Yes.
I was watching a video playlist on applications of binomial coefficents and at the end of one of the videos it asked me to make an equation for them.
This is what I got:
Reasoning: n choose r.
say you have n people and you choose r, you at first have n people to choose from, then you have n-1 people to choose from, then n-2 people to choose from all the way to n-(r-1) because you start with n-0. But you have r! ways to arrange each group of people, so n(n-1)(n-2)...(n-(r-1)) also counts that r! ways to arrange each group of people. So we divide by r!
leaving
This is correct?