Chapter 11 - Counting Methods and Probability Theory - 11.3 Combinations - Exercise Set 11.3 - Page 707: 19

7

${}_{n}C_{r}=\displaystyle \frac{n!}{(n-r)!r!}$. ----------------- ${}_{7}C_{3}=\displaystyle \frac{7!}{(7-3)!3!}=\frac{7\times 6\times 5}{1\times 2\times 3}=35$ ${}_{5}C_{4}=\displaystyle \frac{5!}{(5-4)!4!}=\frac{5}{1}=5$ $\displaystyle \frac{ {}_{7}C_{3}}{ {}_{5}C_{4}}=\frac{35}{5}=7$