Probability with combination problem- hotel room There are 12 people, 6 men and 6 women. You want to assign them randomly into double rooms (2 people per room).
What is the probability that the pairs are formed by people of the same sex?  
So I know that the probability is
P= (number of same sex pairings)/(total number of possible arrangements)  
I've thought of starting by finding the total number of arrangements possible, which is \begin{pmatrix}12\\2\end{pmatrix} which is 66.
The first person can be paired with one of the 11 that are left, 5 of the same sex and 6 of the different sex. But I'm stuck calculating this.
 A: There are 
$$
B=\frac{12!}{(2!)^6}
$$
ways to assign the $12$ people to $6$ different rooms with $2$ people in each room. To find the number of same-sex pairings we proceed as follows. Choose $3$ of the $6$ rooms to be occupied by males and then assign $6$ males to these rooms with $2$ males in each room. Assign the remaining $6$ females to the remaining $3$ rooms with $2$ females in each room. Hence there are
$$
A=\binom{6}{3}\left(\frac{6!}{(2!)^3}\right)\left(\frac{6!}{(2!)^3}\right)
$$
same-sex pairings. The desired probability is $A/B$.  
A: You are correct in that dividing the number of same-sex arrangements by the total number of arrangements gives the probability, but you haven't satrted out right.
$\binom{12}{2}$ is just the number of ways to pick two people.  It's not close to the number of arrangements.  To get the number of arrangements, imagine we place all $12$ people in a line, the assign the first two to the first room, the next two to the second room, and so on.  There are $12!$ ways to form the line.  In each pair, we don't care who comes first, so we have to divide by $2^6$.  Also, we don't really care which pairs are assigned to which rooms, only what people are paired together, so we must divide by $6!$.  That makes $$\frac{12!}{2^6\cdot6!}$$ arrangements.
Now, can you apply the same ideas to figure out how many same-sex arrangements there are? 
Your second approach will work also. Choose any one of the men.  The probability that his roommate is a man is $\frac{5}{11}$ by the reasoning you gave.  Now there are $4$ men and $6$ women left, so if we choose one of the unassigned men, the probability that his roommate is a man is $\frac39$.  Now there are $2$ men and $6$ women left, so the probability that the two men are roommates is $$\frac{5\cdot3\cdot1}{11\cdot9\cdot7}$$ 
Of course, if all the men are paired, so are all the women, so this is the probability.
You can verify that both approaches give the same answer.
