Chapter 11 - Further Topics in Algebra - 11.6 Basics of Counting Theory - 11.6 Exercises - Page 1062: 47

$120$

Arrangements $\rightarrow$ Permutations.$P(n,r)=\displaystyle \frac{n!}{(n-r)!}$. Arranging r=$3$ from n=$6$ available courses, $ P(6,3)=\displaystyle \frac{6!}{(6-3)!}\qquad$ ... $=\displaystyle \frac{6\times 5\times 4\times(3!)}{(3!)}$ $=6\times 5\times 4$ = $120$