<s>[INST] Probability that 2 numbers will occur exactly 4 times. A die is rolled 10 times, what is the probability that 2 numbers will occur exactly 4 times? So I know that the probability of rolling any one number exactly 4 times would be 
$$\frac{10!}{4!6!}\cdot \left(\frac{1}{6}\right)^4 \cdot\left(\frac{5}{6}\right)^6,$$
so would the probability that two numbers occur exactly four times be?
$$\frac{10!}{8!2!}\cdot\left(\frac{1}{6}\right)^4\cdot\left(\frac{1}{6}\right)^4\cdot\left(\frac{4}{6}\right)^2$$
 [/INST] No, you have ignored two effects.  The first is that you seem to be trying to calculate the probability that two specific numbers occur four times each, and the problem asks for any two numbers.  The second is that once one number is given to have occured exactly four times, the odds of the other number also doing so are different that the original odds.
The easy way to do this is to ask how many ways to arrange to get two four-timers are there, and divide by $6^{10}$. So:


*

*There are $\binom{6}{2}$ ways to choose the two numbers.

*There are $\binom{10}{4}$ ways to choose  which of the dice landed on the specific first number.

*There are then $\binom{6}{4}$ ways to choose  which of the dice landed on the specific second number.

*There are then $4^2$ ways to choose the rolls of the remaining two dice.
The net probability is 
$$
\frac{1}{6^{10}} \cdot \frac{6!}{2!4!}\cdot\frac{10!}{6!4!}\cdot \frac{6!}{4!2!}\cdot 4^2
$$
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