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The camera club has 5 members, and the math club has 8 members. There is only one member common to both clubs. In how many ways could a committee of 4 people be formed with at least 1 member from each group?

1. 0
2. 1
1. Let's say Bob is in both clubs. There are 11 other people.

Two cases: Bob is on the committee, or not:

a) Bob included:
First choose him, then choose any 3 of the other 11, since both clubs are represented by Bob:
11C3 ways to do that.

b) Bob excluded:
Now start with 11C4 (committees made up of the other 11 people).
Subtract 4C4 = 1 way with all 4 others from the camera club,
and 7C4 = 35 ways with 4 others from the math club.

11C3 + 11C4 - 1 - 7C4 = 165 + 330 - 1 - 35 = 459

Or we could say:
there are 12C4 total committees ( = 495 = 165 + 300 )
not worrying about Bob, and then just subtract the 1 which is the 4 camera club members
who are not Bob and the 35, which is the same for the math club.

posted by Secret

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