I cannot seem to figure this out:
A 8-Person committee is to be founded by a group of fifteen women and twelve men, how many ways can the committee be chosen if: (Already solved->) a)The committee contains four men and four women b) There must be at least two men? c) There must be more women than men
How would I set up B and C, I cannot seem to find anything useful as far as resources go, so a pointer in the right direction would be very helpful
b) Use complimentary counting. We look for the number of committees with $0$ or $1$ man, then subtract this from the total ways, which is $\dbinom{27}{8}$ ways. This is simple - for $0$ men, we just choose all women, which can be done in $\dbinom{15}{8}$ ways, and for $1$ man, this is just $12 \cdot \dbinom{15}{7}.$ The number of unsuccessful ways is $83,655$. The answer must be $\dbinom{27}{8} - 83,655 = \boxed{2,136,420}.$
c) HINT: Like we did in part (b) do casework for $0,$ $1,$ $2,$ and $3$ men.
