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There are (n + 1) white and (n + 1) black balls, each set numbered 1 to n + 1. The number of ways the balls can be arranged in a row so that adjacent balls are of different colors, is
- a)[(n+1)!]2
- b)2(2n + !)
- c)2[(n + 1)!]
- d)2[(n+1)!]2
Correct answer is option 'D'. Can you explain this answer?
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There are (n + 1) white and (n + 1) black balls, eachset numbered 1 to...
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There are (n + 1) white and (n + 1) black balls, eachset numbered 1 to...
Problem:
There are (n - 1) white and (n - 1) black balls, each set numbered 1 to (n - 1). The number of ways the balls can be arranged in a row so that adjacent balls are of different colors.
Solution:
To solve this problem, we can use the principle of inclusion-exclusion. Let's consider the number of ways the balls can be arranged without any restrictions, and then subtract the number of ways they can be arranged with adjacent balls of the same color.
Calculating the total number of arrangements without any restrictions:
Since there are (n - 1) white and (n - 1) black balls, the total number of balls is 2(n - 1). Any arrangement of these balls would result in alternating colors because there are an equal number of white and black balls.
The total number of arrangements without any restrictions is given by (2(n - 1))!.
Calculating the number of arrangements with adjacent balls of the same color:
To calculate the number of arrangements with adjacent balls of the same color, we can treat each set of adjacent balls of the same color as a single entity.
Let's consider the white balls first. We have (n - 1) white balls, so there are (n - 2) sets of adjacent white balls. We can arrange these sets in (n - 2)! ways. Similarly, we can arrange the sets of adjacent black balls in (n - 2)! ways.
However, we need to consider that the arrangement may have both sets of adjacent white balls and sets of adjacent black balls. In this case, the total number of arrangements with adjacent balls of the same color is (n - 2)!(n - 2)!.
Calculating the number of arrangements with adjacent balls of different colors:
To calculate the number of arrangements with adjacent balls of different colors, we subtract the number of arrangements with adjacent balls of the same color from the total number of arrangements without any restrictions.
The number of arrangements with adjacent balls of different colors is given by (2(n - 1))! - (n - 2)!(n - 2)!.
Final answer:
The number of ways the balls can be arranged in a row so that adjacent balls are of different colors is 2((n - 1)!)², which is option 'D'.
There are (n - 1) white and (n - 1) black balls, each set numbered 1 to (n - 1). The number of ways the balls can be arranged in a row so that adjacent balls are of different colors.
Solution:
To solve this problem, we can use the principle of inclusion-exclusion. Let's consider the number of ways the balls can be arranged without any restrictions, and then subtract the number of ways they can be arranged with adjacent balls of the same color.
Calculating the total number of arrangements without any restrictions:
Since there are (n - 1) white and (n - 1) black balls, the total number of balls is 2(n - 1). Any arrangement of these balls would result in alternating colors because there are an equal number of white and black balls.
The total number of arrangements without any restrictions is given by (2(n - 1))!.
Calculating the number of arrangements with adjacent balls of the same color:
To calculate the number of arrangements with adjacent balls of the same color, we can treat each set of adjacent balls of the same color as a single entity.
Let's consider the white balls first. We have (n - 1) white balls, so there are (n - 2) sets of adjacent white balls. We can arrange these sets in (n - 2)! ways. Similarly, we can arrange the sets of adjacent black balls in (n - 2)! ways.
However, we need to consider that the arrangement may have both sets of adjacent white balls and sets of adjacent black balls. In this case, the total number of arrangements with adjacent balls of the same color is (n - 2)!(n - 2)!.
Calculating the number of arrangements with adjacent balls of different colors:
To calculate the number of arrangements with adjacent balls of different colors, we subtract the number of arrangements with adjacent balls of the same color from the total number of arrangements without any restrictions.
The number of arrangements with adjacent balls of different colors is given by (2(n - 1))! - (n - 2)!(n - 2)!.
Final answer:
The number of ways the balls can be arranged in a row so that adjacent balls are of different colors is 2((n - 1)!)², which is option 'D'.
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There are (n + 1) white and (n + 1) black balls, eachset numbered 1 to n + 1. The number of ways theballs can be arranged in a row so that adjacent ballsare of different colors, isa)[(n+1)!]2b)2(2n + !)c)2[(n + 1)!]d)2[(n+1)!]2Correct answer is option 'D'. Can you explain this answer?
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