## Algebra 2 (1st Edition)

$5400$
We can choose $3$ out of the $10$ tragedies and $2$ of the $10$ histories, thus the number of possible sets of plays: $_{10}C_3\cdot_{10}C_2=\frac{10!}{7!3!}\frac{10!}{8!2!}=\frac{10\cdot9\cdot8\cdot7!}{7!3!}\frac{10\cdot9\cdot8!}{8!2!}=120\cdot45=5400$