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Simple Permutations (Order Matters)

Introduction

Simple permutations deal with arranging objects when order matters. Many real-world problems - seating people in chairs, assigning ranks, or arranging books on a shelf - require counting ordered arrangements, not just selections.

This pattern is important because confusing order vs. selection is a common source of mistakes in aptitude tests; knowing when to use permutations gives the correct answer quickly.

Pattern: Simple Permutations (Order Matters)

Pattern

The key idea: when arranging r items from n distinct items and the order matters, use nPr.

Formula:
nPr = n × (n - 1) × (n - 2) × … × (n - r + 1) = n! / (n - r)!

Step-by-Step Example

Question

Four friends - A, B, C, D - are to sit in 2 distinct chairs in a row. In how many different ordered ways can they be seated?

Solution

  1. Step 1: Identify what is given.

    We have n = 4 distinct people and we want to arrange r = 2 of them in order.

  2. Step 2: Choose the correct formula.

    Since order matters, use permutations: nPr = n! / (n - r)! .

  3. Step 3: Substitute and compute.

    4P2 = 4! / (4 - 2)! = 4! / 2! = (4 × 3 × 2 × 1) / (2 × 1) = 4 × 3 = 12

  4. Final Answer:

    There are 12 ordered seating arrangements.

  5. Quick Check:

    Pick first chair → 4 choices; pick second chair → 3 remaining → 4 × 3 = 12 ✅

Quick Variations

1. r = n → all items arranged: nPn = n! (full permutations).

2. r = 1 → nP1 = n (choosing one item with order trivial).

3. Use when assigning distinct ranks (1st, 2nd, 3rd) or forming ordered codes without repetition.

Trick to Always Use

  • Step 1 → Ask: "Does order matter?" If YES, use nPr. If NO, use nCr.
  • Step 2 → Compute nPr as n × (n - 1) × ... for r factors (stop after r multiplications) - faster than factorials for small r.

Summary

Summary

For ordered arrangements of r distinct items from n:

  • Use nPr = n! / (n - r)!.
  • Compute by multiplying the top r factors: n × (n - 1) × …
  • Always check whether order matters - that decides permutation vs combination.

Practice

(1/5)
1. In how many different ways can 3 students be arranged in a line from a group of 3 students A, B, and C?
easy
A. 3
B. 6
C. 9
D. 12

Solution

  1. Step 1: Identify what is given.

    Total students n = 3; all 3 are to be arranged, so r = 3.

  2. Step 2: Apply permutation formula.

    nPr = n! / (n - r)!.

  3. Step 3: Substitute and compute.

    3P3 = 3! / 0! = 3! = 6.

  4. Final Answer:

    There are 6 ways to arrange 3 students → Option B.

  5. Quick Check:

    List: ABC, ACB, BAC, BCA, CAB, CBA → 6 arrangements ✅
Hint: When all items are arranged, nPn = n!.
Common Mistakes: Using combination instead of permutation or forgetting that 0! = 1.
2. In how many ways can a president and a vice-president (order matters) be chosen from 4 people?
easy
A. 12
B. 8
C. 6
D. 24

Solution

  1. Step 1: Identify what is given.

    n = 4 people; r = 2 positions (president, vice-president) and order matters.

  2. Step 2: Apply permutation formula.

    nPr = n! / (n - r)! = 4! / 2!.

  3. Step 3: Substitute and compute.

    4! / 2! = (4 × 3 × 2 × 1) / (2 × 1) = 4 × 3 = 12.

  4. Final Answer:

    There are 12 ways → Option A.

  5. Quick Check:

    Choose president (4 ways) then vice-president (3 ways) → 4 × 3 = 12 ✅
Hint: For assigning distinct ordered positions, use nPr = n × (n - 1) × ... for r factors.
Common Mistakes: Using combination (nC2) which ignores order.
3. How many 3-digit numbers can be formed using digits 1, 2, 3, and 4 if repetition is not allowed?
easy
A. 12
B. 20
C. 24
D. 36

Solution

  1. Step 1: Identify what is given.

    Available digits n = 4; positions r = 3; repetition not allowed.

  2. Step 2: Apply permutation formula.

    nPr = n! / (n - r)! = 4! / 1!.

  3. Step 3: Substitute and compute.

    4! / 1! = 24 / 1 = 24 (or 4 × 3 × 2 = 24).

  4. Final Answer:

    There are 24 different 3-digit numbers → Option C.

  5. Quick Check:

    First digit 4 choices, second 3, third 2 → 4 × 3 × 2 = 24 ✅
Hint: When repetition is not allowed, multiply decreasing counts for each position.
Common Mistakes: Using n^r instead of nPr when repetition is not allowed.
4. In how many ways can 5 athletes finish first, second, and third in a race (no ties)?
medium
A. 60
B. 80
C. 100
D. 120

Solution

  1. Step 1: Identify what is given.

    Total athletes = 5; positions to fill r = 3; order matters.

  2. Step 2: Apply permutation formula.

    nPr = 5! / (5 - 3)! = 5! / 2!.

  3. Step 3: Substitute and compute.

    5! / 2! = (5 × 4 × 3 × 2 × 1) / (2 × 1) = 5 × 4 × 3 = 60.

  4. Final Answer:

    There are 60 possible finishing orders → Option A.

  5. Quick Check:

    First = 5 choices, second = 4, third = 3 → 5 × 4 × 3 = 60 ✅
Hint: For ranking problems, multiply choices for each position until ranks filled.
Common Mistakes: Using combinations which ignore order.
5. How many 4-letter arrangements can be made from the word ‘BRIGHT’?
medium
A. 120
B. 240
C. 720
D. 360

Solution

  1. Step 1: Identify what is given.

    The word 'BRIGHT' has n = 6 distinct letters; we need r = 4-letter ordered arrangements.

  2. Step 2: Apply permutation formula.

    nPr = n! / (n - r)! = 6! / 2!.

  3. Step 3: Substitute and compute.

    6! / 2! = (720) / 2 = 360 (equivalently 6 × 5 × 4 × 3 = 360).

  4. Final Answer:

    There are 360 possible arrangements → Option D.

  5. Quick Check:

    Multiply top 4 factors of 6!: 6 × 5 × 4 × 3 = 360 ✅
Hint: When n is large and r is smaller, multiply the top r factors (n × (n - 1) × ...).
Common Mistakes: Confusing with combinations or calculating full factorials unnecessarily.

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