Basic probability question - Counting

Question

A closet contains $n$ pairs of shoes. If $2r$ shoes are chosen at random with $2r<n$, what is the probability that there will be no matching pair in the sample?

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!

Answer 1

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 $$\frac{2n(2n-2)(2n-4)\cdots(2n-4r)}{(2r)!}=\binom{n}{2r}2^{2r}.$$

Answer 2

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.