Consider the set $S = \{a,b,c,d,e,f,g,h,i,j\}.$ How many $4$-element subsets of S contain neither a nor b?

Question

I understand most of the concept for set and subsets however, I do not understand the logic here.

Question : Consider the set $S = \{a,b,c,d,e,f,g,h,i,j\}.$ How many $4$-element subsets of S contain neither a nor b?

Solution: $\binom 84= 70$

I don't know where does the 4 comes from...

Answer 1

Combination formula $$C_{p}^{n} = \frac{A_{p}^{n}}{p!} = \frac{n!}{p!(n-p)!}$$ In your situation $p$ is the $4$ element subset, you replace each unknown in the general formula by its value and you get $70$ as an answer.