The product rule for counting (AQA GCSE Further Maths): Flashcards

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The product rule for counting
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Product rule for counting

Multiply the number of options at each step

Factorial n!n! equals

n×(n−1)×(n−2)×...×2×1n \times (n-1) \times (n-2) \times ... \times 2 \times 1

Value of 0!0!

11

Arrangements of 4 distinct cards in a line

4×3×2×1=244 \times 3 \times 2 \times 1 = 24

7!7! equals

5,0405,040

Arrangements of 6 volleyball players in 6 positions

6!=7206! = 720

Strategy for handling counting constraints

Handle the most restrictive conditions first

4-digit numbers from 6 digits (each used once)

6×5×4×3=3606 \times 5 \times 4 \times 3 = 360

Even 3-digit numbers from {3,1,7,9,8,5} (no repeats)

5×4×1=205 \times 4 \times 1 = 20 (last digit = 8)

Key insight about choices in product rule

Each choice is independent

When product rule applies to arrangements

When order matters

Odd numbers > 500,000 from {3,1,7,9,8,5} (no repeats)

408408

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