Permutation

Permutation is a fundamental concept in combinatorics and mathematics, used to count the number of ways elements can be arranged or ordered. Whether you're solving problems in probability, algebra, or computer science, understanding permutations helps in organizing and analyzing data effectively.

1.0Permutation Definition

Permutation refers to the arrangement of a set of objects in a specific order. In mathematics, if you’re asked to arrange things in a line or specific order, you’re working with permutations.

A permutation is an arrangement of objects in a specific order. In mathematics, permutations refer to the different ways in which a set of elements can be ordered or arranged. The order of items matters in permutations, unlike combinations.

For example, the permutations of the set {A, B, C} taken two at a time are: AB, AC, BA, BC, CA, and CB.

2.0Permutation Meaning

The meaning of permutation lies in arranging or reordering items. For instance, the different ways to arrange the letters A, B, and C (like ABC, ACB, BAC...) are all permutations of the three letters. The order matters in permutation — which makes it different from combinations.

3.0Permutation Formula

The general permutation formula to find the number of arrangements of r objects from a total of n is:

$P(n,r)=\frac{n!}{(n-r)!}$

Where:

• n! is the factorial of n
• r is the number of positions to fill

Permutation Without Repetition

This occurs when no item is repeated in the arrangement. Use the formula: $P(n,r)=\frac{n!}{(n-r)!}$

Permutation With Repetition

This happens when items can be repeated. The formula becomes: $n^r$

Where n is the number of things to choose from, and r is how many you're choosing.

4.0Even and Odd Permutation

• Even Permutation: A permutation that can be achieved by an even number of swaps.
• Odd Permutation: A permutation that requires an odd number of swaps.

This concept is especially important in linear algebra, particularly with matrices and determinants.

Also Read: Permutations and Combinations

5.0Solved Example of Permutation

Here are JEE and JEE Advanced level solved examples on Permutation, covering standard to tricky cases including repetition, restrictions, and odd/even permutations:

Example 1: How many different 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5 without repetition?

Solution:
To form a 4-digit number:

• Choose any 4 digits from 5 → ${}^5{P}_4=\frac{5!}{(5-4)!}=120$

Answer: 120

Example 2: How many 3-letter words (real or not) can be formed from the letters A, B, C, D, E if repetition is allowed?

Solution:

Each place (3) can be filled in 5 ways: $5^3=125$

Answer: 125

Example 3: How many even 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5 with no repetition?

Solution:

For an even number, the last digit must be 2 or 4.

• Case 1: Last digit = 2

Remaining digits = 1,3,4,5 → Choose 2 → ${}^4{P}_2=12$

• Case 2: Last digit = 4

Remaining digits = 1,2,3,5 → ${}^4{P}_2=12$

→ Total = 12 + 12 = 24

Answer: 24

Example 4: In how many ways can 5 people be seated around a round table?

Solution:

Circular permutation formula: (n - 1)! → (5 - 1)! = 4! = 24

Answer: 24

Example 5: How many 5-digit numbers can be formed using the digits 0–7 (no repetition) such that the number starts with an odd digit?

Solution:

Odd digits available: 1, 3, 5, 7 → 4 choices for the first digit

Remaining digits: Choose 4 from remaining 7 digits (excluding 0 if not used yet)

• First digit = 4 choices
• Remaining 4 digits = ${}^7{P}_4=\frac{7!}{3!}=840$

→ Total = 4 × 840 = 3360

Answer: 3360

Example 6: How many 5-digit numbers can be formed using digits {1, 2, 3, 4, 5, 6} such that the digit 2 always appears before 5 (not necessarily adjacent), and no digit is repeated?

Solution:

• Step 1: Total 5-digit numbers from 6 digits: ${}^6{P}_5=720$
• Step 2: For each valid number, digit 2 and 5 are both present. We must count only the ones where 2 appears before 5.
• Total arrangements where 2 and 5 are both included:

Select 2 & 5 + 3 other digits from remaining 4 digits

→ ${}^4{C}_3$ ways to choose remaining

→ For each such selection: total permutations = 5! = 120

→ Total with 2 & 5 = 4 × 120 = 480

• Now, out of these, in half of the arrangements 2 will appear before 5.

→ Final answer = $\frac{480}{2}=240$

Answer: 240

Example 7: Find the number of 6-letter words (real or not) that can be formed using the letters of the word ‘SCHOOL’, such that the word starts with a consonant and ends with a vowel.

Solution:

Letters: S, C, H, O, O, L → 6 letters with ‘O’ repeated

• Vowels = {O}, repeated
• Consonants = {S, C, H, L}

Step 1: Fix the first letter = consonant (4 choices), last = vowel (only O)

Step 2: Fix first = one of S/C/H/L (4 ways), last = O (1 way)

Step 3: Remaining 4 letters from remaining 4 (excluding used ones), but we must manage repetition of O carefully.

Let’s consider different cases based on how many Os were used already:

Case 1: First = consonant (say S), last = O

→ Remaining letters = from C, H, L, O (excluding used letters)

Total = 4 letters to permute, with possibly one O → O may still be there

So:

• If O was used at end: 1 O used, 1 left → O still available once
• Remaining = 4 letters, with possibly one O repeated once → 4! /1! = 24

Repeat this for each starting consonant: 4 choices

→ Total = $4\times 24=96$

Answer: 96

Example 8: How many 5-digit numbers formed using digits 1–9 have digits in strictly increasing order?

Solution:

• Digits must be selected in increasing order → e.g., 12349, 13589, etc.
• So we need to select any 5 digits from 1–9 and arrange them in increasing order

→ only one way to arrange 5 digits in increasing order

• Select 5 digits from 9: ${}^9{C}_5=126$

Answer: 126

Example 9: How many different arrangements of the word ‘STATISTICS’ can be made?

Solution:

Letters: S T A T I S T I C S → 10 letters

Repetitions:

• S: 3
• T: 3
• I: 2
• A, C: 1 each

→ Total = $\frac{10!}{3!\times 3!\times 2!}=\frac{3628800}{6\times 6\times 2}=\frac{3628800}{72}=50400$

Answer: 50400

Example 10: 6 boys and 4 girls are to be seated in a row. No two girls can sit together. In how many ways can they be arranged?

Solution:

• Step 1: Place 6 boys first → 6! = 720
• Step 2: There are 7 gaps around boys (between and on both ends): _ B _ B _ B _ B _ B _ B _ → 7 places
• Step 3: Place 4 girls in these 7 gaps (no two girls together) → ${}^7{C}_4=35$
• Step 4: Arrange girls in those 4 chosen gaps → 4! = 24

→ Total = 720 × 35 × 24 = 604800

Answer: 604800

Example 11 (Without Repetition): How many ways can you arrange the letters A, B, and C?
Answer: 3! = 6

Permutations: ABC, ACB, BAC, BCA, CAB, CBA

Example 12 (With Repetition): How many 2-digit numbers can be formed using the digits 1, 2, 3 with repetition?

Answer: $3^2=9$

Permutations: 11, 12, 13, 21, 22, 23, 31, 32, 33

Also Solve : Permutations and Combinations Previous Year Question Papers with Solutions

6.0Applications of Permutation

• Cryptography and password generation
• Scheduling and logistics
• Game theory and decision-making
• Genetic arrangements in biology
• Problem-solving in competitive exams like JEE, CAT, and GRE

7.0Practice Questions on Permutation

1. How many ways can 5 students be arranged in a line?
2. In how many ways can the letters of the word “PINEAPPLE” be arranged?
3. Find the number of 3-letter codes from the word “DIGIT” if repetition is not allowed.
4. In how many ways can a committee of 3 be formed from 7 people?
5. How many 4-digit numbers can be made using digits 1, 2, 3, 5 if repetition is allowed?