Question

Let $X$ be a set containing n elements. If two subsets $A$ and $B$ of $X$ are picked at random, the probability that $A$ and $B$ have the same number of elements is

• A. $\frac{{}^{2n}{C}_n}{2^n}$
• B. $\frac{1}{{}^{2n}{C}_n}$
• C. $\frac{1\cdot 3\cdot 5\cdot \cdot \cdot (2n-1)}{2^n\cdot n!}$
• D. $\frac{3^n}{4^n}$

Answer 1

The correct option is A $\frac{{}^{2n}{C}_n}{2^n}$

We know that the number of subsets of a set containing n elements is $2^n$.
Therefore, the number of ways of choosing A and B is $2^n\cdot 2^n=2^{2n}$.
We also know that the number of subsets (of X) which contain exactly r elements is ${}^n{C}_r$.
Therefore, the number of ways
of choosing A and B so that they have the same number of elements is
${(}^n{C}_0{)}^2+{(}^n{C}_1{)}^2+{(}^n{C}_2{)}^2+....+{(}^n{C}_n{)}^2$
${=}^{2n}{C}_n=\frac{1\cdot 2\cdot 3\cdot \cdot \cdot (2n-1)\left(2n\right)}{n!n!}$
$=\frac{[1\cdot 3\cdot 5\cdot \cdot \cdot (2n-1\left)\right][2\cdot 4\cdot 6\cdot \cdot \cdot (2n\left)\right]}{n!n!}$
$=\frac{2^n(n!)[1\cdot 3\cdot 5\cdot \cdot \cdot (2n-1\left)\right]}{n!n!}=\frac{2^n[1\cdot 3\cdot 5\cdot \cdot \cdot (2n-1\left)\right]}{n!}$
Thus, the probability of the required event is
$=\frac{{}^{2n}{C}_n}{2^{2n}}=\frac{1\cdot 3\cdot 5\cdot \cdot \cdot (2n-1)}{2^n(n!)}$