
Patrick B. answered 08/08/21
Math and computer tutor/teacher
(N choose r) = N! / [ r ! (n - r) ! ]
N for the 1st choice
N-1 for the 2nd choice
N-2 for the 3rd choice
....
N-r+1 for the Rth choice
N(N-1)(N-2)....(N-r+1) =
N(N-1)(N-2)...(N-r+1)(N-r)(N-r-1)...*3*2*1 / (N-r)! <--- multiplies top and bottom by (n-r)!
since there are r items, there are r! ways to arrange and their order does not matter.
So we can remove them by dividing by r! in the denominator.
Therefore (N choose r) = N! / [ r ! (N-r)! ]