Stuck in problem involving combination/ permutation How many ten-digit binary numbers have exactly three 1’s?

What I have find is that some common patterns are they have three 1s and seven 0s. I have an idea by common sense. But, how to prove it mathematically. 

This is what I got

C(7,3) = 35

Not sure if go for permutation or combination
$\frac{10!}{7!3!}=120$  Sorry I am on mobile so I cannot write much. But there $3$ 1's and $7$ 0's. So by the permutation formula. Permutations:$\frac{n!}{r_{1}!r_{2}!...r_{n}!}$
There are 10 ways to put the first 1 in a 10 bit binairy number. Pick any position for the first 1. Then there are 9 slots left for the second 1. Again pick any position... Now there are 8 slots left for the a 1. 

So I think the answer could be 10x9x8=720 combinations of binairy numbers containing exactly three ones.