How many ways to roll a die seven times How many ways to roll a die seven times and obtain a sequence of outcomes with three 1s,  two 5s,  and two 6s? 
Ans: When I was thinking of a way to decompose the problem I first thought that if a die is getting rolled seven times then each roll is independent of the other thus: $6^7$ for all possible outcomes... then I was having trouble with how to take care of the constraint.  The solution given is: $$\frac{7!}{3!2!2!}$$
The section I am covering right now is about repetition,  but I do not see at al how this could be the result.  I thought I am supposed to find the outcomes WITH three 1s,  two 5s,  and two 6s? Isn't dividing those sequemces out doing the exact opposite and now we are finding all the arrangements WITHOUT three 1s,  two 5s,  and two 6s? 
 A: Another way to think about it goes like this:
So you roll seven times. First, choose 3 spots for the 1s,$$\binom{7}{3}.$$
Next, choose 2 spots for the 5s. We already used up 3, so this is $$\binom{4}{2}.$$
Finally, choose the last to spots for the 6s, $$\binom{2}{2}.$$
This gives
$$\binom{7}{3}\binom{4}{2}\binom{2}{2} = \frac{7!\,4!\,2!}{3!\,4!\,2!\,2!\,2!\,0!} = \binom{7}{3,2,2}.$$
A: You want to know in how many ways you can get $1155666$ in any order.
But there are $7!$ possible permutations of these numbers, except that some of the numbers are equal, and you need to compensate for that.
Since there are two $1$'s, two $5$'s, and three $6$'s, I've counted the same permutation $3!2!2!$ times, which is why the answer is
$$\frac{7!}{3!2!2!}.$$
For instance, the following two permutations are really the same: $\color{red}{1}\color{blue}{1}\color{red}{5}\color{blue}{5}\color{red}{6}\color{blue}{6}\color{green}{6}$ and $\color{blue}{1}\color{red}{1}\color{red}{5}\color{blue}{5}\color{red}{6}\color{green}{6}\color{blue}{6}$, but both are contained in the $7!$.
A: What this result does is compute the answer directly: imagine you have $7$ objects, which we will assume to be distinct. We'll say that this is the set $\{1_a,1_b,1_c,5_a,5_b,6_a,6_b\}$. There are $7!$ ways to permute them in a row, which is what you're doing with your rolls of the die. However, we then consider certain objects to be the same, specifically those of the same number, but differing index.
Consider one permutation, $1_a,5_a,6_a,1_b,5_b,6_b,1_c$. There are $3!$ ways to permute the $1$s so that the spaces that contain a $1$ are the same, but the indices differ, $2!$ for the $5$s, and $2!$ for the $6$s. However, by our condition that indices don't matter, these all give the same permutation as above. You can see that for any permutation, there are $3!*2!*2!$ ways to get the same result with different indices, so by the quotient rule, we have that answer.
