Question

1. a basketball coach must choose 5 players from a team of 12 to start a game. in how many ways can the starting lineup be selected?

Explanation:

Step1: Identify the problem type

This is a combination problem where we need to choose 5 players from 12, and the order of selection doesn't matter. The formula for combinations is \( C(n, k)=\frac{n!}{k!(n - k)!} \), where \( n = 12 \) (total number of players) and \( k = 5 \) (number of players to choose).

Step2: Calculate factorials

First, calculate the factorials:

• \( 12! = 12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1 \)
• \( 5! = 5\times4\times3\times2\times1 \)
• \( (12 - 5)! = 7! = 7\times6\times5\times4\times3\times2\times1 \)

Step3: Substitute into the combination formula

Substitute \( n = 12 \) and \( k = 5 \) into the formula:
\[

$$\begin{align*} C(12, 5)&=\frac{12!}{5!(12 - 5)!}\\ &=\frac{12!}{5!7!}\\ &=\frac{12\times11\times10\times9\times8\times7!}{5\times4\times3\times2\times1\times7!}\\ &=\frac{12\times11\times10\times9\times8}{5\times4\times3\times2\times1} \end{align*}$$

\]

Step4: Simplify the expression

Simplify the numerator and denominator:

• Numerator: \( 12\times11\times10\times9\times8 = 95040 \)
• Denominator: \( 5\times4\times3\times2\times1 = 120 \)
• Then, \( \frac{95040}{120}=792 \)

Answer:

The number of ways to select the starting lineup is \(\boldsymbol{792}\).