Combinations and Permutations Question

Suppose a set S has 11 elements. How many subsets of S have an even number of elements? Express your answer as a summation.

I'm not really sure how to approach this. I think the general formula for this type of question is n!/k!(n-k)! where n is 11 and k is the number we're choosing. So would we just take the summation starting at 0 to 5 of 11!/2i!(n-2i)! Or what?

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Your suggestion answers the question as stated. The answer below shows that the sum is 1024. –  Yuval Filmus Nov 8 '10 at 4:46

If you can use the identity:

$\sum_{k = 0}^{n} (-1)^{k} {n \choose k} = 0$

then you can show that the cardinality of the set of subsets with even cardinality is equal to the cardinality of the set of subsets with odd cardinality.

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Wait so is my answer correct? I'm not understanding your answer.. –  maq Nov 8 '10 at 5:10
@fprime: your answer is correct, but Unreasonable Sin is offering you a big shortcut: if the total number of subsets with an even number of elements is the same as the total number of subsets with an odd number of elements, and you know how many subsets there are total, then this gives you a much simpler formula for finding the number you want, instead of having to do a sum of five things. –  Arturo Magidin Nov 8 '10 at 5:16
How do I know that the number of even and odd subsets are equal? –  maq Nov 8 '10 at 5:17
Yeah your answer is correct. What I was pointing out is that, if you have a set with 11 elements then there are $2^{11}$ subsets. Half of those subsets are odd and the other half are even. So the number of even or odd subsets is $2^{10}$. –  Unreasonable Sin Nov 8 '10 at 5:22
Oh ok I see, but how is that we know that half of those subsets are odd and half are even? –  maq Nov 8 '10 at 5:28