# Choosing teams with minimum number of boys and girls.

I need to find different number of teams I can make with 6 people that needs to have at least 2 girls and 2 boys. There are 8 girls and 12 boys. So the way I think is I need to find total number of different teams that I can make first and subtract all boys team, all girls team, one girl 5 boys team, 5 girls one boy team.

$$\binom{20}{6} - \binom{12}{6} - \binom{8}{6} - \left(\binom{8}{5} \binom{12}{1}\right) - \left(\binom{8}{1} \binom{12}{5}\right)$$

$$\binom{20}{6} - \binom{12}{6} - \binom{8}{6} - \left( \binom{8}{5} \binom{12}{1}\right) - \left( \binom{8}{1} \binom{12}{5}\right)$$

-
Andy, there is latex support on this site. You don't need to link to a latex image provider. Type you latex within dollar signs. Example $x^n = y^n$ looks like $x^n = y^n$. – Aryabhata Apr 10 '13 at 9:52
I have tried to fix up what you had, but I might have made a mistake in the process... – Aryabhata Apr 10 '13 at 9:57
Nope you didn't make any mistake. Thanks! I was trying to paste the latex code directly from the latex web editor but couldn't make it work. – Andy Apr 10 '13 at 14:19

## 2 Answers

Just calculate the 3 cases instead. (b,g)=(4,2), (3,3) and (2,4)

Anyways you r also right.

-
Thanks a bunch! – Andy Apr 10 '13 at 9:00

if boys divided into such as Tom,Jack,Jerry....then $$X={20\choose 6} - {8\choose6}-{12\choose6}-{8\choose5}{12\choose1}-{12\choose5}{8\choose1}=30800$$ else it's a naive problem just 3

-