Question

$2 m$ 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

• A. $\frac{(m+n) !}{m ! n !}$
• B. ${ }^{2 m+2 n} C_{2 m}$
• C. $\frac{1}{2} \frac{(m+n) !}{m ! n !}$
• D. none of these

Answer 1

Answer: (A)

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