Combination math

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?

asked by rinchan
  1. How large is the committee ?

    posted by Reiny
  2. oops. It's a committee of 4 people

    posted by rinchan
  3. 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

    posted by Reiny
  4. use the binomial theorem to expand(3x-2y)4

    posted by Anonymous

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