Quantitative Aptitude · Permutations and Combinations

Factorial Notation for CA Foundation Quantitative Aptitude

Updated 1 October 2026 · Fact-checked

Factorial notation n! means the product of all positive integers from n down to 1, defined for whole numbers n ≥ 0, with 0! = 1. To simplify, use n! = n × (n − 1)!, expand the larger factorial only until the smaller one appears, then cancel the common factorial and compute what is left.

Understand Factorial Notation

A factorial is a short way to write a long multiplication. For a whole number n, n! (read "n factorial") is the product of every positive integer from n down to 1. So 5! = 5 × 4 × 3 × 2 × 1 = 120.

Factorials grow very fast. 5! is 120, 8! is 40,320 and 10! is 36,28,800. This is why they appear in counting: the number of ways to arrange n different objects in a row is n!. Permutations and combinations are built on this idea.

The key property is that a factorial contains a smaller factorial inside it. 6! = 6 × 5!. In general, n! = n × (n − 1)!. This one line lets you cancel, expand and solve almost every question.

Why is 0! = 1? Take the property n! = n × (n − 1)! and put n = 1. You get 1! = 1 × 0!. Since 1! = 1, we need 0! = 1. It also matches counting: there is exactly one way to arrange zero objects, which is to do nothing. So 0! = 1 is a definition that keeps every formula consistent. It is not an accident.

Factorials are defined only for whole numbers (0, 1, 2, 3, ...). At Foundation level, negative numbers and fractions do not have a factorial. This matters when you solve equations in n and get a negative answer.

Key formulas to remember

Definition: n! = n × (n − 1) × (n − 2) × ... × 3 × 2 × 1; The product formula is for whole numbers n ≥ 1. 0! = 1 is defined separately, so factorial is defined for all whole numbers n ≥ 0.
Zero factorial: 0! = 1; A definition that keeps n! = n × (n − 1)! true for n = 1. Also 1! = 1.
Recursive property: n! = n × (n − 1)! = n × (n − 1) × (n − 2)!; Use it to expand a large factorial down to a smaller one.
Ratio of factorials: n! ÷ r! = n × (n − 1) × ... × (r + 1), for whole numbers n > r; Only n − r factors remain after cancelling.
Reverse step: (n − 1)! = n! ÷ n; Useful for getting 0! from 1! or checking small values.
Values to memorise: 1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720, 7! = 5,040, 8! = 40,320; Knowing these saves time and lets you match options quickly.

How to solve Factorial Notation questions

Use this method for any factorial question, whether it asks you to simplify, find n or find a missing value.

1Write down which factorials appear and pick the smallest one in the expression.
2Expand every larger factorial only until the smallest factorial appears, using n! = n × (n − 1)!.
3Take the common factorial out as a factor, or cancel it between numerator and denominator.
4Simplify what remains using ordinary arithmetic.
5If the question has an unknown n, form an equation in n (often a quadratic) and solve it.
6Reject any value of n that is negative, a fraction, or makes a factorial in the question undefined.
7Substitute your answer back into the original expression to check it.

Quickest way: Cancel to the smaller factorial and test options

When to use it: Use this in the exam for simplification and 'find n' MCQs, where speed matters and a wrong answer costs 0.25 marks.

For a ratio like n! ÷ r!, just multiply the numbers from n down to r + 1. Do not compute full factorials.
For an equation in n, expand to the smaller factorial and solve the small quadratic. If it looks heavy, substitute each option instead.
When substituting, use small options first. 3!, 4!, 5! are easy to compute.
Eliminate any option that is negative or not a whole number for n.
Remember 0! = 1 and 1! = 1. Many traps rely on these two values.
If a question seems to need factorials above 8! computed fully, look for a cancellation first; it is often available.

Common mistakes in Factorial Notation

Writing 0! = 0.
Students feel that 'nothing' multiplied gives zero.
Fix: Remember 0! = 1 by the rule 1! = 1 × 0!. Check the value every time 0! appears in a combination formula.
Treating (a + b)! as a! + b!, or (a − b)! as a! − b!.
Students assume factorial distributes over addition like a bracket.
Fix: Factorial does not distribute. Compute a + b first, then take its factorial. Example: (2 + 3)! = 120, but 2! + 3! = 8.
Writing (2n)! = 2 × n!.
Students treat the factorial as a factor that can be pulled out of the bracket, so they split (2n)! into 2 × n!.
Fix: Test with n = 3: (6)! = 720, but 2 × 3! = 12. They differ, so never split like this.
Cancelling n! ÷ m! as (n ÷ m)! or as n ÷ m.
Students cancel the factorial sign as if it were a common factor.
Fix: Expand the larger one down to the smaller. Example: 7! ÷ 5! = 7 × 6 = 42, not (7 ÷ 5)!.
Accepting a negative root of n in an equation like (n + 1)! = 20 × (n − 1)!.
The quadratic gives two roots and students keep both.
Fix: n must be a whole number for the factorials to exist. Reject negative roots and check the answer in the original equation.
Expanding too far and making arithmetic errors, such as computing 10! and 8! in full.
Students do not see that the smaller factorial cancels.
Fix: Stop expanding when the smaller factorial appears, then cancel it.

Worked examples

Example 1

The value of 10! ÷ (8! × 2!) is: (a) 40 (b) 45 (c) 90 (d) 180

Show the solution

The smallest large factorial in the denominator is 8!, so expand 10! down to 8!.
10! = 10 × 9 × 8!.
So 10! ÷ (8! × 2!) = (10 × 9 × 8!) ÷ (8! × 2!).
Cancel 8!: we get (10 × 9) ÷ 2!.
2! = 2, so the value is 90 ÷ 2 = 45.

Answer: (b) 45

Example 2

If (n + 1)! = 20 × (n − 1)!, then n equals: (a) 3 (b) 4 (c) 5 (d) 6

Show the solution

Expand the left side down to (n − 1)!: (n + 1)! = (n + 1) × n × (n − 1)!.
So (n + 1) × n × (n − 1)! = 20 × (n − 1)!.
Cancel (n − 1)! from both sides: n(n + 1) = 20.
This gives n² + n − 20 = 0, which factorises as (n + 5)(n − 4) = 0.
So n = 4 or n = −5. A factorial needs a whole number, so reject −5.
Check: 5! = 120 and 20 × 3! = 20 × 6 = 120. This is correct.

Answer: (b) 4

Example 3

If 1/6! + 1/7! = x/8!, then x equals: (a) 56 (b) 63 (c) 64 (d) 72

Show the solution

Multiply both sides by 8! to get x = 8!/6! + 8!/7!.
8!/6! = 8 × 7 = 56.
8!/7! = 8.
So x = 56 + 8 = 64.
Check: 1/720 + 1/5040 = 7/5040 + 1/5040 = 8/5040 = 1/630. Also 64/40320 = 1/630. This matches.

Answer: (c) 64

Exam tips

Most questions are either 'simplify' or 'find n'. In both, the first move is to expand the larger factorial down to the smaller one and cancel.
Learn factorials up to 8! by heart. Options are often built from these values, so you can match an answer without full working.
If a 'find n' question gives numeric options, substituting small values like 3, 4, 5 is often faster than solving the quadratic.
Watch for traps with 0! and 1!. A question that looks hard may collapse once you write both as 1.
If a question needs a full factorial above 8! and no cancellation is visible, re-read it. Skip it if it still looks long, since a wrong answer costs 0.25 marks.

Practice questions from Permutations and Combinations

Factorial Notation: frequently asked questions

Why is 0 factorial equal to 1?

Because n! = n × (n − 1)! must hold for n = 1, which gives 1! = 1 × 0!. Since 1! = 1, 0! has to be 1. It also fits counting: there is exactly one way to arrange zero objects.

Can you find the factorial of a negative number or a fraction in CA Foundation?

No. At this level, factorial is defined only for whole numbers 0, 1, 2, 3 and so on. If an equation gives a negative or fractional n, reject that value.

How do I simplify factorial expressions quickly?

Find the smallest factorial in the expression. Expand the larger ones down to it using n! = n × (n − 1)!, cancel the common factorial, and multiply the few numbers left.

Is (a + b)! the same as a! + b!?

No. Factorial does not distribute over addition. For a = 2 and b = 3, (a + b)! = 5! = 120, while a! + b! = 2 + 6 = 8.