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(3x-2y)4