SOLUTION: If there are 22 students in a frat house. In how many ways can a committee of 9 be selected?

Question 583762: If there are 22 students in a frat house. In how many ways can a committee of 9 be selected?
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
C(n,r) = (n!)/(r!(n-r)!)

C(22,9) = (22!)/(9!*(22-9)!)

C(22,9) = (22!)/(9!*13!)

C(22,9) = (22*21*20*19*18*17*16*15*14*13!)/(9!*13!)

C(22,9) = (22*21*20*19*18*17*16*15*14)/(9!)

C(22,9) = (22*21*20*19*18*17*16*15*14)/(9*8*7*6*5*4*3*2*1)

C(22,9) = (180503769600)/(362880)

C(22,9) = 497420

So there are 497420 different ways.
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