Question

How many ways can 4 baseball players and 3 basketball players be selected from 12 baseball players and 9 basketball players?

Answer 1

$C^4_{12}C^3_9=\frac{12!}{4!8!} \frac{9!}{3!6!}=\frac{9\times10\times11\times12}{2\times3\times4}\frac{7\times8\times9}{2\times3}=495\times84=41580$