Chapter 3 - Probability - Section 3.4 Additional Topics in Probability and Counting - Exercises - Page 174: 14

$\frac{_{7}P_{3}}{_{12}P_{4}}=\frac{7}{396}\approx0.018$

$_{n}P_{r}=\frac{n!}{(n-r)!}$ $_{7}P_{3}=\frac{7!}{(7-3)!}=\frac{7\times6\times5\times4!}{4!}=7\times6\times5=210$ $_{12}P_{4}=\frac{12!}{(12-4)!}=\frac{12\times11\times10\times9\times8!}{8!}=12\times11\times10\times9=11,880$ $\frac{_{7}P_{3}}{_{12}P_{4}}=\frac{210}{11,880}=\frac{7}{396}\approx0.018$