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A closet contains $n$ pairs of shoes. If $2r$ shoes are chosen at random with $2r

The answer is $$\frac{\binom{n}{2r}2^{2r}}{\binom{2n}{2r}}.$$

I can not understand how we get this answer. My question is why the answer is not $$\frac{\binom{n}{2r}}{\binom{2n}{2r}}.$$

This is how I thought about the answer: there are $\binom{2n}{2r}$ ways of choosing $2r$ shoes from a total of $2n$ shoes. There are $\binom{n}{2r}$ ways of choosing $2r$ different shoes. So why and where am I wrong? Can someone explain the solution in plain English please.

Thanks in advance!

  • 0
    Try your argument with $n=3$ and $r=1$. – 2017-01-14

2 Answers 2

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There are
$$\frac{2n(2n-1)(2n-2)\cdots(2n-2r)}{(2r)!}=\binom{2n}{2r}$$ ways of choosing $2r$ shoes from a total of $2n$ shoes.

On the other hand in order to have no pairs we have that the number of choices is $$\frac{2n(2n-2)(2n-4)\cdots(2n-4r)}{(2r)!}=\binom{n}{2r}2^{2r}.$$

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Imagine you have the $n$ pairs of shoes out in front of you, in pairs, each pair of a different type. You are right that there are $n\choose 2r$ different ways to choose $2r$ different types of shoe, but you forget that there is an additional $2^{2r}$ factor from the $2r$ choices for whether you pick the left or right shoe of each type.