Question

A group of $2n$ boys and $2n$ girls is randomly divided into two equal groups. The probability that each group contains the same number of boys and girls is

• A. $1/2$
• B. $1/n$
• C. $1/2n$
• D. none of these

Answer 1

The correct option is D none of these

Total number of ways of choosing a group is ${}^{4n}{C}_{2n}$.
The number of ways in which each group contains equal number of boys and girls is ${(}^{2n}{C}_n\left){(}^{2n}{C}_n\right)$
$\therefore$ Required probability $=\frac{{(}^{2n}{C}_n{)}^2}{{}^{4n}{C}_{2n}}$.