Chapter 3 - Probability - Section 3.4 Additional Topics in Probability and Counting - Exercises - Page 174: 13

$\frac{_{6}P_{2}}{_{11}P_{3}}=\frac{1}{33}\approx0.0303$

$_{n}P_{r}=\frac{n!}{(n-r)!}$ $_{6}P_{2}=\frac{6!}{(6-2)!}=\frac{6\times5\times4!}{4!}=6\times5=30$ $_{11}P_{3}=\frac{11!}{(11-3)!}=\frac{11\times10\times9\times8!}{8!}=11\times10\times9=990$ $\frac{_{6}P_{2}}{_{11}P_{3}}=\frac{30}{990}=\frac{1}{33}\approx0.0303$