Selection Problems in Permutation and Combination Explained
Selection Problems in Permutation and Combination deal with choosing objects or people when order is irrelevant. These questions commonly include conditions such as selecting at least one member, choosing a fixed number from groups, including or excluding particular objects, and selecting items together or separately. Combinations, complementary counting and case-based selection methods solve these problems.
On this page
What Are Selection Problems in Permutation and Combination?
If a committee of 3 is selected from 8 people, the number of selections is 8C3 = 8!/(3!5!) = 56. When a question imposes conditions, first count the valid choices directly or count all choices and subtract the invalid ones. If the selected objects must then be arranged, multiply the selection count by the required permutation count.
Selection Problems in Permutation and Combination Video Lessons
Watch short topic-wise lessons for quick revision.
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.
Selection Problems in Permutation and Combination Formula & Tricks
Important Formulas
Use this when r objects are selected from n distinct objects and order does not matter.
Use this when the selected objects are arranged in an order after or during selection.
Use this when the restriction is easier to handle by counting selections that violate it.
Each object has two choices, selected or not selected; subtract the empty selection.
Use this when a fixed number must be selected independently from each group.
Quick Tricks
If two or more objects must always be selected together, select the complete group as one unit and choose the remaining objects normally.
Count all selections and subtract the cases that fail the condition. This is often shorter than adding several separate cases.
When the question specifies selections from different groups, multiply the combination counts for the groups.
Selection Problems in Permutation and Combination Concepts
Selection Versus Arrangement
Selecting 3 members from 7 people gives 7C3 = 35. Arranging 3 of those 7 people in three distinct posts gives 7P3 = 7C3 × 3! = 210. The same selected group can produce 3! different arrangements when all positions are distinct.
Selection with Compulsory or Excluded Objects
If r objects are selected from n objects and a particular object must be included, the count is n-1Cr-1. If that object must be excluded, the count is n-1Cr. These formulas assume all objects are distinct.
Selection with Together and Separate Conditions
For a committee of r from n objects, if A and B must both be included, choose the remaining r-2 objects from n-2: n-2Cr-2. If A and B cannot both be included, subtract committees containing both from the total: nCr − n-2Cr-2.
Selection from Different Groups
If r1 objects are chosen from a group of n1 and r2 from a group of n2, the number of selections is n1Cr1 × n2Cr2. If the total number is fixed but the group-wise numbers can vary, add the valid cases for each possible distribution.
At Least, At Most and Exactly Conditions
Selecting at least k objects means selecting k, k+1, ..., up to the maximum allowed. Selecting at most k means adding the cases from 0 through k. For a simpler complement, at least one of a specified type equals total selections minus selections containing none of that type.
Selection Problems in Permutation and Combination Data Table
Use the structured data below while following the explanation and examples.
Common Selection Conditions
Formulas for selecting r objects from n distinct objects under common restrictions.
| Condition | Number of selections |
|---|---|
| No restriction | nCr |
| A specified object included | n-1Cr-1 |
| A specified object excluded | n-1Cr |
| Two specified objects both included | n-2Cr-2 |
| Two specified objects not both included | nCr − n-2Cr-2 |
| At least one of two specified objects included | nCr − n-2Cr |
Note: The formulas apply when the specified objects are distinct and 0 ≤ r ≤ n.
Learn more concepts on PrepShots
Start ₹1 Trial →Practice Selection Problems in Permutation and Combination Questions
Practise published questions related to this topic.
Want more practice questions?
Practice in App →Selection Problems in Permutation and Combination Quick Quiz
Attempt 5 questions and check your score instantly.
Keep practising
Practice more Selection Problems in Permutation and Combination questions in the PrepShots app and continue from your current topic.
Practice More Questions - Start ₹1 Trial →Quick Revision Notes
Selection Problems: Revision Points
Use these rules to revise restricted and unrestricted selection questions.
- Use nCr when order does not matter and nPr when order matters.
- For a compulsory object, reduce both n and r by 1: n-1Cr-1.
- For an excluded object, reduce only n by 1: n-1Cr.
- For objects selected together, treat the group as included and choose the remaining objects.
- For objects that cannot be selected together, subtract selections containing both from the total.
- For separate groups, multiply the combination counts for each group.
- For at least or at most conditions, add valid cases or use complementary counting.
- After selection, multiply by r! if all r selected objects are assigned to distinct positions.
Selection Problems in Permutation and Combination FAQs
What is the formula for selecting r objects from n objects?
The formula is nCr = n!/[r!(n-r)!]. It applies when the order of the selected objects does not matter.
How many ways can 3 people be selected from 7 people if one particular person must be included?
Include that person first, then select 2 from the remaining 6: 6C2 = 15 ways.
How many 4-member committees can be formed from 9 people if two particular people cannot both be selected?
Total committees are 9C4 = 126. Invalid committees containing both are 7C2 = 21, so the answer is 126 − 21 = 105.
How do you select at least one object from n distinct objects?
Use 2^n − 1. There are 2^n total subsets, and one of them is the empty selection.
What is the difference between nCr and nPr in selection questions?
nCr counts selections where order is irrelevant, while nPr counts ordered selections or arrangements. They are related by nPr = nCr × r!.
How many ways can 2 men be selected from 4 men and 3 women from 5 women?
Select independently from each group: 4C2 × 5C3 = 6 × 10 = 60 ways.