This page covers IBPS RRB PO Permutation & Combination with complete concept notes, 15 graded practice MCQs, key points and exam-specific tips. Free to study.
Core ConceptRead this first — the foundation of the topic
Core Concept
Permutation deals with ARRANGEMENTS where order matters. If you arrange 3 people in a line, ABC is different from BAC. Combination deals with SELECTIONS where order does not matter. If you select 3 people for a team, ABC is the same as BAC
Permutation Formula
nPr = n!/(n-r)! where n is total items, r is items to arrange
2
Combination Formula
nCr = n!/(r!(n-r)!) where n is total items, r is items to select
3
Factorial
n! = n × (n-1) × (n-2) × ... × 1, and 0! = 1
4. When all items are arranged: nPn = n!
5
Circular permutation
(n-1)! for clockwise and anticlockwise same
Formula BlockMemorise — at least one formula appears in every paper
nPr = n!/(n-r)!
nCr = n!/(r!(n-r)!)
nCr = nC(n-r)
nPr = r! × nCr
Circular arrangement = (n-1)!
Arrangement with repetition = n!/p!q!r! where p,q,r are repeated items
Exam PatternsWhat examiners ask — read before attempting PYQs
SSC CGL typically asks 2-3 questions worth 6-9 marks. Common question types include selecting teams, arranging letters of words, seating arrangements, and forming numbers from given digits.
ShortcutsUse these to save 30–60 seconds per question
Quick nCr calculation
Use nCr = nC(n-r) to reduce calculations. For 10C8, calculate 10C2 instead.
2
Word arrangement shortcut
For repeated letters, use n!/repetition factors
3
Selection with conditions
Use complement method (Total - Unwanted)
Worked ExampleSolve this step-by-step before moving on
1
Step 1
This is arrangement (permutation) as order matters
2
Step 2
All 5 people are being arranged
3
Step 3
Apply formula nPn = n!
4
Step 4
5P5 = 5! = 5 × 4 × 3 × 2 × 1 = 120 ways
Answer: 120 ways
Worked Example 2:
From 8 boys and 6 girls, in how many ways can a committee of 5 be formed with at least 2 girls?
Solution:
1
Step 1
This is selection (combination) as order doesn't matter
2
Step 2
Total people = 14, need 5 with at least 2 girls
3
Step 3
Use complement: Total ways - Ways with 0 girls - Ways with 1 girl
4
Step 4
Total ways = 14C5 = 2002
5
Step 5
Ways with 0 girls = 8C5 = 56
6
Step 6
Ways with 1 girl = 6C1 × 8C4 = 6 × 70 = 420
7
Step 7
Required ways = 2002 - 56 - 420 = 1526
Answer: 1526 ways
Exam Shortcuts:
1. For large factorials, cancel common terms before calculating
2. Use the property nCr × r! = nPr for quick conversion
3. In word problems, identify keywords: 'arrange' means permutation, 'select/choose' means combination
Exam TrapsCommon mistakes students make — avoid these
- The #1 Trap:
Students confuse
When to UseQuickly decide which method to apply in the exam
permutation vs combination. Remember: If the question talks about positions, ranks, or arrangements, use permutation.
If it talks about selection, teams, or groups, use combination. For example, 'selecting 3 students' is combination, but 'arranging 3 students in first, second, third position' is permutation.
Key Points to Remember
Permutation is for arrangements where order matters, combination is for selections where order doesn't matter
Formula shortcut: nPr = n!/(n-r)! and nCr = n!/(r!(n-r)!)
Quick trick: nCr = nC(n-r), so calculate the smaller value
Circular arrangement formula: (n-1)! when clockwise and anticlockwise are same
For repeated items: n!/p!q!r! where p,q,r are repetition counts
Conversion formula: nPr = r! × nCr
Keywords: 'arrange/order' means permutation, 'select/choose' means combination
Complement method: Total - Unwanted cases for complex conditions
0! = 1 and 1! = 1 are important base values
Cancel common factorial terms before calculating to save time
Exam-Specific Tips
0! equals 1 by mathematical definition
nC0 = 1 for any positive integer n
nCn = 1 for any positive integer n
nC1 = n for any positive integer n
Circular permutation of n objects is (n-1)! arrangements
nCr + nC(r-1) = (n+1)Cr Pascal's identity
Maximum value of nCr occurs at r = n/2 when n is even
nPr is always greater than or equal to nCr for same n and r values
Practice MCQs
Permutation & Combination — Practice Questions
15graded MCQs · easy to hard · full solution & trap analysis
In how many ways can 6 students be divided into 2 groups of 3 each?
Practice 2easy
How many ways can 4 red balls and 3 blue balls be arranged in a row?
Practice 3easy
In how many ways can 5 different books be arranged on a shelf?
Practice 4easy
How many ways can a committee of 3 people be selected from a group of 8 people?
Practice 5easy
In how many ways can the letters of the word 'APPLE' be arranged?
Practice 6medium
In how many ways can 10 people be divided into two groups of 5 each?
Practice 7medium
A committee of 5 members is to be formed from a group of 8 men and 6 women. In how many ways can the committee be formed such that it contains at least 2 women?
Practice 8medium
In how many ways can 4 identical red balls and 3 identical blue balls be arranged in a row such that no two blue balls are adjacent?
Practice 9medium
A password consists of 3 digits followed by 2 letters. The digits can be any digit from 0–9 (repetition allowed), and the letters must be distinct and chosen from 26 letters of the English alphabet. How many such passwords are possible?
Practice 10medium
A bookshelf has 5 different English books and 4 different Hindi books. In how many ways can these 9 books be arranged such that all English books are together and all Hindi books are together?
Practice 11hard
A committee of 5 members is to be formed from 8 men and 6 women. In how many ways can this be done such that the committee has at least 2 women and at least 2 men?
Practice 12hard
A committee of 7 members is to be formed from a group of 6 men and 5 women. If the committee must have more women than men, in how many ways can this be done?
Practice 13hard
A password consists of 4 digits followed by 3 letters (uppercase only). How many passwords are possible if the first digit must be non-zero, the last digit must be even, and no letter can be repeated?
Practice 14hard
A team of 6 players is to be selected from 5 batsmen, 4 bowlers, and 3 all-rounders. In how many ways can this be done if the team must have at least 2 batsmen, at least 1 bowler, and at least 1 all-rounder?
Practice 15hard
In how many ways can 12 identical chocolates be distributed among 5 children such that each child gets at least 2 chocolates and at most 4 chocolates?
60-Second Revision — Permutation & Combination
Remember: Order matters = Permutation, Order doesn't matter = Combination
Formula: nPr = n!/(n-r)! and nCr = n!/(r!(n-r)!)
Trick: Use nCr = nC(n-r) to reduce calculation work
Trap: Don't confuse arrangement questions with selection questions
Shortcut: For word arrangements with repetition, divide by repeated letter factorials
Quick check: Permutation answers are always larger than combination answers
Remember: Circular arrangements = (n-1)! when direction doesn't matter