Chapter 13 - Probability - 13-3 Permutations and Combinations - Practice and Problem-Solving Exercises - Page 841: 19

495

Order does not matter, so we use the combination equation. We know that: $_nC_r =\frac{_nP_r}{r!} = \frac{n!}{(n-r)!r!}$ $_{12}C_4 = \frac{12!}{(12-4)!4!}$ $\frac{ 12!}{4!8!}$ We also know that: $x! = x(x-1)(x-2)...(1)$ Thus, we have: 495