Treat vowels (A,I) as one block, giving \(5!\cdot 2!\) arrangements. For together conditions, the block method is most useful.
Step 2
Why this answer is correct
The correct answer is C. (240). Treat vowels (A,I) as one block, giving \(5!\cdot 2!\) arrangements. For together conditions, the block method is most useful.
Step 3
Exam Tip
स्वरों (A,I) को एक खंड मानने पर \(5!\cdot 2!\) व्यवस्थाएं मिलती हैं। साथ वाली शर्त में खंड विधि सबसे उपयोगी है।
The total arrangements are (7!/(3!2!)), and vowel-together cases are (120). For not-together cases, subtract together cases from total.
Step 2
Why this answer is correct
The correct answer is D. (300). The total arrangements are (7!/(3!2!)), and vowel-together cases are (120). For not-together cases, subtract together cases from total.
Step 3
Exam Tip
कुल व्यवस्थाएं (7!/(3!2!)) हैं और स्वर साथ होने पर (120) व्यवस्थाएं हैं। न-साथ के लिए कुल में से साथ वाली गिनती घटाएं।
The first digit must be one of (5,6,7,8), and the last digit must be even, so count by cases. In such problems, fix the first and last positions first.
Step 2
Why this answer is correct
The correct answer is B. (3780). The first digit must be one of (5,6,7,8), and the last digit must be even, so count by cases. In such problems, fix the first and last positions first.
Step 3
Exam Tip
पहला अंक (5,6,7,8) में से और अंतिम अंक सम होना चाहिए, इसलिए केस बनाकर गिनती करें। ऐसे प्रश्नों में पहला और अंतिम स्थान पहले तय करें।
Treat the three people as one block, so (7) units have (6!) circular arrangements and (3!) internal ways. In circular arrangements, rotations are identical.
Step 2
Why this answer is correct
The correct answer is A. (4320). Treat the three people as one block, so (7) units have (6!) circular arrangements and (3!) internal ways. In circular arrangements, rotations are identical.
Step 3
Exam Tip
तीन व्यक्तियों को एक खंड मानने पर (7) इकाइयों की गोल व्यवस्था (6!) है और भीतर (3!) तरीके हैं। गोल व्यवस्था में घूर्णन को समान मानें।
Inside the vowel block there are (4!/3!) arrangements, and outside there are (5!/2!) arrangements. Handle repetition and blocks together.
Step 2
Why this answer is correct
The correct answer is B. (240). Inside the vowel block there are (4!/3!) arrangements, and outside there are (5!/2!) arrangements. Handle repetition and blocks together.
Step 3
Exam Tip
स्वर-खंड के भीतर (4!/3!) और बाहर (5!/2!) व्यवस्थाएं होंगी। दोहराव और खंड दोनों को साथ संभालें।
Subtract the cases where all three are together, \(6!\cdot 3!\), from the total (8!). For not-all-together conditions, subtract bad cases from total.
Step 2
Why this answer is correct
The correct answer is C. (36000). Subtract the cases where all three are together, \(6!\cdot 3!\), from the total (8!). For not-all-together conditions, subtract bad cases from total.
Step 3
Exam Tip
कुल (8!) में से तीनों साथ वाली \(6!\cdot 3!\) व्यवस्थाएं घटती हैं। न-एक-साथ में कुल से खराब केस घटाएं।
In the total (7!) arrangements, divide by (4) because of two independent order conditions. Symmetry gives a quick answer in such problems.
Step 2
Why this answer is correct
The correct answer is A. (1260). In the total (7!) arrangements, divide by (4) because of two independent order conditions. Symmetry gives a quick answer in such problems.
Step 3
Exam Tip
कुल (7!) व्यवस्थाओं में दोनों स्वतंत्र क्रम-शर्तों के कारण (4) से भाग देंगे। ऐसी शर्तों में सममिति से तेज उत्तर मिलता है।
(6) अलग अक्षरों और (10) अंकों से (4) अक्षर के बाद (3) अंक वाला कोड बनाना है। यदि अक्षर और अंक दोहराए नहीं जाते और अंतिम अंक सम है, तो कितने कोड बनेंगे?
Letters can be arranged in \(^{6}P_4\) ways, and digits with an even last digit in \(5\cdot 9\cdot 8\) ways. Count different parts of a code separately.
Step 2
Why this answer is correct
The correct answer is D. (129600). Letters can be arranged in \(^{6}P_4\) ways, and digits with an even last digit in \(5\cdot 9\cdot 8\) ways. Count different parts of a code separately.
Step 3
Exam Tip
अक्षरों के लिए \(^{6}P_4\) और अंकों में अंतिम सम होने पर \(5\cdot 9\cdot 8\) तरीके मिलते हैं। कोड में अलग-अलग भागों को अलग गिनें।
Among the (3!) relative orders of (A,B,C), only one is valid, so the count is (10!/3!). For relative order, divide by the number of possible orders.
Step 2
Why this answer is correct
The correct answer is B. (604800). Among the (3!) relative orders of (A,B,C), only one is valid, so the count is (10!/3!). For relative order, divide by the number of possible orders.
Step 3
Exam Tip
(A,B,C) के (3!) सापेक्ष क्रमों में केवल एक क्रम मान्य है, इसलिए (10!/3!) होगा। सापेक्ष क्रम में कुल क्रमों से भाग दें।
Add arrangements formed by smaller available letters at each position, then add (1). In rank problems, handle repeated letters carefully.
Step 2
Why this answer is correct
The correct answer is C. (35). Add arrangements formed by smaller available letters at each position, then add (1). In rank problems, handle repeated letters carefully.
Step 3
Exam Tip
हर स्थान पर उससे छोटे उपलब्ध अक्षरों से बनने वाली व्यवस्थाएं जोड़कर अंत में (1) जोड़ते हैं। रैंक में दोहराए अक्षरों की गिनती सावधानी से करें।
There are two possible patterns, and in each pattern boys arrange in (5!) ways and girls in (5!) ways. With equal numbers, alternate seating has two starts.
Step 2
Why this answer is correct
The correct answer is A. (28800). There are two possible patterns, and in each pattern boys arrange in (5!) ways and girls in (5!) ways. With equal numbers, alternate seating has two starts.
Step 3
Exam Tip
दो पैटर्न संभव हैं और प्रत्येक में लड़के (5!) तथा लड़कियां (5!) तरीकों से बैठते हैं। बराबर संख्या में वैकल्पिक बैठाने पर दो शुरुआतें होती हैं।
Treat each couple as one block, so (6) blocks have (5!) circular arrangements and each block has (2) internal ways. For couple problems, remember the \(2^n\) internal orders.
Step 2
Why this answer is correct
The correct answer is D. (7680). Treat each couple as one block, so (6) blocks have (5!) circular arrangements and each block has (2) internal ways. For couple problems, remember the \(2^n\) internal orders.
Step 3
Exam Tip
हर जोड़े को एक खंड मानकर (6) खंडों की गोल व्यवस्था (5!) और हर खंड के भीतर (2) तरीके हैं। जोड़ों वाले प्रश्नों में \(2^n\) अंदरूनी क्रम याद रखें।
In the total (8!) arrangements, the (4!) relative orders of consonants are equally likely, so the count is (8!/4!). For original-order conditions, divide by relative orders.
Step 2
Why this answer is correct
The correct answer is B. (1680). In the total (8!) arrangements, the (4!) relative orders of consonants are equally likely, so the count is (8!/4!). For original-order conditions, divide by relative orders.
Step 3
Exam Tip
कुल (8!) व्यवस्थाओं में (4) व्यंजनों के (4!) सापेक्ष क्रम बराबर हैं, इसलिए (8!/4!) है। मूल क्रम वाली शर्त में सापेक्ष क्रमों से भाग दें।
The last two digits can be (25,50,75), and leading-zero cases must be removed separately. For divisibility by (25), check the last two digits first.
Step 2
Why this answer is correct
The correct answer is C. (924). The last two digits can be (25,50,75), and leading-zero cases must be removed separately. For divisibility by (25), check the last two digits first.
Step 3
Exam Tip
अंतिम दो अंक (25,50,75) हो सकते हैं और अग्र शून्य की स्थिति अलग हटानी होगी। (25) से विभाज्यता में अंतिम दो अंक पहले जांचें।
It has (I) four times, (S) four times and (P) twice, so the count is (11!/(4!4!2!)). For long words, make a frequency table.
Step 2
Why this answer is correct
The correct answer is A. (34650). It has (I) four times, (S) four times and (P) twice, so the count is (11!/(4!4!2!)). For long words, make a frequency table.
Step 3
Exam Tip
इसमें (I) चार बार, (S) चार बार और (P) दो बार है, इसलिए संख्या (11!/(4!4!2!)) है। लंबा शब्द हो तो आवृत्ति तालिका बनाएं।
Treat both (P)'s as one block, giving (10) units with (I) and (S) each repeated four times. Hence the count is (10!/(4!4!)).
Step 2
Why this answer is correct
The correct answer is D. (6300). Treat both (P)'s as one block, giving (10) units with (I) and (S) each repeated four times. Hence the count is (10!/(4!4!)).
Step 3
Exam Tip
दोनों (P) को एक खंड मानने पर (10) इकाइयां बनती हैं, जिनमें (I) और (S) चार-चार बार हैं। इसलिए संख्या (10!/(4!4!)) है।
The positions of (A,B) must differ by at least (4), giving (10) position pairs and (2) orders. The remaining (6!) people are arranged afterward.
Step 2
Why this answer is correct
The correct answer is B. (14400). The positions of (A,B) must differ by at least (4), giving (10) position pairs and (2) orders. The remaining (6!) people are arranged afterward.
Step 3
Exam Tip
(A,B) के स्थानों की दूरी कम से कम (4) होनी चाहिए, ऐसे (10) स्थान-जोड़े हैं और क्रम (2) तरीके से होगा। बाकी (6!) लोग व्यवस्थित होंगे।
The first digit must be (1,2,3), and the last digit must be odd, so cases are needed. Count the bound and oddness together.
Step 2
Why this answer is correct
The correct answer is C. (10920). The first digit must be (1,2,3), and the last digit must be odd, so cases are needed. Count the bound and oddness together.
Step 3
Exam Tip
पहला अंक (1,2,3) होना चाहिए और अंतिम अंक विषम होना चाहिए, इसलिए केस बनते हैं। सीमा और विषमता दोनों को साथ गिनें।
Fix (A), then the opposite place for (B) is fixed, and the remaining (6) people arrange in (6!) ways. For opposite seating, fix one person first.
Step 2
Why this answer is correct
The correct answer is A. (720). Fix (A), then the opposite place for (B) is fixed, and the remaining (6) people arrange in (6!) ways. For opposite seating, fix one person first.
Step 3
Exam Tip
(A) को स्थिर करने पर (B) का विपरीत स्थान निश्चित हो जाता है और शेष (6) लोग (6!) तरीकों से बैठते हैं। आमने-सामने में पहले एक व्यक्ति स्थिर करें।
Choose letter positions in \(^{6}C_2\) ways, letters in \(^{5}P_2\) ways and digits in \(^{7}P_4\) ways. It is easier to decide position types first.
Step 2
Why this answer is correct
The correct answer is D. (252000). Choose letter positions in \(^{6}C_2\) ways, letters in \(^{5}P_2\) ways and digits in \(^{7}P_4\) ways. It is easier to decide position types first.
Step 3
Exam Tip
अक्षरों के स्थान \(^{6}C_2\), अक्षर \(^{5}P_2\) और अंक \(^{7}P_4\) तरीकों से आएंगे। पहले स्थानों का प्रकार तय करना आसान होता है।
Subtract the consecutive cases \(2\cdot 8!\) from the total (9!). For non-consecutive conditions, subtract the bad block cases.
Step 2
Why this answer is correct
The correct answer is B. (282240). Subtract the consecutive cases \(2\cdot 8!\) from the total (9!). For non-consecutive conditions, subtract the bad block cases.
Step 3
Exam Tip
कुल (9!) क्रमों में से लगातार आने वाले \(2\cdot 8!\) क्रम घटते हैं। न-लगातार में खंड विधि से खराब केस घटाएं।
Inside the vowel block there are (4!/2!) arrangements, and outside there are (8!/2!) arrangements. Divide for repeated consonants and vowels.
Step 2
Why this answer is correct
The correct answer is C. (241920). Inside the vowel block there are (4!/2!) arrangements, and outside there are (8!/2!) arrangements. Divide for repeated consonants and vowels.
Step 3
Exam Tip
स्वर-खंड के भीतर (4!/2!) और बाहर (8!/2!) व्यवस्थाएं हैं। दोहराए व्यंजन और स्वर दोनों से भाग दें।
Among the (3!) relative orders of the three specified books, only one is valid, so the count is (7!/3!). For relative-order restrictions, divide directly.
Step 2
Why this answer is correct
The correct answer is D. (840). Among the (3!) relative orders of the three specified books, only one is valid, so the count is (7!/3!). For relative-order restrictions, divide directly.
Step 3
Exam Tip
तीन निश्चित पुस्तकों के (3!) सापेक्ष क्रमों में केवल एक क्रम मान्य है, इसलिए (7!/3!) है। सापेक्ष क्रम की शर्त में सीधे भाग दें।
Arrange the boys around the circle in ((4-1)!) ways and place the girls in the (4) gaps in (4!) ways. For circular alternate seating, fix one group first.
Step 2
Why this answer is correct
The correct answer is A. (144). Arrange the boys around the circle in ((4-1)!) ways and place the girls in the (4) gaps in (4!) ways. For circular alternate seating, fix one group first.
Step 3
Exam Tip
लड़कों को गोल मेज पर ((4-1)!) तरीकों से और लड़कियों को बने (4) अंतरालों में (4!) तरीकों से बैठाएं। गोल वैकल्पिक बैठाने में एक समूह पहले स्थिर करें।
Here (E,T,I) are each repeated twice, so the count is (10!/(2!2!2!)). Divide by the factorial of each repeated letter.
Step 2
Why this answer is correct
The correct answer is B. (453600). Here (E,T,I) are each repeated twice, so the count is (10!/(2!2!2!)). Divide by the factorial of each repeated letter.
Step 3
Exam Tip
इसमें (E,T,I) दो-दो बार हैं, इसलिए संख्या (10!/(2!2!2!)) है। हर दोहराए अक्षर के फैक्टोरियल से भाग दें।
Inside the vowel block there are (5!/(2!2!)) arrangements, and outside there are (6!/2!) arrangements. Do not forget repeated letters when forming blocks.
Step 2
Why this answer is correct
The correct answer is D. (10800). Inside the vowel block there are (5!/(2!2!)) arrangements, and outside there are (6!/2!) arrangements. Do not forget repeated letters when forming blocks.
Step 3
Exam Tip
स्वर-खंड के भीतर (5!/(2!2!)) और बाहर (6!/2!) व्यवस्थाएं हैं। खंड बनाते समय दोहराए अक्षरों को न भूलें।
The four odd digits occupy the four odd positions in (4!) ways, and even digits fill the rest in (4!) ways. Identify the type of positions first.
Step 2
Why this answer is correct
The correct answer is C. (576). The four odd digits occupy the four odd positions in (4!) ways, and even digits fill the rest in (4!) ways. Identify the type of positions first.
Step 3
Exam Tip
चार विषम स्थानों पर चार विषम अंक (4!) तरीकों से और बाकी स्थानों पर सम अंक (4!) तरीकों से आएंगे। स्थान की प्रकृति पहले पहचानें।
For a bracelet, the count is ((10-1)!/2) because both rotation and reflection are identical. Remember the difference between bracelets and circular arrangements.
Step 2
Why this answer is correct
The correct answer is A. (181440). For a bracelet, the count is ((10-1)!/2) because both rotation and reflection are identical. Remember the difference between bracelets and circular arrangements.
Step 3
Exam Tip
कंगन में संख्या ((10-1)!/2) होती है क्योंकि घुमाना और पलटना दोनों समान हैं। कंगन और गोल व्यवस्था का अंतर याद रखें।
The last digit is (0) or (5), and if (5) is last, (0) must be included inside. Count divisibility and compulsory digit cases separately.
Step 2
Why this answer is correct
The correct answer is B. (1200). The last digit is (0) or (5), and if (5) is last, (0) must be included inside. Count divisibility and compulsory digit cases separately.
Step 3
Exam Tip
अंतिम अंक (0) या (5) होगा, और (5) अंत में होने पर (0) को भीतर शामिल करना होगा। विभाज्यता और अनिवार्य अंक को अलग केस में गिनें।
Among the (6) relative orders of (A,B,C), (A) is first in (2) orders, so the valid fraction is (4/6). Relative order solves such questions quickly.
Step 2
Why this answer is correct
The correct answer is D. (26880). Among the (6) relative orders of (A,B,C), (A) is first in (2) orders, so the valid fraction is (4/6). Relative order solves such questions quickly.
Step 3
Exam Tip
(A,B,C) के सापेक्ष (6) क्रमों में (A) पहले होने के (2) क्रम हैं, इसलिए मान्य भाग (4/6) है। सापेक्ष क्रम से ऐसे प्रश्न जल्दी हल होते हैं।
Arrange the men in (6!) ways and place the (4) women in \(^{7}P_4\) ways in the (7) gaps. Use the gap method for non-adjacent conditions.
Step 2
Why this answer is correct
The correct answer is C. (604800). Arrange the men in (6!) ways and place the (4) women in \(^{7}P_4\) ways in the (7) gaps. Use the gap method for non-adjacent conditions.
Step 3
Exam Tip
पहले पुरुषों को (6!) तरीकों से बैठाएं और (7) अंतरालों में (4) महिलाओं को \(^{7}P_4\) तरीकों से रखें। न-साथ की शर्त में अंतराल विधि लगाएं।
In increasing order, each chosen set gives only one arrangement, and including (0) would not give a (6)-digit number. So choose (6) digits from (1) to (9).
Step 2
Why this answer is correct
The correct answer is A. (84). In increasing order, each chosen set gives only one arrangement, and including (0) would not give a (6)-digit number. So choose (6) digits from (1) to (9).
Step 3
Exam Tip
बढ़ते क्रम में हर चुने हुए समूह की केवल एक व्यवस्था होती है, और (0) शामिल होने पर संख्या (6)-अंकीय नहीं रहेगी। इसलिए (1) से (9) में से (6) अंक चुनें।
Add arrangements starting with smaller available letters at each position, then add (1). In rank problems, first arrange letters alphabetically.
Step 2
Why this answer is correct
The correct answer is D. (140). Add arrangements starting with smaller available letters at each position, then add (1). In rank problems, first arrange letters alphabetically.
Step 3
Exam Tip
हर स्थान पर छोटे उपलब्ध अक्षरों से शुरू होने वाली व्यवस्थाएं जोड़कर अंत में (1) जोड़ते हैं। रैंक में अक्षरों को पहले वर्णक्रम में रखें।
Choose the (2) wrong letters in \(^{6}C_2\) ways, and they must swap with each other. For exactly (2) wrong, those two letters interchange.
Step 2
Why this answer is correct
The correct answer is C. (15). Choose the (2) wrong letters in \(^{6}C_2\) ways, and they must swap with each other. For exactly (2) wrong, those two letters interchange.
Step 3
Exam Tip
गलत जाने वाले (2) पत्र \(^{6}C_2\) तरीकों से चुनें और वे आपस में बदलेंगे। ठीक (2) गलत होने पर वही दो पत्र अदला-बदली करते हैं।
Treat the three (L)'s as one block, giving (6) units with (A) repeated twice, so the count is (6!/2!). A block of identical letters acts as one unit.
Step 2
Why this answer is correct
The correct answer is D. (360). Treat the three (L)'s as one block, giving (6) units with (A) repeated twice, so the count is (6!/2!). A block of identical letters acts as one unit.
Step 3
Exam Tip
तीनों (L) को एक खंड मानने पर (6) इकाइयां हैं और (A) दो बार है, इसलिए (6!/2!) है। समान अक्षरों का खंड केवल एक इकाई माना जाता है।
There are (2) ends for (C), and for each, (8) ordered position pairs for (A,B), followed by (5!) arrangements. Fix the end and distance first.
Step 2
Why this answer is correct
The correct answer is B. (1920). There are (2) ends for (C), and for each, (8) ordered position pairs for (A,B), followed by (5!) arrangements. Fix the end and distance first.
Step 3
Exam Tip
(C) के लिए (2) सिरे हैं और प्रत्येक में (A,B) के (8) क्रमित स्थान-जोड़े मिलते हैं, फिर बाकी (5!) तरीके हैं। पहले सिरे और दूरी तय करें।
Treat the two specified letters as one block, choose (5) more letters from the remaining (8), and arrange (6) units. Include the (2!) internal orders of the block.
Step 2
Why this answer is correct
The correct answer is C. (80640). Treat the two specified letters as one block, choose (5) more letters from the remaining (8), and arrange (6) units. Include the (2!) internal orders of the block.
Step 3
Exam Tip
दो निश्चित अक्षरों को एक खंड मानें, शेष (8) में से (5) अक्षर चुनें और (6) इकाइयों को सजाएं। खंड के भीतर (2!) क्रम भी जोड़ें।
The three odd positions can be filled from (4) odd digits in \(^{4}P_3\) ways, and the remaining three positions from (5) remaining digits in \(^{5}P_3\) ways. Apply position restrictions first.
Step 2
Why this answer is correct
The correct answer is A. (1440). The three odd positions can be filled from (4) odd digits in \(^{4}P_3\) ways, and the remaining three positions from (5) remaining digits in \(^{5}P_3\) ways. Apply position restrictions first.
Step 3
Exam Tip
तीन विषम स्थानों के लिए (4) विषम अंकों से \(^{4}P_3\) तरीके और शेष तीन स्थानों के लिए (5) बचे अंकों से \(^{5}P_3\) तरीके हैं। स्थान-प्रतिबंध पहले लागू करें।
Total circular arrangements are (8!), and adjacent cases are \(2\cdot 7!\), so the difference is (30240). In circular arrangements, start with ((n-1)!).
Step 2
Why this answer is correct
The correct answer is B. (30240). Total circular arrangements are (8!), and adjacent cases are \(2\cdot 7!\), so the difference is (30240). In circular arrangements, start with ((n-1)!).
Step 3
Exam Tip
कुल गोल व्यवस्थाएं (8!) हैं और साथ वाली \(2\cdot 7!\) हैं, इसलिए अंतर (30240) है। गोल व्यवस्था में कुल ((n-1)!) से शुरू करें।
It has (N) three times, (E) three times and (D) twice, so the count is (11!/(3!3!2!)). Write frequencies before applying the formula.
Step 2
Why this answer is correct
The correct answer is C. (554400). It has (N) three times, (E) three times and (D) twice, so the count is (11!/(3!3!2!)). Write frequencies before applying the formula.
Step 3
Exam Tip
इसमें (N) तीन बार, (E) तीन बार और (D) दो बार है, इसलिए संख्या (11!/(3!3!2!)) है। आवृत्ति लिखकर ही सूत्र लगाएं।
Place the (6) white balls first, creating (7) gaps, and choose (4) of them for black balls. With identical objects, only position selection is needed.
Step 2
Why this answer is correct
The correct answer is D. (35). Place the (6) white balls first, creating (7) gaps, and choose (4) of them for black balls. With identical objects, only position selection is needed.
Step 3
Exam Tip
पहले (6) सफेद गेंदें रखें, उनसे (7) अंतराल बनेंगे और उनमें से (4) में काली गेंदें रखें। समान वस्तुओं में केवल स्थान-चयन करना होता है।
Choose the correctly assigned person in (8) ways, and derange the remaining (7) people with \(D_7=1854\). Exactly one correct means all others are wrong.
Step 2
Why this answer is correct
The correct answer is A. (14832). Choose the correctly assigned person in (8) ways, and derange the remaining (7) people with \(D_7=1854\). Exactly one correct means all others are wrong.
Step 3
Exam Tip
सही कार्य पाने वाले को (8) तरीकों से चुनें और बाकी (7) का पूर्ण विस्थापन \(D_7=1854\) है। ठीक एक सही का मतलब बाकी सभी गलत होंगे।
Choose the even and odd digit groups and count (7!) arrangements, then subtract leading-zero cases. When (0) is included, handle the first digit carefully.
Step 2
Why this answer is correct
The correct answer is B. (230400). Choose the even and odd digit groups and count (7!) arrangements, then subtract leading-zero cases. When (0) is included, handle the first digit carefully.
Step 3
Exam Tip
सम और विषम अंकों के समूह चुनकर (7!) व्यवस्थाएं गिनें, फिर अग्र शून्य वाले केस घटाएं। शून्य शामिल हो तो पहला स्थान सावधानी से देखें।
Arrange consonants in (7!/(2!2!)) ways, then choose (4) of the (8) gaps and arrange vowels in (4!/(2!2!)) ways. Use the gap method for non-adjacent vowels.
Step 2
Why this answer is correct
The correct answer is C. (529200). Arrange consonants in (7!/(2!2!)) ways, then choose (4) of the (8) gaps and arrange vowels in (4!/(2!2!)) ways. Use the gap method for non-adjacent vowels.
Step 3
Exam Tip
पहले व्यंजन (7!/(2!2!)) तरीकों से रखें, फिर (8) अंतरालों में (4) स्वर चुनकर (4!/(2!2!)) तरीकों से सजाएं। न-साथ स्वर में अंतराल विधि लगाएं।
Any chosen set of (5) digits has only one decreasing arrangement, and the first digit will not be zero. So choose (5) digits from (10).
Step 2
Why this answer is correct
The correct answer is D. (252). Any chosen set of (5) digits has only one decreasing arrangement, and the first digit will not be zero. So choose (5) digits from (10).
Step 3
Exam Tip
किसी भी (5) चुने हुए अंकों की घटते क्रम में केवल एक व्यवस्था होती है और पहला अंक शून्य नहीं होगा। इसलिए (10) अंकों में से (5) चुनें।
