In how many ways can 3 seats be occupied if 5 people are willing to sit?

Analysis

The question asks for the number of ways to arrange 3 people out of a group of 5 into 3 specific seats. Since the order in which the people sit matters (sitting in seat A is different from sitting in seat B), this is a permutation problem.

Concept

When we need to select and arrange $r$ items from a total of $n$ items, we use the permutation formula:

Where:

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

Solution

1. Identify the values: $n = 5$, $r = 3$.
2. Apply the formula: $$^5P_3 = \frac{5!}{(5 - 3)!}$$
3. Simplify the expression: $$^5P_3 = \frac{5!}{2!}$$
4. Expand the factorials: $$^5P_3 = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1}$$
5. Calculate: The $(2 \times 1)$ in the numerator and denominator cancel out: $$^5P_3 = 5 \times 4 \times 3 = 60$$

There are 60 different ways the 3 seats can be occupied.

Exam Tip

Whenever you see a problem involving "arranging" or "seating" people where order matters, always think Permutation. If the order did not matter (like choosing a committee), you would use Combination ($^nC_r$). A quick trick for $^nP_r$ is to just multiply $r$ numbers starting from $n$ downwards: for $^5P_3$, just multiply $5 \times 4 \times 3 = 60$.