Number of Optional Pairings?

So I need to calculate the number of ways you can partition a set of $n$ items such that the size of each partition is either 1 or 2.

Let $m = \lfloor \frac{n}{2}\rfloor$

$m$ is the maximum number of pairs (2-partitions).

$$\sum_{i=0}^m{{\prod_{j=0}^{i-1}{n-2j \choose 2}} \over i!}$$

i is the number of pairs to use, and then j is choosing the jth pair from the remaining items, with i! factoring out the irrelevant order selection.

I don't know how to simplify from here, or have I taken a bad turn? Is there a simpler way to approach this problem?

$${{\prod_{j=0}^{i-1}{n-2j \choose 2}} \over i!}$$ $$= {n \choose 2}{n-2 \choose 2}\dots{n-2i-2 \choose 2}$$ $$= {n! \over 2!(n-2)!}{(n-2)! \over 2!(n-4)!}\dots{(n-2i-2)! \over 2!(n-2i-4)!}$$ $$= {n! \over 2^i(n-2i-4)!}$$

Is that right? And now expanding the sum?

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Before you give up, you might try writing out that product.... –  Hurkyl Jun 24 '12 at 1:14

I'm assuming the $n$ items are distinguishable.
I guess then that I mean the same thing you mean. The partition of $\{{a,b,c\}}$ into $\{{a\}}$ and $\{{b,c\}}$ differs from the partition into $\{{b\}}$ and $\{{a,c\}}$. But if you follow the links the meaning should become evident. –  Gerry Myerson Jun 26 '12 at 6:19