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

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