Permutation and Combination Explained with Examples
Permutation and Combination deals with arranging and selecting objects. Permutations count ordered arrangements, while combinations count selections where order does not matter. This chapter covers factorials, the fundamental counting principle, nPr and nCr formulas, arrangements with repeated objects, circular permutations, and standard aptitude methods for solving counting questions.
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What are Permutations and Combinations?
For n distinct objects taken r at a time, the number of permutations is nPr = n!/(n-r)!, and the number of combinations is nCr = n!/[r!(n-r)!]. For example, selecting captain and vice-captain from 5 students gives 5P2 = 20 outcomes because the roles differ. Selecting 2 students for a team gives 5C2 = 10 outcomes because the order is irrelevant.
Permutation and Combination Video Lessons
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Permutation and Combination Selection Problems
This lesson explains how to approach selection-based permutation and combination problems, identify whether order matters, and apply the appropriate counting method.
Permutation and Combination Formula & Tricks
Important Formulas
A factorial is the product of all positive integers up to n. The special value 0! = 1 is used in counting formulas.
If a process has successive independent stages with m₁, m₂, ..., mₖ choices, multiply the choices at each stage.
Use this when r objects are arranged from n distinct objects and changing the order creates a different outcome.
Use this when r objects are selected from n distinct objects and order does not create a new outcome.
First select r objects using nCr and then arrange those selected objects in r! ways.
If n objects contain repeated groups of p, q, r, and so on identical objects, divide by the factorial of each repetition count.
Quick Tricks
Use permutation when positions, ranks, roles or codes are different. Use combination when only the selected group matters.
For questions involving at least one occurrence, count all outcomes and subtract outcomes with none.
Arrange the objects that must be separated first, then place the remaining objects in the available gaps.
For n distinct objects arranged around a circle, rotations are considered identical, so the count is (n − 1)!. If clockwise and anticlockwise arrangements are also identical, divide by 2 for n greater than 2.
Permutation and Combination Concepts
Factorials and the Fundamental Counting Principle
For n distinct objects, the number of linear arrangements is n!. If the first position has n choices, the second has n − 1 choices, and so on, the product is n × (n − 1) × ... × 1 = n!. For a process with separate stages, multiply the number of choices at each stage.
Linear Permutations
The number of ways to arrange r objects chosen from n distinct objects is nPr = n!/(n − r)!. If all n objects are used, the formula becomes n!. When some objects are identical, divide by the factorials of their repetition counts.
Combinations and Selection
The formula is nCr = n!/[r!(n − r)!]. It satisfies nCr = nC(n − r), because selecting r objects is equivalent to leaving out n − r objects. Also, nC0 = nCn = 1.
Arrangements with Repetition and Restrictions
For n objects with repeated groups of p, q and r identical objects, the number of arrangements is n!/(p!q!r!). If specified objects must stay together, treat the complete group as one block and arrange the block with the remaining objects. If objects must not be together, use total arrangements minus arrangements in which they are together.
Circular Permutations
Fixing one object removes the effect of rotation, and the remaining n − 1 objects can then be arranged in (n − 1)! ways. For arrangements where clockwise and anticlockwise orders are identical, the count is (n − 1)!/2 for n greater than 2.
Permutation and Combination Data Table
Use the structured data below while following the explanation and examples.
Permutation and Combination Formula Comparison
Use the formula that matches whether order changes the outcome.
| Counting situation | Formula | Example for n = 5, r = 2 |
|---|---|---|
| Ordered arrangement | nPr = n!/(n − r)! | 5P2 = 5 × 4 = 20 |
| Unordered selection | nCr = n!/[r!(n − r)!] | 5C2 = (5 × 4)/2 = 10 |
| Relation between the two | nPr = nCr × r! | 5P2 = 5C2 × 2! = 20 |
| Circular arrangement of n objects | (n − 1)! | For 5 objects: 4! = 24 |
Note: The circular formula applies to distinct objects when rotations are considered identical.
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Permutation and Combination Revision Points
Key formulas and rules for quick recall.
- 0! = 1 and n! = n × (n − 1) × ... × 1.
- Use nPr when order matters: nPr = n!/(n − r)!.
- Use nCr when order does not matter: nCr = n!/[r!(n − r)!].
- nPr = nCr × r! and nCr = nC(n − r).
- Arrangements with repeated objects: n!/(p!q!r!...).
- Circular arrangements of n distinct objects: (n − 1)! when rotations are identical.
- For identical clockwise and anticlockwise circular arrangements, use (n − 1)!/2 for n > 2.
- For restrictions, use blocks, gap methods, or complementary counting according to the condition.
Permutation and Combination FAQs
What is the difference between nPr and nCr?
nPr counts ordered arrangements, while nCr counts unordered selections. For example, 5P2 = 20 but 5C2 = 10.
How is nCr related to nPr?
The relation is nPr = nCr × r!. After selecting r objects, they can be arranged in r! orders.
How many ways can 4 objects be selected from 9 objects?
Since order does not matter, use 9C4 = 9!/(4!5!) = 126.
How many 3-letter arrangements can be made from 6 distinct letters?
Order matters, so use 6P3 = 6 × 5 × 4 = 120.
How many arrangements can be made from the letters of APPLE?
APPLE has 5 letters, with P repeated twice. The number of arrangements is 5!/2! = 60.
How many ways can 6 people sit around a round table?
Rotations are identical, so the number is (6 − 1)! = 5! = 120.