A class has 40 students. The teacher needs 2 students to solve problems number 2 and number 5 from the exercise book. How many ways can the teacher call the students to solve the exercise? This is a combination problem where we need to choose 2 students from a group of 40.

1. Problem Description

A class has 40 students. The teacher needs 2 students to solve problems number 2 and number 5 from the exercise book. How many ways can the teacher call the students to solve the exercise? This is a combination problem where we need to choose 2 students from a group of

4

0.

2. Solution Steps

We are asked to find the number of ways to choose 2 students out of

4

0. This is a combination problem since the order in which the students are chosen does not matter.

The formula for combinations is:

$C(n, k) = \frac{n!}{k!(n-k)!}$

where $n$ is the total number of items, $k$ is the number of items to choose, and $!$ denotes the factorial function.

In this problem, $n = 40$ (total number of students) and $k = 2$ (number of students to be chosen).

So, we need to calculate $C(40, 2)$.

$C(40, 2) = \frac{40!}{2!(40-2)!} = \frac{40!}{2!38!} = \frac{40 \times 39 \times 38!}{2 \times 1 \times 38!} = \frac{40 \times 39}{2} = 20 \times 39 = 780$

3. Final Answer

The teacher can choose students in 780 ways.