Question

Selecting Players How many ways can 4 baseball

players and 3 basketball players be selected from

12 baseball players and 9 basketball players

Answer 1

$$\dbinom{12}{4}\dbinom{9}{3}=\dfrac{12!}{4!(12-4)!}\cdot \dfrac{9!}{3!(9-3)!}$$

$$=\dfrac{12(11)(10)(9)}{1(2)(3)(4)}\cdot\dfrac{9(8)(7)}{1(2)(3)}$$

$$=495\cdot84=41580$$