# A question about different pairs that are formed from a set of 16 different balls such that...

I got the following problem:

Given a set of 16 different balls, 8 are white and 8 are black.

If we partition the set of balls into pairs of two different balls and let $X$ be a discrete random variable that denotes the number of pairs composed of only white balls.

Find $P\{X=3\}$.

Even though this question sounds simple, I am stuck for at least 2 hours.

What I tried is:

$$\frac{{8\choose 2}{6\choose 2}{4\choose 2}{8\choose 2}{6\choose 2}{4\choose 2}2}{{16\choose 2}{14\choose 2}{12\choose 2}{10\choose 2}{8\choose 2}{6\choose 2}{4\choose 2}}$$

But it doesn't work.

• "If we partition the set to pairs" Which set of pairs? Pairs of what? This is not a well defined question.
– 5xum
Commented Dec 10, 2014 at 13:41
• We partition the set to pairs of two different balls. Commented Dec 10, 2014 at 13:43

The number of ways to partition the balls into eight pairs is ${16\choose 2}{14\choose 2}{12\choose 2}{10\choose 2}{8\choose 2}{6\choose 2}{4\choose 2}\frac 1{8!}$. You choose one pair, then another from the $14$ remaining balls, etc., but the ordering of the pairs is immaterial. To get exactly three pairs of white balls, you also need three pairs of black balls and two mixed pairs. There are ${8\choose 2}{6\choose 2}{4\choose 2}{2 \choose 1}\frac 1{3!}\frac 1{2!}$ ways to split the white balls into three pairs plus two singles, the same number of ways to split up the black balls, and $2$ ways to pair up the singles. In total, the probability is $$\frac {\left ({8\choose 2}{6\choose 2}{4\choose 2}{2 \choose 1}\frac 1{3!}\frac 1{2!}\right)^22}{{16\choose 2}{14\choose 2}{12\choose 2}{10\choose 2}{8\choose 2}{6\choose 2}{4\choose 2}\frac 1{8!}}$$
$${8 \choose 3,3,2} \left[{8 \choose 6} {6 \choose 2,2,2 } 2!\right]^2 = {8 \choose 3}{5 \choose 3} \left[{8 \choose 6} {6 \choose 2 } {4 \choose 2 } \right]^2 4$$
To get the probability you divide that by the total number of pairings: $${16 \choose 2,2,2,2,2,2,2,2}= {16 \choose 2 } {14 \choose 2 } {12 \choose 2 } {10 \choose 2 } {8 \choose 2 } {6 \choose 2 } {4 \choose 2 }{2 \choose 2 }= \frac{16!}{2^8}$$
PS: This gives $P(X=3)=0.174048...$, it coincides with Ross' answer.