It has been explained but still one more explanation will not hurt.
What is this $\binom{m}{3} and \binom{m-1}{3}$, what they are trying to do ?
Say we have 5 elements in our set(this is m=5), now how many subsets can be formed of size 3, it simple combination formula $\binom{m}{3}$=$\binom{5}{3}$,
$\binom{5}{3}$ =$\frac{5!}{3!*2!}$=10 ………….….(this is n)
we want to find out how many of these sets will have element 1 in them.
Which also means that we want sets which are totally made up of elements which are not 1.
we can model this by taking {2,3,4,5} excluding 1 and this is that $\binom{m-1}{c}$ thing we saw earlier.
$\binom{4}{c}=\frac{4!}{3!}$=4
which means that out of all those 10 sets 4 will not contain 1 in them and remaining 6 will contain 1 in them.