How many subsets? How many subsets of size 4 of the set S={1,2,....20} contain at least 1 of the elements 1,2,3,4,5?
$${5 \choose 4}{15 \choose 0}+{5 \choose 3}{15 \choose 1}+{5 \choose 2}{15 \choose 2}+{5 \choose 1}{15 \choose 3} + \binom50 \binom{15}4$$
5 "special elements"
15 "regular elements"  
Is my answer correct?
 A: Your answer is almost correct - the last term is confusing and also wrong. Consider that any subset of size $4$ containing at least one of the $5$ special elements can be partitioned into a subset $S$ of special elements (of size $1$ through $4$) and a subset $R$ of regular elements (of size $0$ through $3$). Thus, the correct answer would simply be:
$${5\choose 4}{15\choose 0}+{5\choose 3}{15\choose 1}+{5\choose 2}{15\choose 2}+{5\choose 1}{15\choose 3}$$
where we take the sum over the possible sizes of $S$ and $R$. This is basically what you have, except without the confusing last term, which would seem to represent the case if $S$ had size $0$ - which is not a case we are interested in - that case represents have no special elements.
A: The complementary subsets – those of size $4$ which contain none of $1,2,3,4,5$ are $\displaystyle \binom{15}{4}$, and there are $\displaystyle \binom{20}{4}$ subsets of size $4$ in all. Hence the number of subsets of size $4$ which contain at least one of these numbers is equal to 
$$\binom{20}{4}-\binom{15}{4}=\frac{20!}{4!\,16!}-\frac{15!}{4!\,11!}=3480.$$
