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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 1024.2010-11-08

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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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    @fprime: When you have a subset with an odd number of elements, the complement has an even number of elements. So there are the same number of odd element subsets and even element subsets. This is just a restatement of Unreasonable Sin's remark 4 ago.2010-11-08
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Here is a very easy way of seeing that the number of sets of even cardinality is equal to the number of sets of odd cardinality: if the total number of elements is odd (like in your case) then the map that sends a set to its complement establishes a bijection between sets of even and of odd cardinality. If the total number of elements is even, then there is the same bijection between the sets of even cardinality not containing the element 1 and the sets of odd cardinality not containing 1. Also, there is the obvious bijection between sets of even cardinality not containing 1 and those of odd cardinality containing 1, so again you have a bijection between odd and even sets.

Now, you just have to compute the total number of subsets, which I am sure you can do.