## Thinking Mathematically (6th Edition)

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