Chapter 11 - Counting Methods and Probability Theory - 11.3 Combinations - Exercise Set 11.3 - Page 707: 23

$\displaystyle \frac{3}{4}$

${}_{n}P_{r}=\displaystyle \frac{n!}{(n-r)!}$ --------------- ${}_{3}P_{2}=\displaystyle \frac{3!}{(3-2)!}=3\times 2=6$ ${}_{4}P_{3}=\displaystyle \frac{4!}{(4-3)!}=4\times 3\times 2=24$ $1-\displaystyle \frac{ {}_{3}P_{2}}{{}_{4}P_{3}}=1-\frac{6}{24}=1-\frac{1}{4}=\frac{3}{4}$