Skip to content

Quantitative Aptitude · Permutations and Combinations

Applications of Combinations: Selection Problems for CA Foundation

Updated 1 October 2026 · Fact-checked

Selection problems ask you to choose items where order does not matter. Use nCr = n! ÷ (r! × (n − r)!). Split the problem into independent groups, multiply the choices for each group, and add the results of separate cases. For 'at least' conditions, use total minus the unwanted cases.

Understand Applications of Combinations: Selection Problems

A combination is a selection where order does not matter. Choosing Ravi and Sita for a committee is the same as choosing Sita and Ravi. The count of ways to choose r items from n different items is nCr.

The first question in any problem is: does order matter? If you are only picking a group, a team, a committee or a set of points, use combinations. If you are arranging, ranking or assigning positions like president and secretary, use permutations.

Most exam problems have conditions. Handle them in three ways. Treat each group separately (men and women) and multiply the choices. Treat mutually exclusive cases separately and add them. For 'at least one' conditions, subtract the unwanted case from the total.

Geometry problems are also selections. A line needs 2 points, a triangle needs 3 points, and a diagonal joins 2 vertices that are not neighbours. You choose the points, and then you correct for special cases such as collinear points.

Key formulas to remember

Combination
nCr = n! ÷ (r! × (n − r)!)
Defined for whole numbers with 0 ≤ r ≤ n.
Symmetry
nCr = nC(n − r)
Use the smaller of r and n − r to save time. If nCx = nCy, then x = y or x + y = n.
Selecting from groups
Ways = (aCp) × (bCq)
Multiply when you must choose p from group A and q from group B together.
Separate cases
Total = Case 1 + Case 2 + ...
Add only when the cases cannot happen together.
At least one
At least one = Total − None
Total ways to choose with no restriction minus ways with none of the wanted type.
Selection of any number of items
Select at least one from n different items = 2ⁿ − 1
Each item is in or out, so 2ⁿ ways, minus the case of choosing nothing.
Lines from points
nC2
For n points with no three collinear. If m points are collinear, lines = nC2 − mC2 + 1.
Triangles from points
nC3
For n points with no three collinear. If m points are collinear, triangles = nC3 − mC3.
Diagonals of a polygon
nC2 − n = n(n − 3) ÷ 2
For a convex polygon with n sides. Subtract the n sides from all pairs of vertices.

How to solve Applications of Combinations: Selection Problems questions

Use this method for any selection question. It keeps the conditions organised and avoids double counting.

  1. 1Decide whether order matters. If you only form a group, use combinations.
  2. 2List the groups and the number of items in each (for example, 6 men and 4 women).
  3. 3Write the condition clearly: exact numbers, 'at least', 'at most', or 'a particular person included or excluded'.
  4. 4If a person must be included, remove them and choose the rest from the remaining items. If excluded, remove them and choose all from the remaining.
  5. 5For exact numbers, write nCr for each group and multiply.
  6. 6For 'at least' or 'at most', either add the valid cases or use total minus the invalid cases, whichever has fewer terms.
  7. 7For geometry, find the total from nCr and subtract the invalid cases, such as collinear points or polygon sides.
  8. 8Calculate with cancellation, then check that the answer is a sensible whole number.

Quickest way: Complement and smaller-r shortcut

When to use it: Use in MCQs when the condition says 'at least one' or 'at most', or when r is large.

  1. Compute the total with no restriction first.
  2. If the condition is 'at least one of a type', compute the 'none of that type' case and subtract.
  3. Use nCr = nC(n − r) so you always expand the smaller number of factors.
  4. For polygon diagonals, go straight to n(n − 3) ÷ 2.
  5. Check the last digit or parity of your answer against the options to eliminate wrong ones before computing fully.
  6. If a problem has more than three cases, mark it and come back later. Wrong answers cost 0.25 marks.

Common mistakes in Applications of Combinations: Selection Problems

  • Using permutations when the question asks for a group.

    Students see 'ways' and start multiplying positions.

    Fix: Ask if swapping two chosen people gives a new outcome. If not, use nCr.

  • Adding group choices instead of multiplying them.

    The words 'and' and 'or' are mixed up.

    Fix: Choosing from group A and from group B together means multiply. Alternative cases mean add.

  • Computing 'at least one' by choosing one first and then the rest.

    It feels natural but it counts the same committee several times.

    Fix: Use total minus none, or add the cases 1, 2, 3 ... separately.

  • Forgetting to subtract sides when counting diagonals.

    Students use nC2 and stop.

    Fix: nC2 counts all segments between vertices, including the n sides. Subtract n.

  • Treating collinear points as if they form triangles or distinct lines.

    The standard formula is applied without reading the condition.

    Fix: For m collinear points, subtract mC3 for triangles. For lines, subtract mC2 and add 1.

  • Handling an included or excluded person wrongly.

    Students forget to reduce both n and r when a person is fixed in.

    Fix: If one person is always included, choose r − 1 from n − 1. If excluded, choose r from n − 1.

Worked examples

Example 1

A committee of 4 is to be formed from 6 men and 4 women, with at least one woman. How many committees are possible? (a) 195 (b) 209 (c) 210 (d) 225

Show the solution
  1. Total committees with no restriction = 10C4 = 210.
  2. Committees with no woman means all 4 are men = 6C4 = 15.
  3. At least one woman = 210 − 15 = 195.

Answer: (a) 195

Example 2

How many diagonals does a convex polygon with 12 sides have? (a) 54 (b) 66 (c) 60 (d) 48

Show the solution
  1. Use diagonals = n(n − 3) ÷ 2 with n = 12.
  2. = 12 × 9 ÷ 2.
  3. = 108 ÷ 2 = 54.
  4. Check: 12C2 = 66 pairs, minus 12 sides = 54.

Answer: (a) 54

Example 3

There are 10 points in a plane, of which 4 are collinear and no other three are collinear. How many triangles can be formed using these points as vertices? (a) 120 (b) 116 (c) 124 (d) 110

Show the solution
  1. Total ways to choose 3 points = 10C3 = 120.
  2. Three points from the 4 collinear points do not form a triangle: 4C3 = 4.
  3. Triangles = 120 − 4 = 116.

Answer: (b) 116

Exam tips

  • Read the condition twice. Words like 'at least', 'exactly', 'particular person' and 'always included' change the whole setup.
  • Memorise the geometry results: nC2 for lines, nC3 for triangles, n(n − 3) ÷ 2 for diagonals.
  • Option traps are common. The total without the condition is usually one of the wrong options, so do not stop early.
  • Work with cancellation. Never expand full factorials for values like 10C4.
  • Skip long multi-case problems on the first pass and return if time remains.

Practice questions from Permutations and Combinations

Applications of Combinations: Selection Problems: frequently asked questions

How do I know if a question needs combination or permutation?

Check whether order matters. Forming a committee, team or set of points is a combination. Arranging people in seats or assigning ranks and posts is a permutation.

How do I solve 'at least one man' committee questions?

Find the total number of committees without any restriction. Subtract the number of committees that contain no man. The remainder is the answer.

Why do we subtract n in the diagonal formula?

nC2 counts every segment joining two vertices. Among them, n segments are the sides of the polygon, which are not diagonals. So diagonals = nC2 − n.

What changes if some points are collinear?

Three collinear points cannot make a triangle, so subtract mC3 from nC3. For lines, the m collinear points give only one line, so use nC2 − mC2 + 1.