Combination math
posted by rinchan .
The camera club has 5 members and the mathematics club has 8. There is only one member common to both clubs. In how many ways could a committee be formed with at least one member from each club?

How large is the committee ?

oops. It's a committee of 4 people

So the only case we DON'T want are committees consisting of all camera members and all math club members.
with out any restrictions we choose 4 from the 13
= C(13,4) = 715
all camera people = C(5,4) = 5
all math club people = C(8,4) = 70
so number with at least one from each club
= 715  5  70 = 640 
use the binomial theorem to expand(3x2y)4
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