There are 10 different people at a party. How many ways are there to pair them off into a collection of five pairings? So my thoughts are 
${10 \choose 2} * {8 \choose 2} * {6 \choose 2} * {4 \choose 2}$
if the first person , has 9 choices 
9 x 7 x 5 x 3 x 1. does this work ? 
cause i cant really make a logical thought about overlapping in chooses
 A: Isn't the double factorial (2m-1)!! the direct answer to this? For 2m people? 
Basically, each person has 9 different other persons to choose from, and after that happens, the next pair's person has 7 other people to choose from (10 minus the first pair and himself) all the way down to the last person, that has only 1 other person to choose from.
so you have 9 * 7 * 5 * 3 * 1  = (9)!! = 945
A: For future reference here is an approach using combinatorial species. We treat the general case of $2m$ people and $m$ pairs. The species $\mathcal{Q}$ that we want to enumerate (using exponential generating functions) has the specification $$\mathcal{Q} = \mathfrak{P}_m(\mathcal{P}(\mathcal{Z}))$$
where $\mathcal{P}$ is the species of pairs.
But we have $$\mathcal{P}(\mathcal{Z}) = \mathfrak{P}_2(\mathcal{Z}).$$
Therefore the generating function $Q(z)$ of $\mathcal{Q}$ is given by
$$Q(z) = \frac{1}{m!} \left(\frac{1}{2!} z^2\right)^m.$$
The number of pairings is $(2m)!$ times the coefficient of $z^{2m}$ of this generating function, giving
$$ (2m)! [z^{2m}] \frac{1}{2^m\times m!} z^{2m} = \frac{(2m)!}{2^m\times m!}.$$
This gives the following sequence of values:
$$1, 3, 15, 105, 945, 10395, 135135, 2027025, 34459425, 654729075,\ldots$$
In particular we get for ten people that there are $945$ arrangements  (fifth term in the sequence with $m=5$ and five pairs for 10 people in total).
This is one of the more popular sequences from the OEIS, where we find this extensive OEIS entry.
A: Just tell your $10$ people to line up such that the first person and the second person make up the first pair, the third and fourth make up the second pair, etcetera. This gives $10!$ pairings.
However, if the first person would swap places with the second person, the pairing would be no different, though we counted it as another pairing! In fact, each pair can have its members swap places while preserving the pairings, so we counted each pairing $2^5$ times. So, that's what we have to divide by.
We're not finished yet, though: what if the first two people swap places with the third and fourth? It would once again preserve the pairings, though we counted it as a different pairing. In fact, the $5$ pairs can swap places with each other all they want, the same pairings would still be preserved. Hence we counted each pairing $5!$ times, and that's what we have to divide by as well.
In conclusion, there are $$\frac{10!}{2^5 \cdot 5!} = 945$$ pairings.
A: Hint:  for the first two pairs, you are counting AB+CD and CD+AB as different.  Shouldn't they be the same?  How does this extend?
Added:  Your first thought was better.  There are indeed $_{10}C_2$ ways to form the first pair.  The problem is that some of the other ways to make pairs will make that same pair later.  For your second thought, if you pick somebody to pair, they will have $9$ choices, but you could have made that pair in the other order.
