Rolling 9 dice probability 
Nine fair dice are rolled simultaneously. What is the probability of getting three pairs?

My attempt:
$$P(A) = \frac{\binom{6}{3}\binom{9}{2}\binom{7}{2}\binom{5}{2}\times3\times2\times1}{6^{9}}$$
We first choose which three numbers will be pairs, and which dice will be pairs, then we are left with three numbers and three positions for them. Is this correct? 
 A: Here's another approach to counting the numerator:
Take the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9.$ Arrange them in any sequence.
Then pick three of the numbers $1, 2, 3, 4, 5, 6$ and replace $7, 8, 9$ with those numbers in increasing order (that is, $7$ is replaced by the smallest number,
$9$ by the largest).
This produces every possible sequence of nine numbers selected from 
$\{1, 2, 3, 4, 5, 6\}$ with three pairs and three singletons, that is, the number of rolls that have exactly three pairs and no other matches, but it produces each such sequence more than once.
In particular, each of the numbers $7, 8,$ and $9$ could originally have appeared either before or after the numbers they eventually are made to duplicate,
so each roll of three pairs is produced exactly eight times.
After correcting for this overcounting, the number of rolls is
$$ \frac{9! \binom63}{8} = \frac52(9!).$$
This turns out to be equal to
$$ \binom{6}{3}\binom{9}{2}\binom{7}{2}\binom{5}{2}\times 6 =
\binom{6}{3} \frac{9!7!5!\times6}{(2!7!)(2!5!)(2!3!)}
= \binom{6}{3} \frac{9!}{(2!)^3},$$
confirming your answer for the probability of exactly three pairs.
