9 students volunteer for a committee. How many different 6-person committees can be chosen?

NetherCraft 0

3 Answers

  • Let the students be A, B, C, D, E, F, G, H and I

    Now, picking any 6 we could have A, B, C, D, E and F, which is the same grouping as B, A, C, D, E and F as order is not important

    => 9C6 = 9!/6!3! = 84 combinations

    However, committees normally consist of specific people, i.e. chair, vice chair, treasurer, e.t.c.

    Hence, this is an example where order matters and we have permutations

    i.e. 9P6 => 9!/(9 – 6)! = 60,480

    Another way of looking at this is to say the following:

    The chair can be chosen in 9 ways, the vice chair in 8 ways, treasurer 7 ways, and so on.

    Hence, 9 x 8 x 7 x 6 x 5 x 4 => 60,480

    Remember, specifying order is important in the whole problem.

    :)>

  • 9C6 = 9!/(6!3!)=84 different committees.

  • C(9,6) = C(9,3) = 9*8*7/(1*2*3) = 3*4*7 = 84

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