How many subsets of set contains at least one even number I have this set: {1, 2, 3, 4, 5, 6, 7, 8, 9}
And I want to know how many subsets of it contains at least one even number (There is no restriction on odd numbers).
I have tried to solve it but I'm not sure if it's correct so I would appreciate some feedback and especially if the answer is correct.
First I thought that the subset with only even numbers is {2, 4, 6, 8} with a total of 4 elements. This can then be formed in $4! = 4 \times 3 \times 2 \times 1 = 24$ ways. We also have the subset of odd numbers {1, 3, 5, 7, 9} which means that there are $2^5=32$ (because $2^n$) ways of forming subsets of that. If we then multiply these two number $24 \times 32 = 768$ and that is our answer. Is this correct? If not can you please explain what im doing wrong.
 A: Your error is thinking the number of sets with only even number (and at least one) is $4!$.  Order doesn't matter.  The number of subsets with only even numbers is $2^4$ as there are $4$ even numbers to choose from and each even number either may or may not be in the subset.  However that includes the empty set with zero even numbers (but no non-even numbers either) so we can't have that as we need at least one even number.  So there are $2^4 -1 = 15$ sets that have at least one even number and only even numbers.
And there are, as you correctly figured, are $2^5$ sets with only odd numbers (this also includes the empty set but that is okay, our final set need not have any odd numbers).
So our sets will be a union of a set of only evens and at least one even, with a set of only odds.  There will be $(2^4-1) \times 2^5 = 480$ such sets.
.....
However I think $90$% of mathematicians, myself included, would first think of mathlander's method first.  Take all the sets, $2^9$ and subtract all the sets with only odds $2^5$.
Note we can prove that $2^m - 2^k; m > k$ will always have $2^m-2^k = (2^k-1)\times 2^{m-k}$ so the two methods are equivalent
A: The way to do this problem is to use complementary counting. Since there are $2^9 = 512$ subsets of this and $2^5 = 32$ of them don't contain an even number, the answer is $512 - 32 = 480$.
