# How many ways to assemble a team of 5 out of 15 girls and 10 boys with limitations?

How many ways to assemble a team of 5 out of 15 girls and 10 boys, if the team must contain at least two boys and two girls?

Why is it wrong to count the following way(?):

1. Choose 2 boys $= {10 \choose 2}$
2. Choose 2 girls $= {15 \choose 2}$
3. Choose one more boy/girl from the rest $= 21$

$${10 \choose 2}{15 \choose 2}21 = 99225$$

I quiet sure that this count has duplicates, if so, how do I eliminate them?

The correct answer is: $${10 \choose 2}{15 \choose 3}+{10 \choose 3}{15 \choose 2}=33075$$

-

The problem is that for the gender of which there are $3$, there are three ways of picking $2$ of the $3$ at first and then one later, so you're counting each team three times. You can either divide by that factor of three, or take the approach in the answer you gave: there must be either $2$ boys and $3$ girls or vice versa, so add the numbers of ways you can make those two selections.