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

My answer is $\frac{\binom{n}{0}^2+\binom{n}{1}^2+\binom{n}{2}^2+...+\binom{n}{n}^2}{2^n.2^n}$ but I cannot simplify it. Answer is given $\frac{1.3.5...(2n-1)}{2^n.n!}$
Please help me in this problem.

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  • $\begingroup$ Can the same subset be picked twice? $\endgroup$
    – zoli
    Commented Oct 29, 2015 at 16:46
  • $\begingroup$ yes same subset can be picked twice. $\endgroup$
    – Arpan Biswas
    Commented Oct 29, 2015 at 16:48
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    $\begingroup$ Actually it must be proved that:$$\binom{n}{0}^{2}+\cdots+\binom{n}{n}^{2}=\binom{2n}{n}$$ Have a look here $\endgroup$
    – drhab
    Commented Oct 29, 2015 at 17:04

1 Answer 1

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Note that

$$\sum_k{n \choose k}^2 = {2n \choose n} = {(2n)! \over (n!)^2}.$$

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