Introduction to the Product Rule for Counting Revision Notes | GCSE AQA Higher Further Maths

1. Why Counting Gets Complicated

Imagine you are creating a PIN code using four different digits. How many possible codes exist? Listing every single one would take hours. There is a far sharper method — and it takes seconds once you know it.

The product rule for counting turns a seemingly overwhelming counting task into a short multiplication. It works whenever you are making a sequence of choices, one after another, and you want to know the total number of possible outcomes.

2. The Product Rule for Counting

The key requirement is that the stages are sequential — you make one choice, then the next, then the next. Each stage's count can depend on what was chosen before (as happens when there is no repetition), and the rule still applies.

Here is a concrete anchor. A cafe offers 3 types of sandwich and 4 types of drink. The number of different meal combinations is $3 \times 4 = 12$. You are not listing every pair — you are multiplying the number of options at each stage.

3. Applying the Rule with No Repetition

When a problem states no repetition, the number of choices at each stage decreases as items get used up. This is where students often lose marks — they forget to reduce the count after each selection.

Consider arranging 4 different letters — say A, B, C, D — into a 4-letter code with no repetition. At the first position you have 4 choices. Once one letter is placed, only 3 remain for the second position, then 2, then 1. The total is $4 \times 3 \times 2 \times 1 = 24$.

Worked example: Counting 3-digit codes from digits 2, 5, 7, 9 with no repetition

How many different 3-digit numbers can be formed using the digits 2, 5, 7, 9 with no repetition of any digit?

1. Stage 1: choose the hundreds digit Stage 2: choose the tens digit Stage 3: choose the units digit — There are 3 positions to fill, so there are 3 stages. [M1 — identifying the structure]
2. 4 digits available: 2, 5, 7, 9 Choices at stage 1 = 4 — No digit has been used yet, so all 4 are available.
3. One digit has been used. Choices at stage 2 = 3 — No repetition means one digit is gone, leaving 3 remaining choices.
4. Two digits have been used. Choices at stage 3 = 2 — Two digits are now used, leaving 2 remaining choices.
5. $4 \times 3 \times 2 = 24$ — Multiply the number of choices at each stage. [A1 — correct answer]

Answer: 24 different 3-digit numbers

4. Handling Restrictions — Deal with Them First

Many exam questions add a restriction: for example, the number must be odd, or must start with a particular digit. The golden rule is to fill the restricted position first, before counting the remaining free positions.

This is where the specification example comes in. The question asks how many 5-digit odd numbers can be formed using the digits 1, 3, 4, 6, 8 with no repetition. A number is odd when its units digit (the last digit) is odd. That restriction pins down the units digit first.

From the available digits 1, 3, 4, 6, 8 — the odd digits are 1 and 3. So there are exactly 2 choices for the units digit. Once that is placed, 4 digits remain for the other 4 positions, with no repetition allowed.

Worked example: 5-digit odd numbers from 1, 3, 4, 6, 8 with no repetition

Work out how many 5-digit odd numbers can be formed using the digits 1, 3, 4, 6, 8 with no repetition of any digit.

1. The number must be odd. A number is odd when its units digit is odd. Odd digits available from {1, 3, 4, 6, 8}: 1 and 3. — Always identify and handle the restriction first. [M1 — recognising the restricted position]
2. Units digit (position 5): 2 choices (either 1 or 3) — Only 1 and 3 are odd, so there are 2 valid choices for this restricted position.
3. 1 digit has been used for the units position. 5 - 1 = 4 digits remain. Position 1: 4 choices — No repetition means the units digit cannot be reused. 4 digits are left for the remaining positions.
4. 2 digits have now been used. Position 2: 3 choices — One more digit is placed, so 3 remain.
5. 3 digits have now been used. Position 3: 2 choices — Two more digits placed, so 2 remain.
6. 4 digits have now been used. Position 4: 1 choice — Only one digit remains — no choice here.
7. Total = (units) $\times$ (pos 1) $\times$ (pos 2) $\times$ (pos 3) $\times$ (pos 4) $= 2 \times 4 \times 3 \times 2 \times 1$ $= 2 \times 24$ $= 48$ — Multiply all stage counts together. $4 \times 3 \times 2 \times 1 = 24$, then $2 \times 24 = 48$. [A1 — correct final answer]

Answer: 48 five-digit odd numbers

Notice the order of working: restriction first, then the free positions in decreasing order of remaining choices. This structure keeps the counting clean and avoids double-counting.

5. Practise the Method

The product rule becomes reliable through repeated application. Try these problems using the same structure: identify any restriction, fill that position first, then count the remaining stages and multiply.

6. Exam Technique and Common Mistakes

Exam questions on the product rule are usually worth 2–3 marks. One mark typically goes to the correct method structure (showing the multiplication), and the remaining marks go to the correct counts at each stage and the final answer. Showing your working at every stage protects your method marks even if you make an arithmetic slip.

The product rule for counting is a compact, reliable tool: identify your stages, count the options at each stage (handling any restriction first), and multiply. Whether the problem involves digits, letters, symbols, or objects, the same structure applies every time.