Question

how many ways can 4 baseball players and 4 basketball players be selected from 15 baseball players and 9 basketball players?

Answer 1

$C^{15}_4 \times C^{9}_4 = \frac{15!}{4!(15-4)!} \times \frac{9!}{4!(9-4)!} \\ = \frac{12 \times 13 \times 14 \times 15 }{2 \times 3 \times 4} \times \frac{6 \times 7 \times 8 \times 9}{2 \times 3 \times 4} \\ = 1365 \times 126 = 171990$