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?
 A: The outcomes counted by "stars and bars" are not equally likely to occur.  There is only one way to obtain a $1$ on every die.  However, there are
$$\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}2!$$
ways to obtain two ones, two twos, two threes, two fours, one five, and one six.  Consequently, if we use "stars and bars", we cannot simply divide the number of favorable outcomes by the number of outcomes in our sample space since those outcomes are not equally likely to occur.
Addendum:  Let's examine the standard approach more carefully.
Since there are $6$ possible outcomes for each of the $10$ rolls, there are $6^{10}$ sequences of rolls in our sample space.  These sequences specify not only how many times each number occurs but also where the numbers appear in the sequence of rolls.
For the favorable cases in which exactly two twos and exactly two fives occurs, two of the ten positions in the sequence are filled by the twos, two of the remaining eight positions in the sequence are filled by the fives, and there are four possible outcomes for each of the other six rolls, which gives
$$\binom{10}{2}\binom{8}{2}4^6$$
favorable outcomes.
Consequently, the probability of obtaining exactly two twos and exactly two fives in ten rolls of a fair die is
$$\frac{\dbinom{10}{2}\dbinom{8}{2}4^6}{6^{10}}$$
When we use "stars and bars" to count the outcomes, we lose information since "stars and bars" counts how many times each number occurs but not where those numbers occur in the sequence.  Therefore, it should not be surprising that we do not get the same result using "stars and bars" as we do when counting sequences of outcomes.
As noted above, the specific reason we do not use "stars and bars" is that the outcomes it counts are not equally likely to occur.  When we calculate probabilities by dividing the number of favorable outcomes by the total number of outcomes, we do so under the assumption that each outcome in our sample space is equally likely to occur.
A: 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 $(4)^6.$
$$N = \frac{10!}{(8!)(2!)} \times \frac{8!}{(6!)(2!)} \times (4)^6 = \frac{10!}{(6!)(2!)(2!)} \times (4)^6.$$
