Subsets of divisors How many subsets of the set of divisors of $72$ (including the empty set) contain only composite numbers? For example, $\{8,9\}$ and $\{4,8,12\}$ are two such sets.
I know $72$ is $2^3\cdot 3^2$, so there are $12$ divisors.  Besides $2$ and $3$, there are $10$ composite divisors.  Would this mean $\binom{10}{1}+\binom{10}{2}+...+\binom{10}{10}$?  If so, how would you calculate this?  Thank you.
 A: There are 10 composite divisors. ($4,8,6,12,24,9,18,36,72$).  Any set of divisors that contains only composite numbers must be a subset of this set of 10.  Conversely, any subset of this set of 10 is a subset that contains only composite numbers.
So your problem is how to count how many subsets there are of this set of 10. Do you know a formula or a way of counting the numbers of subsets of a set?
A: Since $72=2^3 3^2$ as you say, it has 12 positive divisors and therefore (eliminating 1, 2, and 3) has 9 composite positive divisors.
Since the number of subsets of a set with 9 elements is $2^9=512$ (including the empty set), there should be 512 such subsets.
(Notice that this is the same as $\binom{9}{0}+\binom{9}{1}+\binom{9}{2}+\cdots+\binom{9}{9}$.)
A: The $10$ element set has $2^{10}$ subsets.
A: If you have the set of composite divisors, $S$, then $\mathcal{P}(S)$ is the set of subsets of $S$. This is calles "parts of $S$". And $|\mathcal{P}(S)| = 2^{|S|}$. So in your case, there are $2^{10} = 1024$ subsets of the set of divisors of $72$ that only contain composite numbers.
