Year 11: How to solve a word problem using permutations in probability

In Year 11, you often see probability questions where the outcomes are arrangements of people, digits, or objects. When the order matters, we use permutations.

This article will walk you through a step-by-step strategy to solve a typical exam-style word problem where permutations are used inside a probability question.

1. Key ideas you need

1.1 Factorial notation

For a positive integer n, the factorial of n, written as n!, means:

n! = n × (n - 1) × (n - 2) × ... × 3 × 2 × 1

Examples:

3! = 3 × 2 × 1 = 6
4! = 4 × 3 × 2 × 1 = 24
5! = 5 × 4 × 3 × 2 × 1 = 120

1.2 Permutations P(n, r)

A permutation is an arrangement of objects where order matters.

The number of ways to arrange r objects chosen from n distinct objects is:

P(n, r) = n! / (n - r)!

Examples:

P(5, 3) = 5! / (5 - 3)! = 5! / 2! = (5 × 4 × 3 × 2 × 1) / (2 × 1) = 5 × 4 × 3 = 60
The number of ways to arrange 6 people in a row of 6 seats is P(6, 6) = 6! = 720

1.3 Probability and permutations

In many problems, each possible arrangement is assumed to be equally likely. Then:

Probability = (number of favourable permutations) / (number of all possible permutations)

2. Our word problem

Problem:

Six students — Ali, Ben, Chen, Divya, Emma and Faisal — are randomly seated in a row of 6 chairs. Every seating arrangement is equally likely.

How many different seating arrangements are possible?
What is the probability that Ali and Ben sit next to each other?

3. The "how to solve it" plan

We will follow a 4-step problem-solving strategy:

Understand the problem
Plan a strategy
Carry out the plan
Check and reflect

3.1 Step 1: Understand the problem

Key points from the question:

There are 6 distinct students.
They are seated in a row of 6 chairs.
Every arrangement is equally likely (this is important for probability).
We want the probability that Ali and Ben sit next to each other.

This is clearly a question where the order of students matters (Ali sitting on the far left is different from Ali sitting in the middle), so we are dealing with permutations.

3.2 Step 2: Plan a strategy

We need to:

Find the total number of possible seating arrangements using permutations.
Find the number of favourable arrangements where Ali and Ben sit next to each other. This usually involves a clever counting method (such as the "treat them as a block" idea).
Use the probability formula:

Probability = favourable arrangements / total arrangements

4. Step-by-step solution

4.1 Part (1): Total number of seating arrangements

We are seating 6 distinct students in a row of 6 chairs. The number of permutations is:

P(6, 6) = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

So there are 720 possible seating arrangements.

4.2 Part (2): Favourable arrangements (Ali and Ben sit next to each other)

(a) Treat Ali and Ben as a “block”

If Ali and Ben must sit next to each other, it is often easier to treat them as a single combined unit or “block”.

Think of the following:

The pair (Ali and Ben) counts as one unit.
The remaining students are Chen, Divya, Emma, and Faisal (4 students).

Altogether, we have:

1 “Ali–Ben block”
4 individual students

That makes 5 units to arrange in a row.

(b) Arrange the 5 units

The 5 units can be arranged in:

5! = 5 × 4 × 3 × 2 × 1 = 120 ways.

(c) Arrange Ali and Ben inside the block

Inside the block, Ali and Ben can swap places:

Ali then Ben
Ben then Ali

So there are 2! ways to arrange them in the block:

2! = 2 × 1 = 2

(d) Combine the two counts

For each of the 120 ways to arrange the 5 units, there are 2 ways to arrange Ali and Ben inside their block. So the total number of favourable arrangements is:

Favourable arrangements = 5! × 2! = 120 × 2 = 240

4.3 Find the probability

We now know:

Total arrangements = 6! = 720
Favourable arrangements = 240

So the probability that Ali and Ben sit next to each other is:

Probability = favourable / total = 240 / 720

Simplify the fraction:

240 / 720 = (divide top and bottom by 240) = 1 / 3

Therefore, the probability that Ali and Ben sit next to each other is 1/3.

5. Check and reflect

5.1 Does the answer make sense?

The probability 1/3 is between 0 and 1, so it is a valid probability.
It is not too small or too large: it seems reasonable that two students might sit next to each other roughly one-third of the time in random seating.
Our method used clear counting steps without double-counting or missing any cases.

5.2 What did we learn?

From this problem, the key techniques are:

Use factorials to count arrangements.
Recognise when the situation is a permutation (order matters).
For “next to each other” questions, use the block method: treat the pair as one unit, then multiply by the number of internal arrangements.
Use the formula Probability = favourable outcomes / total outcomes.

6. Try these on your own

Practise using the same strategy on these problems:

Seven friends are lining up to buy concert tickets. What is the probability that Sam stands in front of Tom in the line? (Hint: think about symmetry, or use permutations.)
A bike lock uses a 3-digit code where digits can be repeated from 0 to 9. What is the probability that a randomly chosen 3-digit code has all digits different? (Hint: count total 3-digit codes, then count those with all different digits using permutations.)
Five students sit in a row. What is the probability that Emma sits in the middle seat?

Each time, follow the same four steps: understand the problem, plan a strategy, carry out the plan, then check and reflect.