Chapter 11 - Counting Methods and Probability Theory - 11.3 Combinations - Exercise Set 11.3 - Page 707: 25

$-9499$

${}_{n}C_{r}=\displaystyle \frac{n!}{(n-r)!r!}$ -------- $\displaystyle \frac{98!}{96!}=\frac{98\times 97\times 96!}{96!}=98\times 97$ ${}_{7}C_{3}=\displaystyle \frac{7!}{(7-3)!3!}=\frac{7\times 6\times 5}{1\times 2\times 3}=35$ ${}_{5}C_{4}=\displaystyle \frac{5!}{(5-4)!4!}=5$ $\displaystyle \frac{{}_{7}C_{3}}{{}_{5}C_{4}}=\frac{35}{5}=7$ $7-98\times 97=7-9506=-9499$