Answer
126
Work Step by Step
The number of possible combinations if r items are taken from n items is
nCr = $\frac{n!}{r!(n-r)!}$
9C5 = $\frac{9!}{5!(9-5)!}$
=$\frac{9!}{5!4!}$
=$\frac{9*8*7*6*5!}{5!4!}$
Since 5! is identical in both numerator and denominator so cancel 5!
=$\frac{9*8*7*6}{4*3*2*1}$ = 126
