# Number of even/odd subsets.

1. Feb 20, 2010

### xnull

Given a finite set $$S$$ of cardinality m:
Decide how many even or odd subsets there are of S (finding one should give you the other).

Here is what I've done so far.

First, I looked to a multiplication out by hand.

(m * (m-1)) + (m * (m-1) * (m-2) * (m-3)) + (m * (m-1) * (m-2) * (m-3) * (m-4) * (m-5)) + ...

Well, I noticed I could factor (m * (m-1)) out which gives

(m * (m-1)) * (1 + ((m-2) * (m-3)) + ((m-2) * (m-3) * (m-4) * (m-5)) + ...)

Inside the second group of parenteses I noticed I could factor out ((m-2) * (m-3)) and so on

(m * (m-1)) * (1 + ((m-2) * (m-3) * (1 + ((m-4) * (m-5) * (1 + .... )))))))

Because this was a tad ugly I made a recurrence relation from it

$$R_{k} = 0$$
$$R_{i} = (m-2i)(m-2i-1)(1+R_{i+1})$$
$$R_{0}$$ would give the solution for the number of even subsets.

I can't seem to find the closed form.

This is the only direction I've gone with the problem so far.

Does anyone have any directions to go on this or a previously published solution to the problem? I looked all over and couldn't find one but it's the kind of problem I would expect to have already been solved.

2. Feb 20, 2010

### xnull

Solved it.

Heh. Silly to have missed it.

$$2^{m-1}$$ even subsets and odd subsets.

3. Feb 20, 2010

### JSuarez

Remember that the number of subsets (of a set with m elements) with k elements is given by:

$$\binom{m}{k}$$

Now, from the binomial theorem (or the fact that there are as many odd subsets as even ones) and the fact that the total number of subsets is 2m, you should be able to deduce that the number of odd subsets is:

$$\sum_{k\;odd}\binom{m}{k}=2^{m-1}$$