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

Now, selecting any 6 we might have A, B, C, D, E and F, which is similar grouping as B, A, C, D, E and F as order isnt essential

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

Nonetheless, committees usually encompass particular individuals, i.e. chair, vice chair, treasurer, e.t.c.

Therefore, that is an instance the place order issues and we now have permutations

i.e. 9P6 => 9!/(9 6)! = 60,480

One other method of that is to say the next:

The chair will be chosen in 9 methods, the vice chair in 8 methods, treasurer 7 methods, and so forth.

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

Keep in mind, specifying order is essential in the entire drawback.

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

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

This post is last updated on hrtanswers.com at Date : 1st of September – 2022