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SSC MTS Permutation & Combination

Study Material — 18 PYQs (2021–2021) · Concept Notes · Shortcuts

SSC MTS Permutation & Combination is a frequently tested subtopic — 18 previous year questions from 2021–2021 papers are included below with concept notes, key rules and shortcut tricks.

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2021–2021
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Previous Year Questions

SSC MTS Permutation & Combination — Past Exam Questions

18 questions from actual SSC MTS papers · all shown free · click option to reveal solution

Exam Q 12021Previous Year Pattern

How many 2-digit numbers can be formed using the digits 3, 5, 7, and 9 without repetition?

Exam Q 22021Previous Year Pattern

In how many ways can 4 red balls and 3 blue balls be arranged in a line?

Exam Q 32021Previous Year Pattern

A student must answer 5 questions out of 8 questions in an exam. In how many ways can the student choose the questions?

Exam Q 42021Previous Year Pattern

In how many ways can 5 different books be arranged on a shelf?

Exam Q 52021Previous Year Pattern

How many ways can a committee of 3 people be selected from a group of 8 people?

Exam Q 62021Previous Year Pattern

In how many ways can the letters of the word 'APPLE' be arranged?

Exam Q 72021Previous Year Pattern

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?

Exam Q 82021Previous Year Pattern

In how many ways can 5 different books be arranged on a shelf such that two specific books (A and B) are always together?

Exam Q 92021Previous Year Pattern

A committee of 4 people is to be selected from 6 men and 5 women. In how many ways can this be done if the committee must have at least 2 women?

Exam Q 102021Previous Year Pattern

In how many ways can the letters of the word 'MISSISSIPPI' be arranged?

Exam Q 112021Previous Year Pattern

From a group of 8 students, in how many ways can a President, Vice-President, and Secretary be selected such that no student holds more than one position?

Exam Q 122021Previous Year Pattern

A password consists of 2 letters followed by 3 digits. If letters can be repeated but digits cannot, how many different passwords are possible? (Consider 26 letters and 10 digits: 0–9)

Exam Q 132021Previous Year Pattern

A password consists of 3 letters followed by 2 digits. Letters can be repeated but must be from {A, B, C, D, E}, and digits must be distinct and from {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. How many such passwords are possible?

Exam Q 142021Previous Year Pattern

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?

Exam Q 152021Previous Year Pattern

In how many ways can the letters of the word 'MISSISSIPPI' be arranged such that all I's are together and all S's are together?

Exam Q 162021Previous Year Pattern

A committee of 6 members is to be formed from 8 engineers and 5 doctors such that at least 2 engineers and at least 2 doctors are included. In how many ways can this be done?

Exam Q 172021Previous Year Pattern

How many 4-digit numbers can be formed using digits 1, 2, 3, 4, 5, 6 (repetition allowed) such that the number is divisible by 4? (A number is divisible by 4 if its last two digits form a number divisible by 4.)

Exam Q 182021Previous Year Pattern

From a group of 7 boys and 6 girls, a team of 5 is to be selected. In how many ways can this be done such that the team has more girls than boys?

Concept Notes

Permutation & Combination— Rules & Concept

1.Core Concept

Read this first — the foundation of the topic

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

2.Formula Block

Memorise — 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

3.Exam Patterns

What 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.

4.Shortcuts

Use 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)

5.Worked Example

Solve 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

6.Exam Traps

Common mistakes students make — avoid these

- The #1 Trap: Students confuse

7.When to Use

Quickly 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

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
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SSC MTS Permutation & Combination — Study Material, 18 PYQs & Practice MCQs | ZestExam