिलाओं के भीतर (3!) तरीके हैं। गोल खंड विधि में इकाइयों की संख्या ध्यान से लें।
Divisibility by (4) depends on the last two digits; there are (15) valid ordered ending pairs and the first two places fill in \(^{5}P_{2}\) ways. For divisibility by (4), check the last two digits.
Step 2
Why this answer is correct
The correct answer is A. (90). Divisibility by (4) depends on the last two digits; there are (15) valid ordered ending pairs and the first two places fill in \(^{5}P_{2}\) ways. For divisibility by (4), check the last two digits.
Step 3
Exam Tip
अंतिम दो अंकों से (4) से विभाज्यता तय होती है; मान्य क्रमित अंतिम जोड़ों की संख्या (15) है और पहले दो स्थान \(^{5}P_{2}\) तरीकों से भरते हैं। (4) से विभाज्यता में अंतिम दो अंक जांचें।
The two specified pictures occupy the two middle positions in (2!) ways, and the remaining (4) pictures in (4!) ways. Fill restricted positions first.
Step 2
Why this answer is correct
The correct answer is B. (48). The two specified pictures occupy the two middle positions in (2!) ways, and the remaining (4) pictures in (4!) ways. Fill restricted positions first.
Step 3
Exam Tip
दो विशेष चित्र बीच के (2) स्थानों पर (2!) तरीकों से और बाकी (4) चित्र (4!) तरीकों से लगेंगे। प्रतिबंधित स्थान पहले भरें।
This is an ordered selection of (3) from (5), so it is \(^{5}P_{3}\). When there is no repetition and order matters, use permutation.
Step 2
Why this answer is correct
The correct answer is C. (60). This is an ordered selection of (3) from (5), so it is \(^{5}P_{3}\). When there is no repetition and order matters, use permutation.
Step 3
Exam Tip
यह (5) में से (3) का क्रम सहित चयन है, इसलिए \(^{5}P_{3}\) होगा। बिना पुनरावृत्ति और क्रम हो तो क्रमचय लगाएं।
With (P) fixed first, subtract the (4!) arrangements where (L) is last from the remaining (5!). In mixed conditions, apply fixed conditions first.
Step 2
Why this answer is correct
The correct answer is A. (96). With (P) fixed first, subtract the (4!) arrangements where (L) is last from the remaining (5!). In mixed conditions, apply fixed conditions first.
Step 3
Exam Tip
(P) पहले निश्चित है; शेष (5!) में से (L) अंतिम होने वाली (4!) व्यवस्थाएं घटाएं। मिश्रित शर्तों में पहले निश्चित शर्त लगाएं।
First fill the two ends with ordinary students in \(^{5}P_{2}\) ways, then arrange the remaining (5) students in the middle (5) places. For end restrictions, fill the ends first.
Step 2
Why this answer is correct
The correct answer is D. (3600). First fill the two ends with ordinary students in \(^{5}P_{2}\) ways, then arrange the remaining (5) students in the middle (5) places. For end restrictions, fill the ends first.
Step 3
Exam Tip
पहले (5) सामान्य विद्यार्थियों में से सिरों पर \(^{5}P_{2}\) तरीके और बीच के (5) स्थानों पर बाकी (5!) तरीके हैं। सिरों से जुड़ी शर्तों में पहले सिरों को भरना आसान है।
Arrange the four consonants in (4!) ways and place the two vowels in \(^{5}P_{2}\) ways in the gaps. Use the gap method for non-adjacent vowels.
Step 2
Why this answer is correct
The correct answer is C. (480). Arrange the four consonants in (4!) ways and place the two vowels in \(^{5}P_{2}\) ways in the gaps. Use the gap method for non-adjacent vowels.
Step 3
Exam Tip
चार व्यंजन (4!) तरीकों से बैठते हैं और (5) अंतरालों में (2) स्वरों को \(^{5}P_{2}\) तरीकों से रखा जाता है। न-साथ स्वर में अंतराल विधि लगाएं।
Include (1,2), choose (3) more digits from the remaining (7), then arrange the (5) digits in (5!) ways. For compulsory objects, include them first and choose the rest.
Step 2
Why this answer is correct
The correct answer is A. (2520). Include (1,2), choose (3) more digits from the remaining (7), then arrange the (5) digits in (5!) ways. For compulsory objects, include them first and choose the rest.
Step 3
Exam Tip
(1,2) के साथ शेष (7) अंकों में से (3) चुनकर (5!) तरीकों से व्यवस्थित करें। अनिवार्य वस्तुओं को पहले शामिल मानकर बाकी चुनें।
Subtract the (5!) arrangements where that person sits on the fixed chair from the total (6!). With one forbidden seat, subtract the invalid cases from total.
Step 2
Why this answer is correct
The correct answer is A. (600). Subtract the (5!) arrangements where that person sits on the fixed chair from the total (6!). With one forbidden seat, subtract the invalid cases from total.
Step 3
Exam Tip
कुल (6!) में से उस व्यक्ति के निश्चित कुर्सी पर बैठने की (5!) व्यवस्थाएं घटती हैं। एक निषिद्ध स्थान हो तो कुल से अनुकूल-नहीं घटाएं।
