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There are n white and n black balls mark...

There are n white and n black balls marked 1, 2, 3,…..,n. The number of ways in which we can arrange these balls in a row so that neighbouring balls are of different colours, is

A

n!

B

(2n)!

C

`2(n!)^(2)`

D

`((2n)!)/((n!)^(2))`

Text Solution

Verified by Experts

We can arrange n white and n black balls alternately in the following ways:
(i) WBWB….. (ii) BWBW…..
So, required number of ways `n!xxn!+n!xxn!""=2(n!)^(2)`
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