Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

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

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

share|cite|improve this question
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

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

Anyways you r also right.

share|cite|improve this answer
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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.