We know that
• The total is nine starters.
,• The coach wants to choose three.
Given that the player cannot include repetition, we solve this problem as a combination because the order does not matter.
[tex]C^r_n=\frac{n!}{r!(n-r)!}[/tex]Where n = 9, and r = 3. Let's replace these values.
[tex]C^3_9=\frac{9!}{3!(9-3)!}=\frac{9!}{3!\cdot6!}=\frac{9\cdot8\cdot7\cdot6!}{3\cdot2\cdot1\cdot6!}=\frac{504}{6}=84[/tex]