A fair $6$-sided die was rolled $10$ times. Find the probability you rolled exactly two $2$s and exactly two $5$s. 
A fair $6$-sided die was rolled $10$ times. Find the probability you rolled exactly two $2$s and exactly two $5$s.

I know the standard way to calculate this, but wasn't sure why another approach - stars and bars - cannot be applied. If we fix the two $2$s and $5$s to their respective boxes and arrange the rest, we have $nCr(10+6-1-2-2, 6-1)$ ways; If we use stars and bars to calculate the total number of results, we have $nCr(10+6-1, 6-1)$. Setting the former as the numerator and the latter as denominator gives a probability larger than the answer. Why is this the case?

Although a somewhat offbeat approach, Stars and Bars can be used to solve the problem.
You have $10$ slots, $4$ of which are going to be taken up by $2$ 2's and $2$ 5's.  You can construe these as Stars.  You will then have $6$ Bars that are divided into $5$ regions, by the $4$ Stars.
So, you compute the number of solutions to
$$x_1 + x_2 + \cdots + x_5 = 6. \tag1 $$
(1) above has $~\displaystyle \binom{6 + [5-1]}{5-1} = \binom{10}{4}~$ solutions.
Now, for each solution, there are $4$ possibilities for each of the bars.  So, this results in a factor of
$\left(4^6\right).$
Further, for each solution, there are $~\displaystyle \binom{4}{2}$ ways of selecting which $2$ of the $4$ Stars will be 2's, instead of 5's.
So, in computing the combinatoric probability fraction of
$$\frac{N\text{(umerator)}}{D\text{(enominator)}},$$
where $D = 6^{(10)},$ you end with
$$ N = \binom{10}{4} \times \binom{4}{2} \times (4)^6 = \frac{10!}{(6!)(4!)} \times \frac{4!}{(2!)(2!)} \times (4)^6 = \frac{10!}{(6!)(2!)(2!)} \times (4)^6.$$
Alternatively, in the standard approach,

*

*You have to select $2$ slots for the 2's : factor of $\displaystyle \binom{10}{2} = \frac{10!}{(2!)(8!)}.$

*You then have to select $2$ slots for the 5's : factor of $\displaystyle \binom{8}{2} = \frac{8!}{(2!)(6!)}.$

*You then have to select the number of ways of filling the remaining $6$ positions : factor of $(4)^6.$
$$N = \frac{10!}{(8!)(2!)} \times \frac{8!}{(6!)(2!)} \times (4)^6 = \frac{10!}{(6!)(2!)(2!)} \times (4)^6.$$
