# 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"

• Look at the last two terms again, I think you meant something different. – AlexR Feb 2 '15 at 23:29
• ah yea typos! besides that its ok? – Math Major Feb 2 '15 at 23:30
• ... Contain at least one element. You'll want to drop the last term. The rest is fine. – AlexR Feb 2 '15 at 23:31
• oh righ thanks! – Math Major Feb 2 '15 at 23:33
• You'll want to accept Meelo's answer then since there's nothing more to add. Note that {n \choose k} is superseeded by \binom{n}{k} for syntactical reasons ;) – AlexR Feb 2 '15 at 23:34

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.
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
$$\binom{20}{4}-\binom{15}{4}=\frac{20!}{4!\,16!}-\frac{15!}{4!\,11!}=3480.$$