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.
![C^r_n=(n!)/(r!(n-r)!)](https://img.qammunity.org/2023/formulas/mathematics/college/q3923t04vn0fiw4u9vct73rjmu5uwcluy9.png)
Where n = 9, and r = 3. Let's replace these values.
![C^3_9=(9!)/(3!(9-3)!)=(9!)/(3!\cdot6!)=(9\cdot8\cdot7\cdot6!)/(3\cdot2\cdot1\cdot6!)=(504)/(6)=84](https://img.qammunity.org/2023/formulas/mathematics/college/knr7zfkvvp8oqdlooe8ntecpyhdjv2yrke.png)
Hence, there are 84 different ways to choose the first three.