Question

How many four element subsets of \{1, 2, 3, 4, 5, 6, 7\}{1,2,3,4,5,6,7} have 11 as an element but do not have 77 as an element

Answer 1

If we fix 1 to be an element of a subset of size 4, then we can choose from 5 other elements (2, 3, 4, 5, and 6) to fill the other 3 spots in the subset. So there are

$\dbinom53 = \dfrac{5!}{3!(5-3)!} = \boxed{10}$

such subsets.