Arranging 5 pairs of brothers in ten tables Suppose I have $10$ people composed of $5$ pairs of brothers. They are seated randomly around $5$ tables. Each table has only two places.
What is the possibility that exactly $3$ pairs of brothers will be sitting at the same table?
$$\\ \frac{ {5 \choose 3} \cdot 3\cdot 2\cdot {4 \choose 1}{2 \choose 1}}{{10 \choose 2,2,2,2,2} / 5!} $$
What I thought was picking $\ 3  $ pairs out of the $\ 5 $ then seating each of them at a different table. Therefore, $\ {5 \choose 3} \cdot 3 \cdot 2 \cdot 1$ and then I'm left with $\ 4 $ people which is two pairs and picking one of them $\ {4 \choose 1} \cdot {2 \choose 1} $, because after picking one of the people, I have only two options left for the second one and none for the pairing the last two.
The sample space I thought would be $\ {10 \choose 2,2,2,2,2} / 5! $ because it's the number of ways to divide $10$ people into $5$ groups of $2$ with no difference between the groups. 
I really don't know what I am doing wrong here.
 A: Let's take as our sample space the number of ways of placing two people at each table, which is 
$$\binom{10}{2, 2, 2, 2, 2} = \binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
For the favorable cases, choose which three of the five tables will receive a pair of brothers.  Choose a pair of brothers to sit at each of those tables.  That leaves four people and two tables.  Choose which of those two tables the youngest of those four people sits at.  One of the other three people will be his brother.  To ensure that there are exactly three pairs of brothers sitting together, we must choose one of the other two people to sit with the youngest person remaining.  The final two people must sit together at the remaining table.  Hence, the number of favorable cases is 
$$\binom{5}{3} \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 2$$
Thus, the probability that exactly three pairs of brothers will be seated together at the five tables if they are seated randomly is 
$$\frac{\dbinom{5}{3} \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 2}{\dbinom{10}{2}\dbinom{8}{2}\dbinom{6}{2}\dbinom{4}{2}\dbinom{2}{2}}$$
A: First calculate the number of possible arrangements. We place 10 men in 10 places, therefore 
$$|\Omega| = 10!$$
Now we pick 3 pairs out of 5 ($\binom{5}{3}$), pick 3 tables out of 5 ($\binom{5}{3}$), place the pairs in these tables ($3!$), arrange each of these pairs in 2 places ($2!^3$), place first of the remaining men (they can be arranged in a query in one set way, for example alphabetically) in one of the remaining places ($4$), select one of the remaining men, which is not his brother, and sat him in opposite place ($2$), and at last - place the remaining men in 2 places ($2!$)
Therefore we have:
 $$|A|=\binom{5}{3}\cdot \binom{5}{3}\cdot 3!\cdot 2!^3\cdot 4\cdot 2\cdot 2!=\binom{5}{3}^2\cdot 4!\cdot 2^5$$
Of course: 
$$P(A)=\frac{|A|}{|\Omega|}$$
