Why does $\binom{10}{7} = \frac{10!}{(10-7)!7!}$ We just learned that: $\dbinom{10}{7}= \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}$, so that:

If you throw a dice 10 times, the probability of getting $6$, $7$ of the times is: $\dbinom{10}{7} \times {\frac{1}{6}}^7 \times {\frac{5}{6}}^3$, because $\dbinom{10}{7}$ will give us the number of different ways you can get $7$ "correct" out of $10$.

I wonder why this works. Why does $\dbinom{10}{7}$ work as it does? (In my search I stumbled upon this way of writing it: $\dbinom{10}{7} = \frac{10!}{(10-7)!7!}$. It is a little bit different, but maybe it is more correct?
 A: This is quite difficult to communicate with text, but I will try. It's important to understand each point in turn, as each one follows from the previous one. Please comment if you have any questions about anything that isn't clear:


*

*First, think about how many ways there are to arrange ABCDE?
Perhaps it's clear that there are $5\times 4\times 3\times 2\times 1 = 5! = 120$ permutations? If not, we could maybe start with a shorter example, like ABC.

*So then (once you've understood the first point!), how many ways are there to arrange AACDE?
Well, there are still $5! = 120$ permutations. 
But, some of these arrangements are the same. In fact, we are double counting, because the two A-s can appear in either order. So, in fact there are $5!/2 = 60$ combinations.

*How many ways are there to arrange AAADE? There are still $5! = 120$ permutations, but now each arrangement is counted 6 times. Because there are $3! = 6$ ways to arrange the A-s. So, there are $5!/3! = 20$ combinations.

*How many ways are there to arrange AAADD? There are still $5! = 120$ permutations, but there are 6 ways to arrange the A-s (six-times-counting), and 2 ways to arrange the D-s (double counting). In fact, each distinct arrangement is now counted $6\times 2 = 12$ times. In other words, there are $5!/(3!2!)$ combinations here.

*Whether we are interested in permutations or combinations depends on whether we care about the order of the results. If we are just counting the number of 6s, we don't care about the order, we just care about how many 6s we got. So, we should check the combinations.

*If you call A the result of getting a 6, and B the result of getting anything other than a 6, then we are trying to find the number of combinations of A-s and B-s, as in part 4 above. We can extend this idea to any number, like 10 dice and counting 7 sixes.
Hopefully it's clear how this leads to the idea of a binomial coefficient?!
