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We have (n+1) white balls and (n+1) black balls. In each set the balls are numbered from 1 to (n+1). If these balls are to be arranged in a row so that two consecutive balls are of different colours, then the number of these arrangements is
- a)(2n+2)!
- b)2 x (2n+2)
- c)2 x (n+1)!
- d)2[(n+1)!]2
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
We have (n+1) white balls and (n+1) black balls. In each set the balls...
Between (n + 1) white balls there are (n + 2) gaps in which (n + 1) black ball are to arranged.
No. of reqd arrangements = (n + 1)! (n + 1)! = [(n + 1)!]2
Now between (n + 1) black balls (n + 1) white balls are to be filled no. of ways
= (n + 1)! (n + 1)!
∴ Reqd ways = 2[(n + 1)!]2
No. of reqd arrangements = (n + 1)! (n + 1)! = [(n + 1)!]2
Now between (n + 1) black balls (n + 1) white balls are to be filled no. of ways
= (n + 1)! (n + 1)!
∴ Reqd ways = 2[(n + 1)!]2
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We have (n+1) white balls and (n+1) black balls. In each set the balls...
Explanation:
Key Concept:
To ensure that two consecutive balls are of different colors, we need to alternate between white and black balls in the arrangement.
Arranging the White Balls:
- There are (n+1)! ways to arrange the white balls in a row.
- We need to consider the white balls as a single entity, so there are (n+1)! ways to arrange them.
Arranging the Black Balls:
- There are (n+1)! ways to arrange the black balls in a row.
- We need to consider the black balls as a single entity, so there are (n+1)! ways to arrange them.
Arranging the White and Black Balls Alternately:
- Once we have arranged the white and black balls separately, we can alternate between them in 2 ways (white-black or black-white).
Total Number of Arrangements:
- The total number of arrangements will be 2 x [(n+1)!]^2 x 2 = 2 x [(n+1)!]^2 x 2 = 2 x [(n+1)!]^2.
Therefore, the number of arrangements of (n+1) white balls and (n+1) black balls in a row so that two consecutive balls are of different colors is 2 x [(n+1)!]^2, which corresponds to option 'D'.
Key Concept:
To ensure that two consecutive balls are of different colors, we need to alternate between white and black balls in the arrangement.
Arranging the White Balls:
- There are (n+1)! ways to arrange the white balls in a row.
- We need to consider the white balls as a single entity, so there are (n+1)! ways to arrange them.
Arranging the Black Balls:
- There are (n+1)! ways to arrange the black balls in a row.
- We need to consider the black balls as a single entity, so there are (n+1)! ways to arrange them.
Arranging the White and Black Balls Alternately:
- Once we have arranged the white and black balls separately, we can alternate between them in 2 ways (white-black or black-white).
Total Number of Arrangements:
- The total number of arrangements will be 2 x [(n+1)!]^2 x 2 = 2 x [(n+1)!]^2 x 2 = 2 x [(n+1)!]^2.
Therefore, the number of arrangements of (n+1) white balls and (n+1) black balls in a row so that two consecutive balls are of different colors is 2 x [(n+1)!]^2, which corresponds to option 'D'.
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We have (n+1) white balls and (n+1) black balls. In each set the balls are numbered from 1 to (n+1). If these balls are to be arranged in a row so that two consecutive balls are of different colours, then the number of these arrangements isa)(2n+2)!b)2 x (2n+2)c)2 x (n+1)!d)2[(n+1)!]2Correct answer is option 'D'. Can you explain this answer?
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We have (n+1) white balls and (n+1) black balls. In each set the balls are numbered from 1 to (n+1). If these balls are to be arranged in a row so that two consecutive balls are of different colours, then the number of these arrangements isa)(2n+2)!b)2 x (2n+2)c)2 x (n+1)!d)2[(n+1)!]2Correct answer is option 'D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about We have (n+1) white balls and (n+1) black balls. In each set the balls are numbered from 1 to (n+1). If these balls are to be arranged in a row so that two consecutive balls are of different colours, then the number of these arrangements isa)(2n+2)!b)2 x (2n+2)c)2 x (n+1)!d)2[(n+1)!]2Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for We have (n+1) white balls and (n+1) black balls. In each set the balls are numbered from 1 to (n+1). If these balls are to be arranged in a row so that two consecutive balls are of different colours, then the number of these arrangements isa)(2n+2)!b)2 x (2n+2)c)2 x (n+1)!d)2[(n+1)!]2Correct answer is option 'D'. Can you explain this answer?.
We have (n+1) white balls and (n+1) black balls. In each set the balls are numbered from 1 to (n+1). If these balls are to be arranged in a row so that two consecutive balls are of different colours, then the number of these arrangements isa)(2n+2)!b)2 x (2n+2)c)2 x (n+1)!d)2[(n+1)!]2Correct answer is option 'D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about We have (n+1) white balls and (n+1) black balls. In each set the balls are numbered from 1 to (n+1). If these balls are to be arranged in a row so that two consecutive balls are of different colours, then the number of these arrangements isa)(2n+2)!b)2 x (2n+2)c)2 x (n+1)!d)2[(n+1)!]2Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for We have (n+1) white balls and (n+1) black balls. In each set the balls are numbered from 1 to (n+1). If these balls are to be arranged in a row so that two consecutive balls are of different colours, then the number of these arrangements isa)(2n+2)!b)2 x (2n+2)c)2 x (n+1)!d)2[(n+1)!]2Correct answer is option 'D'. Can you explain this answer?.
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