#### Answer

7

#### Work Step by Step

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