Probability: If you roll 6 fair dice, what is the probability that you roll exactly 4 different numbers? I know that the probability will be:
number of outcomes with 4 different numbers/ number of total possible outcomes
Total outcomes: 6^6
I am unsure of how to find the number of outcomes with 4 numbers, my attempt is below:
Outcomes with exactly $4$ numbers being the same: $C(6,4)*C(4, 2)*C(4,2)*C(4,1)*C(3,1)$
I got this from a textbook example, I know that C(6,4) provides the number of combinations (without repetition) of 6 elements into groups of 4, but I am lost beyond that.
 A: We count the "favourables." The numbers are small enough that we can break up the calculation into cases.
The collection of $4$ numbers we get can be chosen in $\binom{6}{4}$ ways. Now we count the number of ways our sequence of tosses can be made up of say $1,2,3,4$.
The $6$ tosses can yield the numbers $1,2,3,4$ is the following ways:
(i) One number occurs $3$ times, and the others once each. I would call this Type $3$-$1$-$1$-$1$. The popular number can be chosen in $\binom{4}{1}$ ways. Its location can be chosen in $\binom{6}{3}$ ways. And then the rest of the positions can be filled  in $3!$ ways, for a total of $\binom{4}{1}\binom{6}{3}3!$.
(ii) Two numbers occur twice each, and the other two once each. We can call this Type $2$-$2$-$1$-$1$.  The popular numbers can be chosen in $\binom{4}{2}$ ways. For each such way, the locations of the smaller popular number can be chosen in $\binom{6}{2}$ ways, and then the locations of the other popular number can be chosen in $\binom{4}{2}$ ways. The remaining positions can then be filled in $2!$ ways, for a total of $\binom{4}{2}\binom{6}{2}\binom{4}{2}2!$.
For the number of favourables, add up (i) and (ii), and multiply by $\binom{6}{4}$. For the probability, divide by $6^6$. 
A: The probability of getting exactly $k$ different values when you throw $n$ fair $r$-sided dice is $${{r\choose k}{n\brace k}k!\over r^n},$$ where $n\brace k$ is a Stirling number of the second kind. With $n=6$, $r=6$, and $k=4$ this gives $325/648\approx .50154.$
A: For small problems, you could uniformly formulate favorable ways as
$3-1-1-1-0-0 : \binom{6}{3,1,1,1,0,0}\cdot\frac{6!}{3!2!}= 7200$
$2-2-1-1-0-0 : \binom{6}{2,2,1,1,0,0}\cdot\frac{6!}{2!2!2!}= 16,200$
$Pr = \dfrac{23,400}{6^6} = \dfrac{325}{648}$
PS:
The formula can also be  written as the product of two multinomial coefficients, representing choose  and place, e.g., as $\binom{6}{3,2,1}\binom{6}{3,1,1,1,0,0}$
