Question

1. a committee of 3 students is to be chosen from 10 students. how many possible committees are there?

options: 120, 60, 240, 720

Explanation:

Step1: Identify the problem type

This is a combination problem since the order of choosing students for the committee does not matter. The formula for combinations is \( C(n, k)=\frac{n!}{k!(n - k)!} \), where \( n = 10 \) (total number of students) and \( k=3 \) (number of students to be chosen).

Step2: Calculate factorial values

First, calculate the factorials:

• \( n!=10! = 10\times9\times8\times7\times6\times5\times4\times3\times2\times1 \)
• \( k!=3! = 3\times2\times1=6 \)
• \( (n - k)!=(10 - 3)!=7! = 7\times6\times5\times4\times3\times2\times1 \)

Step3: Substitute into the combination formula

Substitute \( n = 10 \) and \( k = 3 \) into the combination formula:
\[

$$\begin{align*} C(10, 3)&=\frac{10!}{3!(10 - 3)!}\\ &=\frac{10!}{3!7!}\\ &=\frac{10\times9\times8\times7!}{3\times2\times1\times7!}\\ &=\frac{10\times9\times8}{3\times2\times1}\\ &=\frac{720}{6}\\ & = 120 \end{align*}$$

\]

Answer:

120