Combinatorics Probability: No Socks are a matched Pair? So suppose you have $16$ socks ($8$ distinct pairs) and you pull out $5$ of these without looking. What is the probability that none of the socks create a matched pair?
I know the answer is $8\over12$ by using a tree diagram, but I want to know how to do this with combinatorics, but I can't seem to get it.
We have $16\choose5$ ways of drawing $5$ socks from the $16$ socks, this is our sample space! But how do you calculate the numerator of this equation? I supposed it was $16 \cdot 14 \cdot 12 \cdot 10 \cdot 8$, but obviously this yields a probability > $1$ and so I don't know how to proceed.
Can someone please demonstrate this to me? Thanks!
 A: You took order into account in your numerator but not your denominator.
Method 1:  Without taking order into account.
It is true that you can select $5$ of the $16$ socks in $\binom{16}{5}$ ways.  Notice that you have not taken the order of selection into account.
You then say that you can pick socks from five different pairs in $16 \cdot 14 \cdot 12 \cdot 10 \cdot 8$ ways.  However, this time you have taken order into account.  For instance, if each pair of socks has a different color, choosing blue, black, grey, brown, and red in that order results in the same selection of five socks as choosing blue, brown, grey, black, and brown in that order.  Since there are $5!$ orders in which you could pick the same socks, if we do not take order into account, the number of favorable cases is 
$$\frac{1}{5!} \cdot 16 \cdot 14 \cdot 12 \cdot 10 \cdot 8 = \frac{16 \cdot 14 \cdot 12 \cdot 10 \cdot 8}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 8 \cdot 14 \cdot 2 \cdot 8$$
Dividing by 
$$\binom{16}{5} = \frac{16!}{5!11!} = \frac{16 \cdot 15 \cdot 14 \cdot 13 \cdot 12}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 8 \cdot 3 \cdot 14 \cdot 13$$
yields the probability
$$\Pr(\text{five different pairs}) = \frac{8 \cdot 14 \cdot 2 \cdot 8}{8 \cdot 3 \cdot 14 \cdot 13} = \frac{2 \cdot 8}{3 \cdot 13} = \frac{16}{39}$$
Method 2:  Taking order into account.
We know that there are $16 \cdot 14 \cdot 12 \cdot 10 \cdot 8$ favorable cases when order is taken into account.
If we also take the order of selection into account in counting all ways of selecting five of the sixteen socks, we obtain $16 \cdot 15 \cdot 14 \cdot 13 \cdot 12$ possible selections.  
Hence,
$$\Pr(\text{five different pairs}) = \frac{16 \cdot 14 \cdot 12 \cdot 10 \cdot 8}{16 \cdot 15 \cdot 14 \cdot 13 \cdot 12} = \frac{10 \cdot 8}{15 \cdot 13} = \frac{2 \cdot 8}{3 \cdot 13} = \frac{16}{39}$$

Notice that these answers both agree with Sam's answer.  Sam did not take order of selection into account, so there are $\binom{16}{5}$ of selecting five socks.  For the favorable cases, Sam chose which five of the eight pairs of socks from which socks are drawn and one of the two socks from each of those pairs, giving 
$\binom{8}{5}2^5$ favorable cases.  Therefore,
$$\Pr(\text{five different pairs}) = \frac{\dbinom{8}{5}2^5}{\dbinom{16}{5}} = \frac{16}{39}$$
Personally, I prefer Sam's method, but I wanted to point out how you could correct your answer.
A: Total sample space $={16\choose 5}$
Ways to pick 5 different socks from 8 pairs $={8\choose 5}$
The socks picked from these pairs may be either of the two identical socks. Hence we multiply $2$ for each of the pairs ($5$)
$$P={{{8\choose 5}\times 2^5}\over {16\choose 5}} = {16\over 39}$$
