Question

Assertion :A set $X$ contains $n$ elements. Two subsets $A$ and $B$ of $X$ are chosen at random. The probability that $A$ and $B$ have same number of elements is $\frac{\mathrm{1.3.5...}(2n-1)}{2^n(n!)}$. Reason: ${\left({}^n{C}_0\right)}^2+{\left({}^n{C}_1\right)}^2+...+{\left({}^n{C}_n\right)}^2=\frac{2^n\left(\mathrm{1.3.5...}(2n-1)\right)}{n!}$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

We know that the number of subsets that can be formed of $X$, is $2^n$.
$\therefore n\left(S\right)=2^n.2^n=2^{2n}$
Now, the number of subsets of $X$ which contain exactly $r$ elements, is ${}^n{C}_r$.
Thus, two subsets both containing $r$ elements can be chosen in ${}^n{C}_r{.}^n{C}_r={\left({}^n{C}_r\right)}^2$ ways.
$\therefore$ If $E=$ choosing thw subsets both containing same number of elements, then
$n\left(E\right)={\left({}^n{C}_0\right)}^2+{\left({}^n{C}_1\right)}^2+...+{\left({}^n{C}_n\right)}^2{=}^{2n}{C}_n$
$=\frac{\mathrm{1.2.3.4....}(2n-1)\left(2n\right)}{n!n!}=\frac{\left(\mathrm{1.3.5...}(2n-1)\right)\left(\mathrm{2.4.6...}\left(2n\right)\right)}{{(n!)}^2}$
$=\frac{\left(\mathrm{1.3.5...}(2n-1)\right)2^n\left(\mathrm{1.2.3...}n\right)}{{(n!)}^2}=\frac{2^n\left(\mathrm{1.3.5...}(2n-1)\right)}{n!}$
Hence, the required probability is $P\left(E\right)=\frac{n\left(A\right)}{n\left(S\right)}\frac{\left(\mathrm{1.3.5...}(2n-1)\right)}{2^{n}n!}$