Chapter 11 - Counting Methods and Probability Theory - 11.3 Combinations - Exercise Set 11.3 - Page 707: 20

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${}_{n}C_{r}=\displaystyle \frac{n!}{(n-r)!r!}$. ----------------- ${}_{10}C_{3}=\displaystyle \frac{10!}{7!\cdot 3!}=\frac{10\times 9\times 8}{1\times 2\times 3}=120$ ${}_{6}C_{4}=\displaystyle \frac{6!}{2!\cdot 4!}=\frac{6\times 5}{1\times 2}=15$ $\displaystyle \frac{ {}_{10}C_{3}}{ {}_{6}C_{4}}=\frac{120}{15}=8$