There are 5 boys and 5 girls on a co-ed basketball team. A team plays 5 players on the court at one time. We can play at most 3 boys. Position doesn’t matter.

Question

dan815 · Answer

okay so consider when you have 0,1,2,3 boys

dan815 · Answer

0 boys = 5 girls = 5C5= 1
1 boy = 4 girls = 5C4 * 5C1=5
2 boys =3 girls = 5C3*5C2=
3 boys =2 girls = 5C2*5C3=

anonymous · Answer

I think you should first choose 2 girls, then chose 3 players from both girls and boys, somehow.

anonymous · Answer

It is apparently okay for the team to have no boys.

anonymous · Answer

I wonder if ${5\choose 2}{3+5\choose 3}$ is equal to what dan's method.

zarkon · Answer

they are not the same...you would be double counting

anonymous · Answer

$$
{10 \choose 5} - {5 \choose 5} - {5 \choose 4}{5 \choose 1}
$$Hmm, I got to figure out where my double counting is coming from.

anonymous · Answer

that is just counting by complement. The double counting came from the fact that the set of 5 girls and the set of 8 people remaining are not disjoint

anonymous · Answer

Hmm, so the same girl can be chosen twice?

anonymous · Answer

yes. (5C2) (8C3) would count at least 1 girl (twice?) maybe more, but counting some girls again is certain