Concept-wise Practice

gap-method MCQ Questions for Class 11

gap-method se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.

Practice Questions

30 questions tagged with gap-method.

(6) boys और (5) girls को row में बैठाना है ताकि कोई दो girls साथ न बैठें। Girls placement factor क्या होगा?

(6) boys and (5) girls are seated in a row so that no two girls sit together. What is the girls placement factor?

Explanation opens after your attempt
Correct Answer

B. \(^{7}C_5\cdot5!\)

Step 1

Concept

After (6) boys, (7) gaps are formed, and (5) girls are arranged in them. In exams use the gap method for no two together.

Step 2

Why this answer is correct

The correct answer is B. \(^{7}C_5\cdot5!\). After (6) boys, (7) gaps are formed, and (5) girls are arranged in them. In exams use the gap method for no two together.

Step 3

Exam Tip

(6) boys के बाद (7) gaps बनते हैं, जिनमें (5) girls arrange होती हैं। परीक्षा में no two together में gap method लगाएं।

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(1) से (15) तक की संख्याओं में से (5) संख्याएं चुननी हैं, जिनमें कोई दो लगातार न हों। कितने तरीके हैं?

Choose (5) numbers from (1) to (15) such that no two are consecutive. How many ways are possible?

Explanation opens after your attempt
Correct Answer

A. (462)

Step 1

Concept

By the gap method, the count is \(^{15-5+1}C_{5}=^{11}C_{5}=462\). In such problems identify the gap condition first.

Step 2

Why this answer is correct

The correct answer is A. (462). By the gap method, the count is \(^{15-5+1}C_{5}=^{11}C_{5}=462\). In such problems identify the gap condition first.

Step 3

Exam Tip

अलगाव विधि से संख्या \(^{15-5+1}C_{5}=^{11}C_{5}=462\)। ऐसी समस्याओं में पहले अंतर की शर्त पहचानें।

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(5) अलग-अलग गणित और (4) अलग-अलग भौतिकी पुस्तकों को शेल्फ पर रखना है। कोई दो भौतिकी पुस्तकें साथ न हों, तो व्यवस्थाओं की संख्या कितनी है?

(5) distinct mathematics books and (4) distinct physics books are arranged on a shelf. If no two physics books are together, how many arrangements are possible?

Explanation opens after your attempt
Correct Answer

A. (86400)

Step 1

Concept

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।

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(7) अलग-अलग लड़कों और (3) अलग-अलग लड़कियों को पंक्ति में बैठाना है। सभी लड़कियां अलग-अलग रहें, तो व्यवस्थाओं की संख्या कितनी है?

(7) distinct boys and (3) distinct girls are to be seated in a row. If all girls must be separated, how many arrangements are possible?

Explanation opens after your attempt
Correct Answer

B. (1693440)

Step 1

Concept

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।

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(6) पुरुष और (4) महिलाएं पंक्ति में इस तरह बैठें कि कोई दो महिलाएं साथ न बैठें, तो व्यवस्थाओं की संख्या क्या होगी?

(6) men and (4) women are seated in a row so that no two women sit together. How many arrangements are possible?

Explanation opens after your attempt
Correct Answer

A. (604800)

Step 1

Concept

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।

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शब्द (ENGINEERING) के अक्षरों से कितनी व्यवस्थाएँ बनेंगी जिनमें कोई भी दो स्वर साथ-साथ न आएँ?

How many arrangements of the letters of (ENGINEERING) can be formed such that no two vowels are adjacent?

Explanation opens after your attempt
Correct Answer

D. (12600)

Step 1

Concept

Arrange the consonants in \(\frac{6!}{3!2!}\) ways and place the vowels in (7) gaps in \(\binom{7}{5}\frac{5!}{3!2!}\) ways. Use the gap method when vowels must not be adjacent.

Step 2

Why this answer is correct

The correct answer is D. (12600). Arrange the consonants in \(\frac{6!}{3!2!}\) ways and place the vowels in (7) gaps in \(\binom{7}{5}\frac{5!}{3!2!}\) ways. Use the gap method when vowels must not be adjacent.

Step 3

Exam Tip

पहले व्यंजनों को \(\frac{6!}{3!2!}\) तरीकों से सजाएँ और (7) खाली स्थानों में स्वरों को \(\binom{7}{5}\frac{5!}{3!2!}\) तरीकों से रखें। स्वर साथ न हों तो gap method लगाएँ।

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(5) अलग-अलग गणित की और (4) अलग-अलग भौतिकी की पुस्तकों को शेल्फ पर सजाना है। किसी भी दो भौतिकी पुस्तकों के बीच कम से कम एक गणित पुस्तक हो, ऐसे कितने क्रम हैं?

(5) distinct mathematics books and (4) distinct physics books are to be arranged on a shelf. How many orders have at least one mathematics book between any two physics books?

Explanation opens after your attempt
Correct Answer

A. (86400)

Step 1

Concept

Arrange mathematics books in (5!) ways and place (4) physics books in (6) gaps in (6P4) ways. Use the gaps method for separation.

Step 2

Why this answer is correct

The correct answer is A. (86400). Arrange mathematics books in (5!) ways and place (4) physics books in (6) gaps in (6P4) ways. Use the gaps method for separation.

Step 3

Exam Tip

पहले गणित पुस्तकों को (5!) तरीकों से रखें और (6) खाली स्थानों में (4) भौतिकी पुस्तकें (6P4) तरीकों से रखें। बीच में अलगाव के लिए gaps method लगाएं।

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(6) समान लाल, (5) समान नीली और (4) समान हरी गेंदों को पंक्ति में कितने तरीकों से रखा जा सकता है यदि कोई दो हरी गेंदें साथ न हों?

In how many ways can (6) identical red, (5) identical blue and (4) identical green balls be arranged in a row if no two green balls are adjacent?

Explanation opens after your attempt
Correct Answer

B. (228690)

Step 1

Concept

First arrange red and blue balls in \(^{11}C_5\) ways, then choose (4) of the (12) gaps for green balls. With identical objects, position selection is the key.

Step 2

Why this answer is correct

The correct answer is B. (228690). First arrange red and blue balls in \(^{11}C_5\) ways, then choose (4) of the (12) gaps for green balls. With identical objects, position selection is the key.

Step 3

Exam Tip

पहले लाल और नीली गेंदों को \(^{11}C_5\) तरीकों से रखें, फिर बने (12) अंतरालों में से (4) में हरी गेंदें रखें। समान वस्तुओं में स्थान-चयन प्रमुख होता है।

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शब्द (INSTITUTE) की व्यवस्थाओं में कोई दो (T) साथ न हों, ऐसी कितनी भिन्न व्यवस्थाएं हैं?

In arrangements of (INSTITUTE), how many distinct arrangements have no two (T)'s adjacent?

Explanation opens after your attempt
Correct Answer

B. (12600)

Step 1

Concept

Arrange non-(T) letters in (6!/2!) ways and choose (3) of the (7) gaps for the (T)'s. Since the (T)'s are identical, only choose gaps.

Step 2

Why this answer is correct

The correct answer is B. (12600). Arrange non-(T) letters in (6!/2!) ways and choose (3) of the (7) gaps for the (T)'s. Since the (T)'s are identical, only choose gaps.

Step 3

Exam Tip

पहले गैर-(T) अक्षरों को (6!/2!) तरीकों से रखें और (7) अंतरालों में से (3) में (T) रखें। समान (T) होने के कारण केवल अंतराल चुनें।

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(7) लड़के और (5) लड़कियां गोल मेज पर कितने तरीकों से बैठ सकते हैं यदि कोई दो लड़कियां साथ न बैठें?

In how many ways can (7) boys and (5) girls sit around a circular table if no two girls sit together?

Explanation opens after your attempt
Correct Answer

A. (181440)

Step 1

Concept

Arrange the boys around the circle in (6!) ways and place (5) girls in the (7) gaps in \(^{7}P_5\) ways. For circular gaps, arrange one group first.

Step 2

Why this answer is correct

The correct answer is A. (181440). Arrange the boys around the circle in (6!) ways and place (5) girls in the (7) gaps in \(^{7}P_5\) ways. For circular gaps, arrange one group first.

Step 3

Exam Tip

पहले लड़कों को गोल मेज पर (6!) तरीकों से बैठाएं और बने (7) अंतरालों में (5) लड़कियां \(^{7}P_5\) तरीकों से बैठेंगी। गोल अंतराल में पहले एक समूह व्यवस्थित करें।

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शब्द (DAUGHTER) की व्यवस्थाओं में कोई दो स्वर साथ न आएं, ऐसी कितनी व्यवस्थाएं हैं?

In arrangements of (DAUGHTER), how many have no two vowels adjacent?

Explanation opens after your attempt
Correct Answer

C. (14400)

Step 1

Concept

Arrange the (5) consonants in (5!) ways and place the vowels in (3) of the (6) gaps in \(^{6}P_3\) ways. If vowels are distinct, count both positions and order.

Step 2

Why this answer is correct

The correct answer is C. (14400). Arrange the (5) consonants in (5!) ways and place the vowels in (3) of the (6) gaps in \(^{6}P_3\) ways. If vowels are distinct, count both positions and order.

Step 3

Exam Tip

पहले (5) व्यंजनों को (5!) तरीकों से रखें और (6) अंतरालों में से (3) में स्वरों को \(^{6}P_3\) तरीकों से रखें। स्वर अलग हों तो स्थान और क्रम दोनों गिनें।

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(10) व्यक्तियों की पंक्ति में (A,B,C) में से कोई भी दो साथ न बैठें, ऐसी कितनी व्यवस्थाएं हैं?

In a row of (10) people, how many arrangements have no two of (A,B,C) sitting together?

Explanation opens after your attempt
Correct Answer

A. (1693440)

Step 1

Concept

Arrange the other (7) people in (7!) ways and place (A,B,C) in the (8) gaps in \(^{8}P_3\) ways. The gap method is powerful for non-adjacent conditions.

Step 2

Why this answer is correct

The correct answer is A. (1693440). Arrange the other (7) people in (7!) ways and place (A,B,C) in the (8) gaps in \(^{8}P_3\) ways. The gap method is powerful for non-adjacent conditions.

Step 3

Exam Tip

बाकी (7) लोगों को (7!) तरीकों से बैठाकर (8) अंतरालों में (A,B,C) को \(^{8}P_3\) तरीकों से रखें। न-साथ की शर्त में अंतराल विधि शक्तिशाली होती है।

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शब्द (EXCELLENCE) की व्यवस्थाओं में कोई दो (E) साथ न हों, ऐसी कितनी भिन्न व्यवस्थाएं हैं?

In arrangements of (EXCELLENCE), how many distinct arrangements have no two (E)'s adjacent?

Explanation opens after your attempt
Correct Answer

A. (6300)

Step 1

Concept

First arrange the non-(E) letters in (6!/(2!2!)) ways and place (E)'s in (4) of the (7) gaps. With identical letters, gap selection is enough.

Step 2

Why this answer is correct

The correct answer is A. (6300). First arrange the non-(E) letters in (6!/(2!2!)) ways and place (E)'s in (4) of the (7) gaps. With identical letters, gap selection is enough.

Step 3

Exam Tip

पहले गैर-(E) अक्षरों को (6!/(2!2!)) तरीकों से रखें और (7) अंतरालों में से (4) में (E) रखें। समान अक्षर हों तो अंतराल चयन काफी होता है।

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शब्द (ARRANGEMENT) की व्यवस्थाओं में कोई दो स्वर साथ न हों, ऐसी कितनी भिन्न व्यवस्थाएं हैं?

In arrangements of (ARRANGEMENT), how many distinct arrangements have no two vowels adjacent?

Explanation opens after your attempt
Correct Answer

C. (529200)

Step 1

Concept

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।

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(6) सफेद और (4) काले समान गेंदों को पंक्ति में कितने तरीकों से रखा जा सकता है यदि कोई दो काली गेंदें साथ न हों?

In how many ways can (6) identical white balls and (4) identical black balls be arranged in a row if no two black balls are adjacent?

Explanation opens after your attempt
Correct Answer

D. (35)

Step 1

Concept

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।

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(6) पुरुष और (4) महिलाएं एक पंक्ति में कितने तरीकों से बैठ सकते हैं यदि कोई दो महिलाएं साथ न बैठें?

In how many ways can (6) men and (4) women sit in a row if no two women sit together?

Explanation opens after your attempt
Correct Answer

C. (604800)

Step 1

Concept

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।

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शब्द (RANDOM) की व्यवस्थाओं में स्वर कभी साथ न आएं, ऐसी कितनी व्यवस्थाएं हैं?

In arrangements of the word (RANDOM), how many have no two vowels adjacent?

Explanation opens after your attempt
Correct Answer

C. (480)

Step 1

Concept

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।

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(5) लड़के और (4) लड़कियां एक पंक्ति में कितने तरीकों से बैठ सकते हैं यदि कोई दो लड़कियां साथ न बैठें?

In how many ways can (5) boys and (4) girls sit in a row if no two girls sit together?

Explanation opens after your attempt
Correct Answer

A. (86400)

Step 1

Concept

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।

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(6) पुरुष और (4) महिलाएं पंक्ति में ऐसे बैठें कि कोई दो महिलाएं साथ न बैठें। कितनी व्यवस्थाएं होंगी?

In how many ways can (6) men and (4) women sit in a row so that no two women sit together?

Explanation opens after your attempt
Correct Answer

A. (604800)

Step 1

Concept

Arrange the men first in (6!) ways and place (4) women in (7) gaps in \(^{7}P_4\) ways. The total is \(6!\cdot{}^{7}P_4=604800\).

Step 2

Why this answer is correct

The correct answer is A. (604800). Arrange the men first in (6!) ways and place (4) women in (7) gaps in \(^{7}P_4\) ways. The total is \(6!\cdot{}^{7}P_4=604800\).

Step 3

Exam Tip

पहले पुरुषों को (6!) तरीकों से बैठाएं और (7) gaps में (4) महिलाएं \(^{7}P_4\) तरीकों से रखें। कुल \(6!\cdot{}^{7}P_4=604800\) है।

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शब्द (ARRANGE) के अक्षरों की कितनी व्यवस्थाएं होंगी जिनमें दोनों (R) साथ न आएं?

How many arrangements of the letters of (ARRANGE) are possible in which the two (R)'s are not together?

Explanation opens after your attempt
Correct Answer

A. (900)

Step 1

Concept

First arrange the non-(R) letters in \(\frac{5!}{2!}=60\) ways and place two (R)'s in \(\binom{6}{2}\) gaps. The total is \(60\cdot15=900\).

Step 2

Why this answer is correct

The correct answer is A. (900). First arrange the non-(R) letters in \(\frac{5!}{2!}=60\) ways and place two (R)'s in \(\binom{6}{2}\) gaps. The total is \(60\cdot15=900\).

Step 3

Exam Tip

पहले बिना (R) के अक्षरों की \(\frac{5!}{2!}=60\) व्यवस्थाएं बनती हैं और (6) gaps में दो (R) \(\binom{6}{2}\) तरीकों से रखे जाते हैं। कुल \(60\cdot15=900\) है।

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(6) अलग-अलग हिंदी और (5) अलग-अलग अंग्रेजी पुस्तकों को पंक्ति में ऐसे रखा जाए कि कोई दो अंग्रेजी पुस्तकें साथ न हों। कितनी व्यवस्थाएं होंगी?

In how many ways can (6) distinct Hindi books and (5) distinct English books be arranged in a row so that no two English books are together?

Explanation opens after your attempt
Correct Answer

A. (3628800)

Step 1

Concept

Arrange Hindi books first in (6!) ways, then place (5) English books in (7) gaps in \(^{7}P_5\) ways. The total is \(6!\cdot^{7}P_5=1814400\).

Step 2

Why this answer is correct

The correct answer is A. (3628800). Arrange Hindi books first in (6!) ways, then place (5) English books in (7) gaps in \(^{7}P_5\) ways. The total is \(6!\cdot^{7}P_5=1814400\).

Step 3

Exam Tip

पहले हिंदी पुस्तकें (6!) तरीकों से रखें, फिर (7) gaps में (5) अंग्रेजी पुस्तकें \(^{7}P_5\) तरीकों से रखें। कुल \(6!\cdot^{7}P_5=1814400\) है।

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(5) पुरुष और (4) महिलाएं पंक्ति में ऐसे बैठें कि कोई दो महिलाएं साथ न बैठें। व्यवस्थाएं कितनी होंगी?

In how many ways can (5) men and (4) women sit in a row so that no two women sit together?

Explanation opens after your attempt
Correct Answer

A. (86400)

Step 1

Concept

Arrange the men first in (5!) ways, then place (4) women in (6) gaps in \(^{6}P_4\) ways. The total is (86400).

Step 2

Why this answer is correct

The correct answer is A. (86400). Arrange the men first in (5!) ways, then place (4) women in (6) gaps in \(^{6}P_4\) ways. The total is (86400).

Step 3

Exam Tip

पहले पुरुषों को (5!) तरीकों से बैठाएं, फिर (6) gaps में (4) महिलाएं \(^{6}P_4\) तरीकों से बैठेंगी। कुल (86400) है।

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(5) अलग-अलग लाल और (4) अलग-अलग नीली गेंदों को ऐसे पंक्ति में रखें कि कोई दो नीली गेंदें साथ न हों। तरीकों की संख्या क्या है?

In how many ways can (5) distinct red balls and (4) distinct blue balls be arranged in a row so that no two blue balls are together?

Explanation opens after your attempt
Correct Answer

B. (86400)

Step 1

Concept

Arrange the red balls first in (5!) ways, then place (4) blue balls in (6) gaps in \(^{6}P_4\) ways. The total is \(5!\cdot^{6}P_4=86400\).

Step 2

Why this answer is correct

The correct answer is B. (86400). Arrange the red balls first in (5!) ways, then place (4) blue balls in (6) gaps in \(^{6}P_4\) ways. The total is \(5!\cdot^{6}P_4=86400\).

Step 3

Exam Tip

पहले लाल गेंदें (5!) तरीकों से रखें, फिर (6) gaps में (4) नीली गेंदें \(^{6}P_4\) तरीकों से रखें। कुल \(5!\cdot^{6}P_4=86400\) है।

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(5) अलग-अलग लड़कों और (3) अलग-अलग लड़कियों को एक पंक्ति में कितने तरीकों से बैठाया जाए ताकि कोई दो लड़कियां साथ न बैठें?

In how many ways can (5) distinct boys and (3) distinct girls sit in a row so that no two girls sit together?

Explanation opens after your attempt
Correct Answer

A. (14400)

Step 1

Concept

Arrange boys first in (5!) ways, then place (3) girls in (6) gaps in \(^{6}P_3\) ways. The total is \(5!\cdot^{6}P_3=14400\).

Step 2

Why this answer is correct

The correct answer is A. (14400). Arrange boys first in (5!) ways, then place (3) girls in (6) gaps in \(^{6}P_3\) ways. The total is \(5!\cdot^{6}P_3=14400\).

Step 3

Exam Tip

पहले लड़कों को (5!) तरीकों से बैठाएं, फिर (6) खाली स्थानों में (3) लड़कियां \(^{6}P_3\) तरीकों से बैठेंगी। कुल \(5!\cdot^{6}P_3=14400\) है।

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(4) अलग-अलग हिंदी और (4) अलग-अलग अंग्रेजी पुस्तकों को ऐसे कितने तरीकों से रखा जाए कि कोई दो हिंदी पुस्तकें साथ न हों?

In how many ways can (4) distinct Hindi books and (4) distinct English books be arranged so that no two Hindi books are together?

Explanation opens after your attempt
Correct Answer

A. (2880)

Step 1

Concept

Arrange English books first in (4!) ways, then place (4) Hindi books in (5) gaps in \(^{5}P_4\) ways. The total is \(4!\cdot^{5}P_4=2880\).

Step 2

Why this answer is correct

The correct answer is A. (2880). Arrange English books first in (4!) ways, then place (4) Hindi books in (5) gaps in \(^{5}P_4\) ways. The total is \(4!\cdot^{5}P_4=2880\).

Step 3

Exam Tip

पहले अंग्रेजी पुस्तकें (4!) तरीकों से रखें, फिर (5) gaps में (4) हिंदी पुस्तकें \(^{5}P_4\) तरीकों से रखें। कुल \(4!\cdot^{5}P_4=2880\) है।

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लंबाई (7) की कितनी द्विआधारी श्रेणियाँ बनेंगी जिनमें ठीक (4) शून्य हों और लगातार (3) शून्य कहीं न आएँ?

How many binary strings of length (7) have exactly (4) zeros and no occurrence of (3) consecutive zeros?

Explanation opens after your attempt
Correct Answer

C. (19)

Step 1

Concept

Place the (4) zeros in the (4) gaps around the (3) ones, with no gap receiving more than (2) zeros. Control forbidden consecutiveness by gap limits.

Step 2

Why this answer is correct

The correct answer is C. (19). Place the (4) zeros in the (4) gaps around the (3) ones, with no gap receiving more than (2) zeros. Control forbidden consecutiveness by gap limits.

Step 3

Exam Tip

(3) एकों के आसपास बने (4) खाली स्थानों में (4) शून्य रखें और किसी स्थान पर (2) से अधिक शून्य न रखें। प्रतिबंधित लगातारपन को खाली स्थान सीमा से नियंत्रित करें।

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(6) पुरुष और (4) महिलाएँ गोल मेज पर बैठते हैं। कोई दो महिलाएँ साथ-साथ न बैठें। कुल कितनी व्यवस्थाएँ होंगी?

(6) men and (4) women sit around a round table. No two women should sit together. How many arrangements are possible?

Explanation opens after your attempt
Correct Answer

C. (43200)

Step 1

Concept

First seat the men around the circle in ((6-1)!) ways and place the (4) women in the (6) gaps. Use the gap method for circular separation.

Step 2

Why this answer is correct

The correct answer is C. (43200). First seat the men around the circle in ((6-1)!) ways and place the (4) women in the (6) gaps. Use the gap method for circular separation.

Step 3

Exam Tip

पहले पुरुषों को वृत्त में ((6-1)!) तरीकों से बैठाएँ और उनके (6) खाली स्थानों में (4) महिलाओं को रखें। वृत्तीय अलगाव में खाली स्थान विधि प्रयोग करें।

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शब्द (BANANA) के अक्षरों से कितने भिन्न क्रम बनेंगे जिनमें कोई दो (A) साथ-साथ न हों?

How many distinct arrangements of the letters of (BANANA) are possible if no two (A)'s are adjacent?

Explanation opens after your attempt
Correct Answer

C. (12)

Step 1

Concept

First arrange (B,N,N), then place the (3) identical (A)'s in the (4) gaps formed. Handle repeated letters and gaps carefully.

Step 2

Why this answer is correct

The correct answer is C. (12). First arrange (B,N,N), then place the (3) identical (A)'s in the (4) gaps formed. Handle repeated letters and gaps carefully.

Step 3

Exam Tip

पहले (B,N,N) को सजाएँ और बने (4) खाली स्थानों में (3) समान (A) रखें। समान अक्षरों और खाली स्थानों को ध्यान से संभालें।

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शब्द (STATISTICS) के अक्षरों से कितने भिन्न क्रम बनेंगे जिनमें कोई दो स्वर साथ-साथ न हों?

How many distinct arrangements of the letters of (STATISTICS) are possible if no two vowels are adjacent?

Explanation opens after your attempt
Correct Answer

C. (23520)

Step 1

Concept

First arrange the consonants and place vowels in the gaps formed. Do not forget division due to repeated letters.

Step 2

Why this answer is correct

The correct answer is C. (23520). First arrange the consonants and place vowels in the gaps formed. Do not forget division due to repeated letters.

Step 3

Exam Tip

पहले व्यंजनों को सजाकर बने खाली स्थानों में स्वरों को रखें। समान अक्षरों के कारण विभाजन करना न भूलें।

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(4) अलग गणित और (3) अलग अंग्रेजी की पुस्तकों को पंक्ति में सजाना है। कोई भी दो अंग्रेजी पुस्तकें साथ न हों। कुल व्यवस्थाएँ कितनी होंगी?

(4) distinct mathematics books and (3) distinct English books are to be arranged in a row. No two English books should be together. How many arrangements are possible?

Explanation opens after your attempt
Correct Answer

B. (1440)

Step 1

Concept

First arrange the mathematics books in (4!) ways, then place the (3) English books in (5) gaps with order. The gap method is useful for separation restrictions.

Step 2

Why this answer is correct

The correct answer is B. (1440). First arrange the mathematics books in (4!) ways, then place the (3) English books in (5) gaps with order. The gap method is useful for separation restrictions.

Step 3

Exam Tip

पहले गणित की पुस्तकें (4!) तरीकों से रखें और बने (5) खाली स्थानों में (3) अंग्रेजी पुस्तकें क्रम सहित रखें। अलग-अलग रखने में खाली स्थान विधि उपयोगी है।

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