Q) A committee of 11 members is to be formed from 8 males and 5 females. The number of ways the committee is formed with at least 6 males is?
The general solution I see everywhere is to take 3 cases where there are 6 males, 7 males and 8 males.
What I want to know is why the following logic is wrong:
- Select 6 males from 8: $\binom{8}{6}$
- Now since all conditions are satisfied select the remaining members for the committee($ 11-6=5 $) from the remaining total($(8+5)-6 = 7$): $\binom{7}{5}$
This gives the answer $$\binom{8}{6} * \binom{7}{5} = 588$$ but the required answer is 68.
when you choose 6 out of 8 males, this means you mark every make either as "Yes" or "No". if you then choose 5 out of the remaining 7 (2 males and 5), each of them gets another label "Yes" or "No", so essentially some males will get multiple labels - "No+No" or "No+Yes". So your approach distinguishes between males with "Yes" and males with "No-Yes"