Permutations

Master the art of arranging objects in different orders

Table of Contents

Simplified Quantitative Formulas: Permutations & Combinations

  • Factorial (n!): Product of all positive integers up to n. 0! = 1.
  • Permutation (nPr): Number of ways to choose and arrange r items out of n distinct items: nPr = n! / (n-r)!. Order matters.
  • Combination (nCr): Number of ways to choose r from n: nCr = n! / [r! * (n-r)!]. Order does not matter.
  • Symmetry: nCr = nC(n-r).
  • Sum of combinations: Σ nCk = 2^n.
  • Partitions (Distribution): Distributing n identical items into r distinct boxes: If boxes can be empty, ways = C(n+r-1, r-1). If each must get ≥1, ways = C(n-1, r-1).
  • Arrangements with Repetition: If n items with some identical, divide by factorial of repeats: n!/(a!b!...).
  • Rank of a Word: Position in sorted list of all permutations. Count smaller letters at each position, multiply by (remaining)!.
  • Combinatorial Identities: Pascal's triangle: nCk = (n-1)C(k-1) + (n-1)Ck. nC0 + nC1 + ... + nCn = 2^n.
  • Integral Solutions: Number of positive integer solutions to x₁ + x₂ + ... + x_n = S is C(S-1, n-1). Non-negative: C(S+n-1, n-1).
  • Circular Arrangements: n distinct items in a circle: (n-1)!. If reflections are same, divide by 2.
  • Derangements: Number of ways no item is in its original place: !n = n! * (1 – 1/1! + 1/2! – ... + (-1)^n/n!).

What do these mean? (Super Simple Explanations & Examples)

  • Permutation: Arranging 3 books A, B, C: 3! = 6 ways (ABC, ACB, BAC, BCA, CAB, CBA).
  • Combination: Choosing 2 out of 4: 4C2 = 6 ways.
  • Partitions: 5 identical balls into 3 bins (can be empty): C(5+3-1,3-1)=C(7,2)=21 ways.
  • Arrangements with Repetition: "BALLOON": 7!/(2!2!) = 840 ways.
  • Rank of a Word: "CAB": C before A and B, so 2×2! = 4 permutations before CAB.
  • Integral Solutions: x+y+z=10, x,y,z>0: C(9,2)=36 ways.
  • Circular Arrangements: 4 people around a table: 3! = 6 seatings.
  • Derangements: For 3 items: !3 = 2 ways (CBA, BAC).

Introduction to Permutations

Permutations are arrangements of objects in a specific order. The order matters in permutations, making them different from combinations.

Key Concepts

  • Permutation: Arrangement of objects in a specific order
  • Factorial: Product of all positive integers up to n (n!)
  • Linear Permutation: Arrangement in a line
  • Circular Permutation: Arrangement in a circle

Basic Permutation Formula

nPr = n!/(n-r)!

where:

  • n is the total number of objects
  • r is the number of objects to arrange
  • n! is the factorial of n

Basic Concepts

Factorial

n! = n × (n-1) × (n-2) × ... × 2 × 1

Examples:

  • 5! = 5 × 4 × 3 × 2 × 1 = 120
  • 0! = 1 (by definition)

Common Factorial Values:

n n!
0 1
1 1
2 2
3 6
4 24
5 120
6 720
7 5,040
8 40,320
9 362,880
10 3,628,800

Important Notes:

  • 0! = 1 (by definition)
  • 1! = 1
  • n! grows very rapidly with increasing n
  • For n > 10, use calculator or approximation methods

Permutation with Repetition

n^r

where:

  • n is the number of different objects
  • r is the number of positions

Example

How many 3-digit numbers can be formed using digits 1-5?

5^3 = 125 numbers

Linear Permutations

Permutation of n Different Objects

n! = n × (n-1) × (n-2) × ... × 2 × 1

Permutation of n Objects with Repetition

n!/(n₁! × n₂! × ... × nₖ!)

where:

  • n is the total number of objects
  • n₁, n₂, ..., nₖ are the frequencies of repeated objects

Example

How many different words can be formed using all letters of "MISSISSIPPI"?

11!/(4! × 4! × 2!) = 34,650 words

Circular Permutations

Circular Permutation of n Different Objects

(n-1)!

Circular Permutation with Repetition

(n-1)!/(n₁! × n₂! × ... × nₖ!)

Example

In how many ways can 5 people sit around a circular table?

(5-1)! = 4! = 24 ways

Restricted Permutations

Permutation with Fixed Positions

(n-r)! × r!

where:

  • n is the total number of objects
  • r is the number of fixed positions

Permutation with Restrictions

Use the following steps:

  1. Count total permutations
  2. Count restricted permutations
  3. Subtract restricted from total

Example

How many ways can 5 people sit in a row if two specific people must sit together?

4! × 2! = 48 ways

Practice Questions

Basic Permutations

Question 1: Simple Permutation
In how many ways can 5 different books be arranged on a shelf?
Solution:

5! = 5 × 4 × 3 × 2 × 1 = 120 ways

Question 2: Permutation with Repetition
How many different 4-letter words can be formed using the letters of "BOOK"?
Solution:

4!/(2! × 1! × 1!) = 12 words

Question 3: Partial Permutation
In how many ways can 3 students be selected from a class of 10 to fill the positions of president, vice-president, and secretary?
Solution:

10P3 = 10!/(10-3)! = 720 ways

Question 4: Permutation with Conditions
How many 4-digit numbers can be formed using digits 1-5 if no digit can be repeated?
Solution:

5P4 = 5!/(5-4)! = 120 numbers

Circular Permutations

Question 1: Simple Circular
In how many ways can 6 people sit around a circular table?
Solution:

(6-1)! = 5! = 120 ways

Question 2: Circular with Repetition
In how many ways can 8 beads of different colors be arranged in a necklace if 3 beads are identical?
Solution:

(8-1)!/(3! × 1! × 1! × 1! × 1! × 1!) = 840 ways

Question 3: Circular with Fixed Position
In how many ways can 7 people sit around a circular table if one person must sit in a specific chair?
Solution:

6! = 720 ways

Question 4: Circular with Restrictions
In how many ways can 5 couples sit around a circular table if each couple must sit together?
Solution:

(5-1)! × 2⁵ = 384 ways

Restricted Permutations

Question 1: Fixed Positions
In how many ways can 6 people sit in a row if two specific people must sit at the ends?
Solution:

2! × 4! = 48 ways

Question 2: Adjacent Objects
How many different words can be formed using all letters of "MATHEMATICS" if all vowels must be together?
Solution:

8! × 4!/(2! × 2!) = 241,920 words

Question 3: Non-adjacent Objects
In how many ways can 5 boys and 3 girls sit in a row if no two girls can sit together?
Solution:

5! × 6P3 = 14,400 ways

Question 4: Multiple Restrictions
How many 4-digit numbers can be formed using digits 1-5 if the first digit must be odd and no digit can be repeated?
Solution:

3 × 4P3 = 72 numbers

Practice Questions

Test your understanding of Permutations with 20 fully solved, step-by-step questions designed for beginners.

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