Fundamentals of Business Mathematics and Statistics · Permutation and Combinations

Committee Selection and Grouping Using Combinations

Updated 10 October 2026 · Fact-checked

Use combinations when you only choose people or objects and the order does not matter. Split the condition into cases, find nCr for each case, multiply the choices within a case, and add the cases. For "at least" questions, subtract the unwanted cases from the total. For groups, divide by the factorials of group sizes.

Understand Applications of Combinations: Selection and Grouping

A combination is a selection where order does not matter. Picking Asha, Ravi and Meena for a committee is the same committee as Meena, Asha and Ravi. So you use nCr, not nPr.

Most questions add a condition. Examples: "at least 2 women", "a particular person must be included", or "two particular people cannot both be on the committee". The idea is always the same. Turn the condition into separate cases that do not overlap, count each case, and add them. For the last kind of condition, it is often quicker to take the total and subtract the cases where both people are included.

Within one case, you pick from different pools, such as men and women. Picking from one pool AND another pool means you multiply. Choosing between different cases (OR) means you add. This AND-multiply, OR-add rule is the backbone of every problem here.

Grouping questions ask you to divide objects into groups. First check whether the groups are labelled (for example, given to named people) or unlabelled (just groups). Labelled groups use n! ÷ (a! b! c!). If some groups are the same size and unlabelled, you also divide by the factorial of the number of such equal groups, because swapping them gives the same division.

Key formulas to remember

Combination formula: nCr = n! ÷ [r! × (n − r)!]; Valid for whole numbers with 0 ≤ r ≤ n. Order does not matter.
Symmetry rule: nCr = nC(n − r); Use it to shorten work. For example, 10C8 = 10C2 = 45.
Special values: nC0 = nCn = 1 and nC1 = n; Choosing none or all can be done in exactly one way.
Selections of any number from n distinct items: nC1 + nC2 + ... + nCn = 2ⁿ − 1; This counts selecting at least one item. Each item is either in or out (2 choices), giving 2ⁿ, and you remove the empty selection.
Particular items included or excluded: Particular person included: (n − 1)C(r − 1). Particular person excluded: (n − 1)Cr; For k particular persons always included: (n − k)C(r − k).
Dividing into labelled groups: n! ÷ (a! × b! × c!), where a + b + c = n; Use when the groups are different in identity, such as given to different people.
Dividing into equal unlabelled groups: n! ÷ [(m!)ᵏ × k!], where n = m × k; k groups of m items each, with groups not named. Divide by k! to remove repeated divisions.

How to solve Applications of Combinations: Selection and Grouping questions

This method works for committee, team, selection and grouping questions.

1Read the question and confirm order does not matter. If it does, this is a permutation problem.
2List the pools: for example, 6 men and 5 women, or 4 teachers and 7 students.
3Write the condition in plain words: at least, at most, exactly, always included, never included.
4Split the condition into non-overlapping cases. For "at least 2 women" in a team of 4, the cases are 2, 3 and 4 women.
5For each case, write nCr for every pool and multiply them.
6Add the case totals. If the cases are many, try the complement: total ways minus the unwanted ways.
7For grouping, decide whether the groups are labelled, then apply the division formula and divide by k! for equal unlabelled groups.
8Check that the answer is a sensible whole number and fits one of the four options.

Quickest way: Complement and include-exclude shortcut

When to use it: Use it when the condition says "at least one", "not all", or fixes particular persons, and the case list would be long.

For "at least" conditions, compute total selections minus the selections that break the condition. Often only one or two cases need subtracting.
If a person must be in the team, remove that person and choose the remaining places from the remaining people: (n − 1)C(r − 1).
If a person must be out, remove that person and choose from the rest: (n − 1)Cr.
Use nCr = nC(n − r) so you always compute the smaller r.
Compare your result with the four options. Eliminate those that are larger than the unrestricted total nCr, as they cannot be right.

Common mistakes in Applications of Combinations: Selection and Grouping

Adding the pool selections within one case instead of multiplying.
Students see "men and women" and add the two counts.
Fix: Choosing from one pool AND from another pool means multiply. Add only across different cases.
Treating "at least 2" as just "exactly 2".
The words are read quickly and the higher cases are forgotten.
Fix: List every case from the minimum up to the largest possible, or subtract the cases below the minimum from the total.
Using permutations (nPr) for a committee.
Students remember nPr more strongly and apply it without checking order.
Fix: Ask yourself: does swapping two members make a new outcome? If not, use nCr.
Forgetting to divide by k! for equal unlabelled groups.
Students apply n! ÷ (m!)ᵏ and stop there.
Fix: If the groups are the same size and not named, also divide by k!. If groups are named or given to different people, do not divide.
Wrong handling of "a particular person is included".
Students reduce n but forget to reduce r, or the reverse.
Fix: Included: reduce both n and r by 1. Excluded: reduce only n by 1.
Counting the empty selection in "at least one" questions.
Students use 2ⁿ directly.
Fix: Use 2ⁿ − 1 whenever at least one item must be chosen.

Worked examples

Example 1

A committee of 4 is to be formed from 6 men and 5 women so that it has at least 2 women. In how many ways can this be done?

Show the solution

Cases: 2 women and 2 men, 3 women and 1 man, 4 women and 0 men.
2 women, 2 men: 5C2 × 6C2 = 10 × 15 = 150.
3 women, 1 man: 5C3 × 6C1 = 10 × 6 = 60.
4 women, 0 men: 5C4 × 6C0 = 5 × 1 = 5.
Add: 150 + 60 + 5 = 215.
Check by complement: total 11C4 = 330. Zero women: 6C4 = 15. One woman: 5C1 × 6C3 = 5 × 20 = 100. So 330 − 15 − 100 = 215.

Answer: 215 ways

Example 2

In how many ways can 9 different books be divided equally into three unnamed piles of 3 books each?

Show the solution

The piles are not labelled, so the division must not count the same piles in different orders.
First treat the piles as labelled: 9! ÷ (3! × 3! × 3!) = 362880 ÷ 216 = 1680.
Three piles are of equal size, so divide by 3! = 6 to remove repeated arrangements of the piles.
1680 ÷ 6 = 280.

Answer: 280 ways (if the piles were given to three named students, the answer would be 1680)

Exam tips

Read for the key word first: at least, at most, exactly, always, never. It decides your method.
If the number of cases is three or more, try the complement and see which route is shorter.
Check for groups of equal size. That is where the extra division by k! hides, and the wrong option usually matches the missing division.
Use nCr = nC(n − r) to keep numbers small and save time on the OMR paper.
There is no negative marking, so if time runs out, still mark an option after eliminating any answer larger than the unrestricted total.

Practice questions from Permutation and Combinations

Applications of Combinations: Selection and Grouping in other exams

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

Applications of Combinations: Selection and Grouping: frequently asked questions

How do I form a committee using combinations?

Identify the pools, such as men and women, and the condition on the committee. Write nCr for each pool and multiply within a case. Then add all valid cases.

How do I solve "at least one" selection problems?

Count all possible selections and subtract the selections that have none. If you can pick any number from n distinct items, the count of at least one is 2ⁿ − 1.

When do I divide by k! in grouping problems?

Divide by k! when k groups have the same size and are not labelled. If the groups are named or go to different people, do not divide.

How is a combination different from a permutation in these problems?

A combination only chooses items and ignores order. A permutation also arranges them. Committees, teams and selections use combinations.