Chapter 13 - Probability - 13-3 Permutations and Combinations - Practice and Problem-Solving Exercises - Page 841: 17

a) 4060 b) 142,506

We know that: $_nC_r =\frac{_nP_r}{r!} = \frac{n!}{(n-r)!r!}$ $_{30}C_3 = \frac{30!}{(30-3)!3!}$ We also know that: $x! = x(x-1)(x-2)...(1)$ Thus, we have: 4060 b) We know that: $_nC_r =\frac{_nP_r}{r!} = \frac{n!}{(n-r)!r!}$ $_{30}C_5 = \frac{30!}{(30-5)!5!}$ We also know that: $x! = x(x-1)(x-2)...(1)$ Thus, we have: 142,506