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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 in which the balls can be arranged in a row so that adjacent balls are of different colors is

A. $(2n + 2)!$

B. $2(2n + 2)!$

C. $2(n + 1)!$

D. $2((n + 1)!)^2$

Nimmi M
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  • This is like alternate boys and girls in a row question.Also you should provide your inputs if you don't want too many downvotes –  Jan 08 '18 at 11:30
  • For $n=1$, the answer is easily seen to be $8$. That rules out options A, B, and C. – Barry Cipra Jan 08 '18 at 14:21

2 Answers2

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There are two ways for the balls to be arranged (white first or black first). For each color, there are $(n+1)!$ ways to arrange them. So there are $2 * (n + 1)!*(n + 1)!$ ways to arrange them.

Azlif
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The correct answer is $ 2((n + 1)!)^2.$

Let us start with a black ball. You have n+1 choices for the first ball. Now you go to the white balls and you have n+1 choices. At this point both your sets have n balls. You have n choices for the third and the fourth ball, n-1 choices for the fifth and the sixth ball. Continue until you are done with the balls. Therefore with the first ball being black your total number of choices is (n+1)!. Similarly you get (n+1)! choices for the case that your first ball is white. Thus the total number of choices is $ 2((n + 1)!)^2.$

Mohammad Riazi-Kermani
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  • $n + 1$ white and $n + 1$ for black means $(n + 1)(n + 1)$ for the first and second ball. Let us start with $n = 1$, $$B_1W_1B_2W_2, B_1W_2B_2W_1, B_2W_1B_1W_2,B_2W_2B_1W_2$$ there are $4$ for black first, which mean the total number of choices is $8$. That's not $2(2)!$. – Azlif Jan 08 '18 at 12:42
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    I fixed my error. Thanks for your comment. – Mohammad Riazi-Kermani Jan 08 '18 at 13:06