Chapter 13 - Probability - 13-3 Permutations and Combinations - Practice and Problem-Solving Exercises - Page 841: 16

a) 1320 b) 95,040

a) We use the permutation equation: We know that: $_nP_r = \frac{n!}{(n-r)!}$ $_{12}P_3 = \frac{12!}{(12-3)!}$ $\frac{12!}{9!}$ We also know that: $x! = x(x-1)(x-2)...(1)$ Thus, we have: 1320 b) We know that: $_nP_r = \frac{n!}{(n-r)!}$ $_{12}P_5 = \frac{12!}{(12-5)!}$ $\frac{12!}{7!}$ We also know that: $x! = x(x-1)(x-2)...(1)$ Thus, we have: 95,040