In how many ways can 2 students be selected from a group of 5 students in a debating competition?

Analysis

The question asks for the number of ways to select 2 students out of a group of 5. In this scenario, the order in which the students are selected does not matter (selecting Student A then Student B is the same as selecting Student B then Student A). Therefore, we use the concept of Combinations.

Concept

When selecting $r$ items from a total of $n$ items where order does not matter, we use the combination formula:

Where:

• $n$ = total number of items (5 students)
• $r$ = number of items to be selected (2 students)
• $!$ = factorial (e.g., $3! = 3 \times 2 \times 1 = 6$)

Solution

1. Identify the values: $n = 5$ $r = 2$

2. Apply the formula: $$^5C_2 = \frac{5!}{2!(5 - 2)!}$$

3. Simplify the expression: $$^5C_2 = \frac{5!}{2! \times 3!}$$

4. Expand the factorials: $$^5C_2 = \frac{5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1) \times (3 \times 2 \times 1)}$$

5. Cancel out common terms: $$^5C_2 = \frac{5 \times 4}{2 \times 1}$$ $$^5C_2 = \frac{20}{2}$$ $$^5C_2 = 10$$

There are 10 different ways to select the students.

Exam Tip

Whenever you see the word "selected," "chosen," or "committee," think Combination ($^nC_r$). If the question implies a specific order or position (like "President" and "Secretary"), then you should use Permutation ($^nP_r$) instead.