Study Material — 8 PYQs (2018–2020) · Concept Notes · Shortcuts
NDA Permutation & Combination is a frequently tested subtopic — 8 previous year questions from 2018–2020 papers are included below with concept notes, key rules and shortcut tricks.
NDA Permutation & Combination — Past Exam Questions
8 questions from actual NDA papers · all shown free · click option to reveal solution
Exam Q 12019Previous Year Pattern
In how many ways can 5 different books be arranged on a shelf such that a specific book (say, Book A) is always at one of the end positions?
Exam Q 22020Previous Year Pattern
In how many ways can 5 different books be arranged on a shelf such that a specific book (say, Book A) is always at one of the end positions?
Exam Q 32018Previous Year Pattern
In how many ways can 5 different books be arranged on a shelf such that a specific book always remains in the middle position?
Exam Q 42020Previous Year Pattern
A committee of 5 members is to be formed from a group of 6 men and 4 women. In how many ways can the committee be formed such that it contains at least 2 women?
Exam Q 52019Previous Year Pattern
A committee of 5 members is to be formed from a group of 6 men and 4 women. In how many ways can the committee be formed such that it contains at least 2 women?
Exam Q 62018Previous Year Pattern
In how many ways can the letters of the word 'GARDEN' be arranged such that the vowels always occupy the odd positions?
Exam Q 72019Previous Year Pattern
A committee of 5 members is to be formed from a group of 6 men and 5 women. In how many ways can the committee be formed such that it contains at least 2 women and at least 1 man, but the number of women must not exceed the number of men?
Exam Q 82020Previous Year Pattern
A committee of 5 members is to be formed from a group of 6 men and 5 women. In how many ways can the committee be formed such that it contains at least 2 women and at least 1 man, but the number of women must not exceed the number of men?
Concept Notes
Permutation & Combination— Rules & Concept
Core ConceptRead this first — the foundation of the topic
Simple Rule
• Permutation = Arrangement (ORDER matters)
• Combination = Selection (ORDER does NOT matter)
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KEY RULES & PROPERTIES
1
Factorial
n! = n × (n−1) × (n−2) × ... × 1
Example: 5! = 5 × 4 × 3 × 2 × 1 = 120
Special rule: 0! = 1 (memorise this — it is always asked)
2
Permutation Formula
nPr = n! / (n−r)!
This gives arrangements of r items from n items.
3
Combination Formula
nCr = n! / (r! × (n−r)!)
This gives selections of r items from n items.
4
Key Relationship
nPr = nCr × r
Meaning
Permutation = Combination × arrangements of selected items
5
Mirror Property of Combination
nCr = nC(n−r)
Example: 10C3 = 10C7 — Use this to reduce calculations!
6. nC0 = nCn = 1 (selecting none or selecting all = 1 way)
7. nC1 = n (selecting 1 from n = n ways)
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Formula BlockMemorise — at least one formula appears in every paper
nPr = n! / (n−r)!
nCr = n! / [r! × (n−r)!]
Total subsets of n items = 2^n
Arrangements of n items where one repeats 'p' times = n! / p!
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Exam PatternsWhat examiners ask — read before attempting PYQs
— WHAT GETS ASKED
• How many ways to arrange letters of a word (with/without repetition)
• Selecting a committee of X people from Y people
• Forming numbers using given digits
• Sitting arrangements in a row or circle
• Handshakes / matches in a tournament (always nC2)
Circular Arrangement Formula: (n−1)! for n people sitting in a circle
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SHORTCUTS & TRICKS
Trick 1 — Handshakes/Matches Shortcut:
If n people shake hands with each other, total handshakes = nC2 = n(n−1)/2
Example: 10 people = 10×9/2 = 45 handshakes. Fast and clean.
Trick 2 — Vowel/Consonant Grouping:
To keep vowels together in a word arrangement — treat all vowels as ONE unit. Then arrange (remaining letters + 1 unit) and multiply by arrangements of vowels within the unit.
Trick 3 — Complement Method:
When 'at least one' is mentioned:
At least 1 = Total ways − Ways with NONE selected
This avoids long addition and saves time in exam.
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Worked ExampleSolve this step-by-step before moving on
Q: In how many ways can 5 students be arranged in a row?
Solution:
This is a pure arrangement — use Permutation.
5P5 = 5! = 5 × 4 × 3 × 2 × 1 = 120
Answer: 120 ways
WORKED EXAMPLE 2
Q: A committee of 3 is to be selected from 6 people. How many ways?
Solution:
Order does not matter in a committee — use Combination.
6C3 = 6! / (3! × 3!) = (6 × 5 × 4) / (3 × 2 × 1) = 120 / 6 = 20
Answer: 20 ways
WORKED EXAMPLE 3
Q: How many 3-digit numbers can be formed from digits 1, 2, 3, 4, 5 without repetition?
Solution:
First digit: 5 choices
Second digit: 4 choices (one used)
Third digit: 3 choices
Total = 5 × 4 × 3 = 60
OR use 5P3 = 5!/(5−3)! = 120/2 = 60
Answer: 60 numbers
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Exam TrapsCommon mistakes students make — avoid these
— #1 TRAP
Students mix up
When to UseQuickly decide which method to apply in the exam
Permutation vs Combination.
The golden test: Ask yourself — Does changing the ORDER give a NEW outcome?
Yes → Permutation. No → Combination.
Example: Selecting a President AND Secretary from 5 people = ORDER matters (different roles) = Permutation = 5P2 = 20.
Selecting 2 members for a team = ORDER does NOT matter = Combination = 5C2 = 10.
Many students use Combination here and get 10 instead of 20.
This is the most common wrong answer in SSC CGL.
Key Points to Remember
Permutation = Arrangement where ORDER matters; Formula: nPr = n! / (n−r)!
Combination = Selection where ORDER does NOT matter; Formula: nCr = n! / [r! × (n−r)!]
Always remember: 0! = 1 — this is frequently tested directly
Mirror property shortcut: nCr = nC(n−r), so 10C7 = 10C3 — reduces big calculations fast
Handshakes or matches between n people = nC2 = n(n−1)/2
Circular arrangement of n people = (n−1)! ways
nPr = nCr × r! — Permutation is always greater than or equal to Combination
Total number of subsets possible from a set of n items = 2^n
For words with repeated letters, arrangements = n! divided by factorial of each repeated letter count
At least one selected = Total selections minus (no item selected) — use complement method to save time
Exam-Specific Tips
0! = 1 — this is a fixed mathematical fact tested as a direct MCQ option
nC0 = 1 and nCn = 1 for any value of n
The number of diagonals in a polygon of n sides = nC2 − n = n(n−3)/2
Number of handshakes when n persons each shake hands with every other = nC2 = n(n−1)/2
For circular arrangement of n distinct objects, the number of arrangements = (n−1)!
nC1 = n for any value of n
The total number of ways to select one or more items from n distinct items = 2^n − 1
nPn = n! and nP0 = 1 — both are standard exam facts
Practice MCQs
Permutation & Combination — Practice Questions
50graded MCQs · easy to hard · full solution & trap analysis · showing 20 of 50
How many 2-digit numbers can be formed using the digits 3, 5, 7, and 9 without repetition?
Practice 2easy
In 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 2 red balls and 3 blue balls be arranged in a row?
Practice 6easy
A committee of 4 people is to be formed from 6 men and 4 women. In how many ways can this be done if the committee must have at least 2 women?
Practice 7easy
How many ways can 3 students be selected from a group of 7 students?
Practice 8easy
How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5 without repetition?
Practice 9easy
In how many ways can the letters of the word 'BOOK' be arranged?
Practice 10easy
In how many ways can the letters of the word 'BOOK' be arranged?
Practice 11easy
In how many ways can 5 different books be arranged on a shelf?
Practice 12easy
A committee of 3 members is to be formed from 6 men and 4 women. In how many ways can this be done if the committee must have at least 1 woman?
Practice 13easy
In how many ways can the letters of the word 'LETTER' be arranged?
Practice 14easy
How many ways can 6 students be divided into two groups of 3 each?
Practice 15easy
How many ways can a committee of 3 members be selected from a group of 8 people?
Practice 16easy
A student must answer 4 questions out of 7 questions in an exam. In how many ways can the student select the questions?
Practice 17medium
How many 4-digit numbers can be formed using the digits 2, 3, 5, 7, 8, 9 without repetition, such that the number is even?
Practice 18medium
A committee of 6 people is to be formed from 8 engineers and 5 doctors such that the committee contains at least 2 engineers and at least 2 doctors. In how many ways can this be done?
Practice 19medium
In a group of 8 people, how many ways can we select a president, a vice-president, and a treasurer such that no person holds more than one position?
Practice 20medium
In how many ways can 5 men and 4 women be arranged in a row such that no two women sit adjacent to each other?
30 more practice questions in the Study Panel
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