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

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 isn’t 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.

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

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

Leave a Comment