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 ?
Best Answer
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.