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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)2x(2n+2)
- c)2x(n+1)!
- d)2[(n+1)!]2
Correct answer is option 'D'. Can you explain this answer?
Verified 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
Most Upvoted Answer
We have (n+1) white balls and (n+1) black balls. In each set the balls...
Given: (n-1) white balls and (n-1) black balls
To find: Number of arrangements such that two consecutive balls are of different colors
Approach:
- We can arrange the white balls and black balls separately
- The first ball can be of any color, so we have 2 choices
- After the first ball, the color of the next ball is fixed, so we have (n-2) choices for each set
- We can then arrange the balls in each set in (n-1)! ways
- Finally, we can combine the arrangements for the white balls and black balls to get the total number of arrangements
Solution:
- Number of choices for the first ball: 2
- Number of choices for the remaining (n-2) balls in each set: (n-2)
- Number of ways to arrange (n-1) balls in each set: (n-1)!
- Total number of arrangements for one set: 2 x (n-2) x (n-1)!
- Total number of arrangements for both sets: 2 x (n-2) x (n-1)! x 2 x (n-2) x (n-1)!
- Simplifying the expression: 2 x [(n-1)!]^2 x (n-2)^2
- Option D is the correct answer: 2[(n-1)!]^2
To find: Number of arrangements such that two consecutive balls are of different colors
Approach:
- We can arrange the white balls and black balls separately
- The first ball can be of any color, so we have 2 choices
- After the first ball, the color of the next ball is fixed, so we have (n-2) choices for each set
- We can then arrange the balls in each set in (n-1)! ways
- Finally, we can combine the arrangements for the white balls and black balls to get the total number of arrangements
Solution:
- Number of choices for the first ball: 2
- Number of choices for the remaining (n-2) balls in each set: (n-2)
- Number of ways to arrange (n-1) balls in each set: (n-1)!
- Total number of arrangements for one set: 2 x (n-2) x (n-1)!
- Total number of arrangements for both sets: 2 x (n-2) x (n-1)! x 2 x (n-2) x (n-1)!
- Simplifying the expression: 2 x [(n-1)!]^2 x (n-2)^2
- Option D is the correct answer: 2[(n-1)!]^2
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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)2x(2n+2)c)2x(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)2x(2n+2)c)2x(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)2x(2n+2)c)2x(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)2x(2n+2)c)2x(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)2x(2n+2)c)2x(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)2x(2n+2)c)2x(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)2x(2n+2)c)2x(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)2x(2n+2)c)2x(n+1)!d)2[(n+1)!]2Correct answer is option 'D'. Can you explain this answer?, a detailed solution 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)2x(2n+2)c)2x(n+1)!d)2[(n+1)!]2Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of 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)2x(2n+2)c)2x(n+1)!d)2[(n+1)!]2Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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