Chapter 11 - Counting Methods and Probability Theory - 11.3 Combinations - Exercise Set 11.3 - Page 707: 24

$\displaystyle \frac{83}{84}$

${}_{n}P_{r}=\displaystyle \frac{n!}{(n-r)!}$ --------------- ${}_{5}P_{3}=\displaystyle \frac{5!}{(5-3)!}=5\times 4\times 3$ ${}_{10}P_{4}=\displaystyle \frac{10!}{(10-4)!}=10\times 9\times 8\times 7$ $1-\displaystyle \frac{ {}_{5}P_{3}}{{}_{10}P_{4}}=1-\frac{5\times 4\times 3}{10\times 9\times 8\times 7}$ $=1-\dfrac{1}{3\times 4\times 7}$ $=1-\displaystyle \frac{1}{84}=\frac{83}{84}$