Question

n men and n women are seated at round table in random order. The probability that they can be divided into n non-interrecting pairs so that each pair consists of a man and a women is

• A. 1/2n
• B. $2(2^n-1){/}^{2n}{C}_n$
• C. $2n{/}^{2n}{C}_n$
• D. $1/{(}^n{C}_n{)}^2$

Answer 1

The correct option is B $2(2^n-1){/}^{2n}{C}_n$

The women can be seated in ${}^{2n}{C}_n$ ways. Let A denote the event that pairs occupy the seats (1, 2), (3, 4), ...., (2n-1, 2n)
and B denote the event that pairs occupy the seats (2, 3), (4, 5), (6, 7), ...., (2n-2, 2n-1), (2n-1).
The number of cases favourable to A (B) is $2^n$. (For each man in the pair there are two choices.)
However, there are just two cases common to A and B. One is the case $\left\{\right(M,W),(M,W),...(M,W\left)\right\}$ of A(B) and $\left\{\right(W,M),(W,M),...,(W,M\left)\right\}$ of B(A).
Therefore, $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cup B)$
$=\frac{2^n+2^n-2}{{}^{2n}{C}_n}=\frac{2^{n+1}-2}{{}^{2n}{C}_n}$.