I'm really bad at counting problems, so I'm not surprised I couldn't get this one.
The Sagebrush student council has 6 boys and 6 girls as class
representatives. Two different subcommittees, each consisting of 2
boys and 2 girls are to be created. If no student can be in both
subcommittees, how many different combinations of subcommittees are
possible?
So I first find the number of ways to make the first subcommittee, which I think is $2\cdot(6\operatorname{C}2)=30$(6 choose 2 because we are choosing 2 people out of 6 original; times 2 for both girl and boy choices). For the second subcommittee, I have $2\cdot(4\operatorname{C}2)=12$(4 choose two because we are choosing two people out of the now limited 4; times 2 for both girl and boy choices). Adding them together, I have a total of $42$. This is wrong.
What did I do wrong? How should I solve this problem?
Thanks! Your help is appreciated!
Max0815
Best Answer
Hint:
Since for the first subcommittee you are choosing boys and choosing girls, the number of ways of doing that has to be multiplied: $$\binom{6}{2} \times \binom{6}{2} \ .$$
The second subcommittee follows the same reasoning, but now we have $4$ boys and girls left.
As for the total number of combinations of the subcommittees, we will need to have the first subcommittee and the second subcommittee.
Can you compute everything now?