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CAPF AC Permutation & Combination

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This page covers CAPF AC Permutation & Combination with complete concept notes, 36 graded practice MCQs, key points and exam-specific tips. Free to study.

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Concept Notes

Permutation & Combination— Rules & Concept

1.Core Concept

Read this first — the foundation of the topic
It answers two basic questions

'In how many ways can we arrange objects?' (Permutation) and 'In how many ways can we select objects?' (Combination). Understanding the difference is crucial for NDA success. PERMUTATION deals with ARRANGEMENT. Order matters here. If you have 3 books A, B, C, then ABC and BAC are different arrangements.

The formula is nPr = n!/(n-r)! where n is total objects and r is objects being arranged. COMBINATION deals with SELECTION. Order does not matter. Selecting books A, B is same as selecting B, A. The formula is nCr = n!/[r!(n-r)!]

KEY RULES

(1) 0! = 1 always (2) nPn = n! (3) nCr = nC(n-r) (4) nC0 = 1 (5) When arrangement matters, use P. When selection matters, use C.

2.Formula Block

Memorise — at least one formula appears in every paper
• nPr = n!/(n-r)!
• nCr = n!/[r!(n-r)!]
• Circular arrangement = (n-1)!
• Arrangement with repetition = n!/[n1! × n2! × n3!...]

3.Exam Patterns

What examiners ask — read before attempting PYQs

NDA frequently asks arrangement of people in rows, selection of teams, circular arrangements, and word formation problems. Questions often mix permutation-combination concepts to test understanding.

4.Shortcuts

Use these to save 30–60 seconds per question

#1: Quick nCr calculation. For nC2, use n(n-1)/2. For nC3, use n(n-1)(n-2)/6.

This saves factorial calculation time.

5.Worked Example

Solve this step-by-step before moving on
1
Step 1

This is arrangement, so use permutation

2
Step 2

All 5 people are being arranged, so n=5, r=5

3
Step 3

5P5 = 5!/(5-5)! = 5!/0! = 5!/1 = 120 Answer: 120 ways WORKED EXAMPLE 2: From 8 students, select 3 for a committee.

1
Step 1

This is selection, order doesn't matter, so use combination

2
Step 2

n=8, r=3

3
Step 3

8C3 = 8!/(3! × 5!) = (8×7×6)/(3×2×1) = 336/6 = 56 Answer: 56 ways SHORTCUT TRICK #2: Circular arrangement shortcut. For n objects in circle, answer is (n-1)!. For clockwise/anticlockwise being same, divide by 2. SHORTCUT TRICK #3: Word problems - if word has repeated letters, use n!/[repetitions!]. For COMMITTEE with 2 M's and 2 E's: 9!/(2!×2!).

6.Exam Traps

Common mistakes students make — avoid these

#1: Students confuse

7.When to Use

Quickly decide which method to apply in the exam

P vs C. Remember: If question asks 'arrange', 'permute', 'different ways to sit/stand', use P.

If question asks 'select', 'choose', 'committee formation', use C. The word 'different' is tricky - it usually means arrangement unless context suggests selection. Another frequent error is forgetting that 0! = 1.

This appears in many calculations and wrong value leads to incorrect answers. For NDA preparation, master these problem types: linear arrangements, circular arrangements, selection problems, word formation, and problems involving restrictions (like specific people sitting together or apart). Practice identifying keywords that indicate permutation vs combination.

Key Points to Remember

  • Permutation is for arrangement (order matters), Combination is for selection (order doesn't matter)
  • Formula: nPr = n!/(n-r)! and nCr = n!/[r!(n-r)!]
  • Remember: 0! = 1 always, this is frequently tested
  • Shortcut: nC2 = n(n-1)/2 and nC3 = n(n-1)(n-2)/6
  • Circular arrangement of n objects = (n-1)! ways
  • For words with repeated letters: total letters!/[repetition1! × repetition2!...]
  • Keywords 'arrange', 'permute', 'sit/stand' indicate Permutation
  • Keywords 'select', 'choose', 'committee' indicate Combination
  • Property: nCr = nC(n-r), useful for quick calculations
  • Restriction problems: calculate total cases minus restricted cases

Exam-Specific Tips

  • 0! equals 1 by mathematical definition
  • nC0 = 1 for any positive integer n
  • Circular permutation of n distinct objects is (n-1)! ways
  • For identical circular arrangements (clockwise = anticlockwise), divide by 2
  • Number of ways to arrange n objects with r identical = n!/r!
  • Sum of all combinations: nC0 + nC1 + nC2 + ... + nCn = 2^n
  • Maximum value of nCr occurs at r = n/2 (when n is even)
  • Pascal's triangle property: nCr + nC(r+1) = (n+1)C(r+1)
Practice MCQs

Permutation & Combination — Practice Questions

36graded MCQs · easy to hard · full solution & trap analysis · showing 20 of 36

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Practice 1easy

The number of ways to arrange the letters of the word 'MISSISSIPPI' is:

Practice 2easy

From a group of 6 men and 4 women, in how many ways can a committee of 5 be formed such that it contains at least 2 women?

Practice 3easy

How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5 (without repetition) such that the number is even?

Practice 4easy

A committee of 3 members is to be selected from a group of 5 men and 4 women. In how many ways can the committee be formed such that it has at least 1 woman?

Practice 5easy

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

Practice 6easy

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

Practice 7easy

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

Practice 8easy

A team of 5 players is to be selected from 6 men and 4 women such that the team has at least 2 women. In how many ways can this be done?

Practice 9easy

How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6 (without repetition) such that the number is even?

Practice 10easy

In how many ways can 5 distinct books be arranged on a shelf such that a specific book (say Book A) is always at one of the two ends?

Practice 11easy

In how many ways can a committee of 3 members be selected from a group of 8 people such that a particular person (say Person X) is always included?

Practice 12easy

How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6 without repetition, such that the number is even?

Practice 13medium

The number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5, 6 (with repetition allowed) such that the digits are in strictly increasing order is:

Practice 14medium

The number of ways to arrange the letters of the word 'PERMUTATION' such that no two vowels are adjacent is:

Practice 15medium

A committee of 5 people is to be formed from 6 men and 4 women. The number of ways to form the committee such that it contains at least 2 women is:

Practice 16medium

The number of ways to distribute 10 identical balls into 3 distinct boxes such that each box contains at least 2 balls is:

Practice 17medium

In how many ways can 5 distinct books be arranged on a shelf such that a specific book (say Book A) is never at either end?

Practice 18medium

How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6 (without repetition) such that the number is divisible by 5?

Practice 19medium

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?

Practice 20medium

The number of ways to arrange the letters of the word 'MISSISSIPPI' is:

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60-Second Revision — Permutation & Combination

  • Remember: Arrangement = Permutation, Selection = Combination
  • Formula: nPr = n!/(n-r)!, nCr = n!/[r!(n-r)!]
  • Quick trick: nC2 = n(n-1)/2, saves time in calculations
  • Trap: Don't forget 0! = 1, appears in most problems
  • Circular arrangement = (n-1)!, not n!
  • Word problems with repetition: divide by factorial of repeated elements
  • Check question keywords carefully - 'arrange' vs 'select' determines method
←Quadratic Equations & PolynomialsBinomial Theorem→
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