$2 m$ white counters and $2 n$ red counters are arranged in a straight line with $(m + n)$ counters on each side of a central mark. The number of ways of arranging the counters, so that the arrangements are symmetrical with respect to the central mark, is

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$\frac{1}{2} \frac{(m + n)!}{m! n!}$

$^{2 m + 2 n} C_{2 m}$

none of these

$\frac{(m + n)!}{m! n!}$

answer is A.

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Detailed Solution

Arrange $m$ white and $n$ red counters on one side of the central mark. This can be done in $\frac{(m + n)!}{m! n!}$

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