Fundamentals of Business Mathematics and Statistics · Permutation and Combinations

Combinations and the Formula nCr: Meaning, Properties and Solved Questions

Updated 10 October 2026 · Fact-checked

A combination is a selection of r objects from n distinct objects where order does not matter. Its count is nCr = n! ÷ [r! × (n − r)!]. To solve a question, decide if order matters, apply nCr, and multiply the counts of independent selections. Use nCr = nC(n−r) to shorten the work.

Understand Combinations and the Formula nCr

A combination is a way of choosing objects when the order of choosing does not matter. Picking a team of 3 from 10 students is a combination. The team {Asha, Ravi, Meena} is the same team however you list the names.

A permutation is an arrangement where order matters. Choosing a captain, vice-captain and secretary from 10 students is a permutation, because the same three people in different posts give different outcomes.

The link is simple. Arrange r objects chosen from n in nPr ways. Each selection of r objects can be arranged in r! ways. So nPr = nCr × r!, which gives nCr = nPr ÷ r! = n! ÷ [r! × (n − r)!]. Combinations are fewer than permutations because you divide out the r! orderings.

Choosing r objects to take is the same as choosing n − r objects to leave behind. That is why nCr = nC(n − r). Use this to make the calculation smaller, for example 50C48 = 50C2.

In this topic the objects are distinct and each is used at most once, unless the question says otherwise.

Key formulas to remember

Combination formula: nCr = n! ÷ [r! × (n − r)!]; Valid for whole numbers with 0 ≤ r ≤ n.
Link with permutation: nCr = nPr ÷ r!, so nPr = nCr × r!; Use when a question gives one and asks for the other.
Complementary property: nCr = nC(n − r); Choosing r to take equals choosing n − r to leave.
Special values: nC0 = 1, nCn = 1, nC1 = n; Exactly one way to choose none or all.
If nCx = nCy: Either x = y or x + y = n; Use to solve equations in r or n.
Pascal's rule: nCr + nC(r − 1) = (n + 1)Cr; Handy for adding two consecutive combinations.
Ratio of consecutive terms: nCr = [(n − r + 1) ÷ r] × nC(r − 1); Useful for ratio-based equations.
Fast expansion: nCr = [n × (n − 1) × … (r factors)] ÷ r!; Write r factors from n downwards on top, r! below.

How to solve Combinations and the Formula nCr questions

Follow this routine for any combination question. It keeps you from mixing up selection and arrangement.

1Read the question and decide whether order matters. Words like select, choose, committee, team, group point to combinations.
2Identify n (total distinct objects) and r (number to choose).
3Note any conditions: some objects must be included, excluded, or chosen from separate groups.
4Handle conditions by reducing n and r. If k objects must be included, choose r − k from n − k. If k are excluded, choose r from n − k.
5For separate groups, find nCr for each group and multiply. For 'at least' or 'at most' cases, add the separate cases or subtract the unwanted case from the total.
6Compute using the fast expansion, switching to nC(n − r) if r is large.
7Check that the answer is a positive whole number and that it matches one option.

Quickest way: Shortcut: small r and cancel early

When to use it: Use for any numerical nCr in an MCQ where you must compute quickly without a calculator.

If r is more than half of n, replace r with n − r.
Write r factors starting from n going down on top, and r! on the bottom.
Cancel common factors before multiplying.
Example: 10C3 = (10 × 9 × 8) ÷ (3 × 2 × 1) = 120.
For 'at least one' questions, use total − none, as it is often faster than adding cases.
Test options: the answer must be a whole number, which often removes two choices.

Common mistakes in Combinations and the Formula nCr

Using permutation when order does not matter
Students see numbers and apply nPr by habit.
Fix: Ask: if I swap two chosen items, is the outcome the same? If yes, use nCr.
Forgetting to divide by r!
Confusing nCr with nPr.
Fix: Remember nCr = nPr ÷ r!. Always keep r! in the denominator.
Computing with large r instead of n − r
Students do not use the property nCr = nC(n − r).
Fix: For 12C10, compute 12C2 = 66. It is shorter and less error-prone.
Adding instead of multiplying for separate groups
Mixing up the 'and' and 'or' rules.
Fix: Choosing from group A and group B together means multiply. Alternative cases (either this or that) mean add.
Wrong handling of 'must include' conditions
Students reduce n but forget to reduce r.
Fix: If k given objects are included, use (n − k)C(r − k). Reduce both.
Solving nCx = nCy and keeping only x = y
The second possibility x + y = n is forgotten.
Fix: Check both x = y and x + y = n, then reject any answer that makes the combination undefined.

Worked examples

Example 1

A company has 8 managers and 5 engineers. A committee of 4 is to be formed with exactly 2 managers and 2 engineers. In how many ways can this be done?

Show the solution

Order does not matter, so use combinations.
Choose 2 managers from 8: 8C2 = (8 × 7) ÷ 2 = 28.
Choose 2 engineers from 5: 5C2 = (5 × 4) ÷ 2 = 10.
Both choices are made together, so multiply: 28 × 10 = 280.

Answer: 280 ways

Example 2

If nC8 = nC12, find the value of 22Cn.

Show the solution

Since nC8 = nC12 and 8 ≠ 12, we must have n = 8 + 12 = 20.
Now find 22C20.
Use the property: 22C20 = 22C2.
22C2 = (22 × 21) ÷ 2 = 231.

Answer: 231

Exam tips

Spend the first few seconds deciding order matters or not. This one step decides the whole question.
Always reduce nCr using nC(n − r) before multiplying. It saves time on a one-hour paper.
For 'at least' questions, try total minus the opposite case first.
Check the options after calculating. A non-whole number means you made an error.
There is no negative marking, so attempt every question. If unsure, eliminate options that are not whole numbers or that look like permutation answers.

Practice questions from Permutation and Combinations

Combinations and the Formula nCr in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

CA FoundationCombinations (nCr) and Its Properties

Combinations and the Formula nCr: frequently asked questions

What is the difference between permutation and combination?

A permutation counts arrangements where order matters. A combination counts selections where order does not matter. They are linked by nPr = nCr × r!.

Why is nCr equal to nC(n − r)?

Choosing r objects to include automatically decides the n − r objects left out. Each selection of r matches exactly one selection of n − r, so the counts are equal. You can also see it from the formula, as the denominator r! × (n − r)! is the same.

What is the value of nC0 and nCn?

Both equal 1. There is exactly one way to choose no objects and exactly one way to choose all n objects.

How do I solve combination problems with conditions?

Convert each condition into a smaller selection. Included objects reduce both n and r. Excluded objects reduce n only. For separate groups, multiply the counts of each group.