A club with 32 boys and 36 girls needs to form a committee of size nine. How many committees with
at least three girls and at least three boys can be formed?
The source is a professor's assignment in a Discrete Math course.
At first this appeared simple.
$$(_{32}C_{3})*(_{36}C_{3})*(_{62}C_{3})$$
with 62 coming from 29 boys left + 33 girls left to choose from.
This is choose 3 boys * choose 3 girls * choose 3 of either.
However, this has duplicates. Let B1 represent 1st boy & G1 represent the 1st girl.
B1, B2, B3, G1, G2, G3, B4, B5, B6 is the same committee as
B1, B5, B6, G1, G2, G3, B4, B2, B3
A. How do I compensate for duplicates like these and any other duplicates?
B. If an entirely other approach is required, please share.
Best Answer
Just add the $4$ cases where $0,1,2,3$ boys are from the last case (or similar, $b = 3$ up to $6$ boys in the set of $9$, complemented with $9-b$ girls.
$$\sum_{b=3}^{6}C(32, b)*C(36, 9-b) $$
Such formulas are perfectly fine as an answer. A single term does not exist, and the resulting exact value is optional: it does not show your understanding or how you came to the solution.
Final value:
Yes, I admit, I used Excel: