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

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) ((m+n) !/m ! n !) (B)  2 m+2 n C2 m (C) (1/2) ((m+n) !/m ! n !) (D) none of these

Tardigrade Admin · Accepted Answer

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

Q. $2 m$ 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

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

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

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

none of these

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

Q. 2m white counters and 2n 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

m!n!(m+n)!​

2m+2nC2m​

21​m!n!(m+n)!​