Chapter 11 - Further Topics in Algebra - 11.6 Basics of Counting Theory - 11.6 Exercises - Page 1062: 48

$\approx 2.05\times 10^{10} $

Arrangements, schedules $\rightarrow$ Permutations.$P(n,r)=\displaystyle \frac{n!}{(n-r)!}$. Arranging r=4 from n=380 available courses, $ P(380,4)=\displaystyle \frac{380!}{(380-4)!}\qquad$ $=\displaystyle \frac{380\times 379\times 378\times 377\times(376!)}{(376!)}$ $=380\times 379\times 378\times 377$ $\approx 2.052371412\times 10^{10} $