divide 6 people in group of 2 in same size Exercise: divide 6 people in group of 2 in same size.
My solution:
The exercise tells us to calculate the combination without repetition. If I start by calculating the number of ways to select how many ways that one can divide into $3$ of $6$ people. In group 1 , we get that the number of ways to divide $3$ of $6$ people in a group of size $3$ is
$$\frac{6!}{3!3!}=\frac{6 \cdot 5 \cdot 4}{6} = 20$$
Because we "used" $3$ of $6$ subjects in group 1 , we have $6-3$ people left to divide in group 2. This can be done(calculated in the same way as above) ​​in one way.
From this we get that the number of ways to divide the six people in two groups of the same size is equal to
$$20 \cdot 1=20$$
but the solution in the book says $10$. What am I doing wrong?
EDIT:
Shall I skip the second calculation and divide the first one with $2!$ since the dividing in group 1 and 2 are same since don't care about the order in which the groups appear because they are same size ?
 A: 20 is the correct answer for "How many ways to choose 3 people from a group of 6", but the question is slightly different. The remaining three people who are not chosen form the second group - and they will also be chosen as a group in one of the other combinations.
So there are a few ways to fix this up. One way - the quickest way - is to observe the symmetry and divide by two to account for each group being chosen twice. Another way might be to fix one of the people onto a team and then chose two people to go with that fixed person, leaving the remainder in the opposite team. That would give ${5 \choose 2}$ options $= 10$ again. An elaborate way might be to model the schoolyard system - pick two captains arbitrarily then assign two people to one team and two to the other, ${4\choose 2}{2\choose 2} = 6$, then add in the ${4\choose 1}=4$ other teams possible when the two "captains" are assigned to the same team.
A: @Vikram's answer
To find the number of ways in which m+n things can be divided into two groups containing m and n things respectively. This is clearly equivalent to finding the number of combi nations of $m+n$ things m at a time, for every time we select one group of m things we leave a group of n things behind. Thus the required number = $\frac{(m+n)!}{m!.n!}$. If $n=m$, the groups are equal, and in this case the number of different ways of subdivision is $\frac{(2m)!}{m!m!2!}$ for in any one way it is possible to interchange the two groups without obtaining a new distribution.
