Use factorial notation and the zero factorial
Evaluate factorials and simplify ratios by cancelling common factors, including the special value 0! = 1.
Section 11 · Permutations and combinations
Cambridge IGCSE Additional Mathematics 0606, 2028–2030 syllabus, Version 1. Paper 1 is non-calculator; Paper 2 requires a scientific calculator. These original lessons provide self-checked topic practice, not a Cambridge grade or a timed paper. Full-paper assessment and human specialist review are still outstanding.
Before you begin
- Cambridge assumes knowledge of IGCSE Mathematics or an equivalent course. Check the domain before using an inverse or logarithm, and retain exact values where the question asks for them.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which product is equal to 4!?
What you will learn
- Interpret n! as the product of the positive integers from n down to 1
- Use 0! = 1
- Simplify factorial ratios before evaluating them
A factorial multiplies every positive integer down to one
For a positive integer n, n! = n(n − 1)(n − 2)…2 × 1. Thus 5! = 5 × 4 × 3 × 2 × 1 = 120. The exclamation mark is a notation for this product, not punctuation.
By definition, 0! = 1. This value keeps the permutation and combination formulas valid at their boundary cases and represents the one way to arrange an empty set of objects.
Cancel matching factorial factors before multiplying
In 8!/6!, write 8! = 8 × 7 × 6!. The 6! factors cancel, leaving 8 × 7 = 56; evaluating both large factorials first is unnecessary.
For a quotient such as 8!/(6!2!), cancel 6! first, then divide 8 × 7 by 2 × 1. Keep brackets around every factorial so that the denominator is interpreted correctly.
Pause and explain
What is 0!?
Worked example
Evaluate 7!/5! and 0!.
Show the worked solution
- Write 7! = 7 × 6 × 5!.
- Cancel the common 5! to leave 7 × 6 = 42.
- Use the defined boundary value 0! = 1.
Answer 7!/5! = 42 and 0! = 1.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Simplify 9!/7!.
Give me a hint
Expand only until the denominator's 7! appears.
Compare my reasoning
- Write 9! = 9 × 8 × 7!.
- Cancel 7! from numerator and denominator.
- Multiply 9 × 8.
9!/7! = 72.
Look for these in your work
- Factorial expanded correctly
- Common factor cancelled
- Product evaluated
Evaluate 8!/(6!2!).
Give me a hint
Cancel 6! before multiplying the remaining factors.
Compare my reasoning
- Write 8! = 8 × 7 × 6!.
- Cancel 6!; the expression becomes (8 × 7)/(2 × 1).
- Compute 56/2.
28.
Look for these in your work
- Factorial cancellation valid
- 2! retained in the denominator
- Final quotient correct
For integer n ≥ 2, simplify n!/(n − 2)! and explain which factors remain.
Give me a hint
Expand the numerator down to (n − 2)!.
Compare my reasoning
- Write n! = n(n − 1)(n − 2)!.
- Cancel the common (n − 2)!.
- The remaining product is n(n − 1).
n!/(n − 2)! = n(n − 1), for integer n ≥ 2.
Look for these in your work
- Only required factors expanded
- Common factorial cancelled
- Boundary condition stated
Common mistakes
- Treating 0! as zero
- Writing n! as n multiplied by itself n times
- Expanding huge factorials before cancelling
- Dropping a factorial bracket in a denominator
What can you explain now?
What is 1!?
Compare with the explanation
1
The product from 1 down to 1 contains only the factor 1.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
What does n! mean, what is 0!, and why should you cancel before calculating?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.