If you have 2n socks in a drawer, n white and n black, and you reach in to choose 2 socks at random, how many ways are there to choose a pair? If you have 2n socks in a drawer, n white and n black, and you reach in to choose 2 socks at random,
How many ways are there to choose? For this I got 2(n^2-n)
How many of these ways result in getting a pair of the same color?
Write a simple closed form formula in terms of n for the chance choosing a matching pair of socks from a drawer with n white and n black socks.
 A: *

*The number of ways to pick two socks is
$$\binom{2n}{2}.$$

*Well, the long way is 
$$\binom{n}{2}\binom{n}{0}+\binom{n}{0}\binom{n}{2} = 2\binom{n}{2}.$$
In other words, the first terms says there are $n$ white socks and I need to choose two of those to make a match. Then I choose zero of the black ones. Similar for the second term. Yes, $\binom{n}{0} = 1$, but I choose to write it for completeness.

*The answer is

 $\frac{\text{Number of matches}}{\text{Number of pairs}} =\frac{2\binom{n}{2}}{\binom{2n}{2}}.$

A: 
How many ways are there to choose? For this I got 2(n^2-n)

There are $2n$ socks. You have a choice of $2n$ for the first and then $2n-1$ for the second, but order doesn't matter.   That's $\frac{2n (2n-1)}{2}$, which is $\dbinom{2n}{2}$.

How many of these ways result in getting a pair of the same color?

For each colour count the ways to select two socks. That is $\dbinom{\Box}{2}+\dbinom{\Box}{2}$, or $\underline{\qquad}$

Write a simple closed form formula in terms of n for the chance choosing a matching pair of socks from a drawer with n white and n black socks.

Divide and simplify.
A: I'm a little surprised nobody has pointed out: To get a matching pair the second sock must match the first.  There'll be 2n-1 socks left and n -1 of them will match the first.  So the probability is $\frac {n -1}{2n - 1}$
And
$\frac{2\binom{n}{2}}{\binom{2n}{2}}=$ $\frac {2*n!*(2n-2)!*2!}{2n!(n-2)!2!} =$ $\frac {(n-1)n*2!}{(2n-1)(2n)} =$$\frac {n-1}{2n-1} $
