Permutation (original) (raw)

Last Updated : 27 Apr, 2026

Permutation determines the number of possible arrangements for a specific set of elements. Therefore, it plays a big role in computer science, cryptography, and operations research.

**Note: In Permutations, order matters; for example, (2, 1) and (1, 2) are counted as different.

For example, take a set {1, 2, 3}:

Calculating permutations involves figuring out how many different ways you can arrange a set of items where the order matters.

**Permutation Formula

The permutation formula is used to calculate the number of ways to arrange a subset of objects from a larger set where the order of selection matters.

The formula for Permutation is given as follows,

Permutation

Some of the most common representations or notations are as follows:

**Derivation of Permutation Formula

To derive the formula for permutation, we can use the first principle of counting

If an event can occur in m different ways, and another event can occur in n different ways, then the total number of occurrences of the events is m × n.

By the definition of permutation and the principle of counting, we know

nPr = n × (n - 1) . . . (n - r + 1)

This product is exactly:

P(n, r) = n! / (n−r)!

Note that there are n ways to pick an item for the first position, (n - 1) ways to pick the second and so on

Multiplying and dividing by (n - r)! on the LHS, we get

nPr = n × (n - 1) × (n - 2) × . . . × (n - r + 1) × (n - r)! / (n - r)!

⇒ nPr = n × (n - 1) × (n - 2) × . . . × (n - r + 1) × (n - r) × (n - r - 1) × . . . × 1 / (n - r)!

⇒ nPr = n! / (n - r)! where 0 ≤ r ≤ n

**Properties of Permutations

Some of the common properties of permutations are listed as follows:

Types of Permutation

In the study of permutation, there are some cases such as:

**1. Permutation With Repetition

This is the simplest of the lot. In such problems, the objects can be repeated. Let's understand these problems with some examples.

**Example: How many 3-digit numbers greater than 500 can be formed using 3, 4, 5, and 7?

Since a three-digit number greater than 500 will have either 5 or 7 at its hundredth place, we have **2 choices for this place.

There is no restriction on the repetition of the digits hence, for the remaining 2 digits****;** we have **4 **choices for each

So the total permutations are,

2 × 4 × 4 = 32

**2. Permutation Without Repetition

In this class of problems, the repetition of objects is not allowed. Let's understand these problems with some examples.

**Example: How many 3-digit numbers divisible by 3 can be formed using the , digits 2, 4, 6, and 8 without repetition?

For a number to be divisible by 3, the sum of it digits must be divisible by 3

From the given set, various arrangements like 444 can be formed but since repetition isn't allowed we won't be considering them.

We are left with just 2 cases i.e. 2, 4, 6 and 4, 6, 8

Number of arrangements are 3! in each case

Hence the total number of permutations are: 3! + 3! = 12

**3. Permutation of Multi-Sets

A permutation of a multiset is used when the objects are not distinct

**Example: Find the number of arrangements of the word BALLOON.

Number of distinct permutations: \frac{7!}{2! \cdot 2!}

**Relation Between **n P r and **n C r

We can understand nCr through the following analogy. Consider that we have n distinct boxes and r identical balls. (n > r)

The task is to place all the r balls into boxes such that no box contains more than 1 ball.

As all r objects are the same, the r! Ways of arranging them can be considered as a single way.

To group all r! ways of arranging, we divide nPr by r!

**nCr = nPr/r! = n!/{(n - r)! × r!}

Hence, the relation betweenn P rand**n** C r is,

**nCr = nPr/r!

**Permutation vs Combination

The key differences between permutation and combination, some of those differences are listed as follows:

Permutations Combinations
Arrangements of elements in a specific order. Selections of elements without considering the order.
nPr = n!​/(n−r)! nCr = n!​/[(n−r)! × r!]
nPr OR P(n, r) nCr OR C(n, r)
Yes, order matters. No, the order doesn't matter.
Arranging books on a shelf. Selecting members for a committee.
How many ways to arrange 3 books out of 5? How many ways to choose 2 fruits from a basket of 7?
Permutations are used when order matters, such as when arranging items in a sequence or forming a code. Combinations are used when order doesn't matter, like selecting a group of people or choosing items without caring about their order.

Solved Problems

**Problem **1. Find 6P3

**Solution

As per the formula,

nPr = n! / (n - r)!
6P3 = 6! / (6−3)!
= 6 × 5 × 4 = 120

**Problem 2. Find n if nP2 = 12

**Solution

nPr = n! / (n - r)!

⇒ nP2 = n! / (n - 2)!
⇒ nP2 = n × (n - 1) × (n - 2)! / (n - 2)!
⇒ nP2 = n × (n - 1)
⇒ nP2 = n2 - n
∴ n2 - n = 12

Solving the equation,

n2 - n - 12 = 0
⇒ n (n - 4) + 3 (n - 4) = 0
⇒ (n + 3) (n - 4) = 0
∴ n = -3 or n = 4
∵ n ≥ 0

Thus, n = 4

**Problem 3. How many 4-letter words, with or without meaning? Can be formed out of the letters of the word, 'SATURDAY' if repetition of letters is not allowed?

**Solution:

Word SATURDAY has 8 letters i.e. S, A, T, U, R, D, A, and Y

To form 4-letter words, we first have to select 4 letters from these 8 letters

The ways of selecting 4 letters from 8 letters disregarding the order is 8C4 .

After selection, there are 4! arrangements.

Hence, total number of words formed are: 8C4 × 4!

**Note: Selecting r objects out of n objects and then arranging them is same as r-permutation of n objects.

**Problem 4. Find the number of ways of selecting 6 balls from 4 red, 6 blue, and 5 white given that the selection must have 2 balls of each color.

**Solution:

We need to select 2 balls each of color red, blue and white as per the given condition.

Number of ways of selecting 2 red balls is 4C2

Number of ways of selecting 2 blue balls is 6C2

Number of ways of selecting 2 white balls is 5C2

Hence, the total ways of selection are 4C2 × 6C2 × 5C2 = 900

**Problem 5. A class has just 3 seats vacant. Three people, P, A, and R, arrive at the same time. In how many ways can P, A, and R be arranged on those 3 vacant seats?

**Solution:

For the very first seat, we have 3 choices i.e. P, A and R.

Let us randomly select A for the first seat.

For the second seat, we have 2 choices i.e. P and R

Let us randomly select R for the second seat.

For the third seat, we have 1 choice i.e. P

To summarize, we did the following:

Placed a person on seat 1 and placed a person on seat 2 and placed a person on seat 3.

Usage of and comes from the fact that occupation of all 3 seats was mandatory.

In mathematics, and is related with multiplication, hence we can say that total choices = 3 × 2 × 1 = 3!

If we change the seating order to P on the first seat, A on the second seat, and R on the third, does that change the total number of choices?

No, it does not. This is because equal importance is given to all three P, A, and R.

**Problem 6. Find the number of ways of arranging 5 people if 2 of them always sit together.

**Solution:

Let us consider the 2 people as a unit and the remaining 3 person as 3 separate units, So we have total 4 units.

Number of ways of arranging these 4 units is 4!

(just the way we proved in previous problem)

Number of ways of arranging the 2 person amongst themselves is 2!

In conclusion, the number of ways of arranging the 4 units and 2 person amongst themselves is 4! × 2!

**Problem 7. 10 Olympians are running a race. Find the different arrangements of 1st, 2nd, and 3rd place possible?

**Solution:

We have to find different arrangements of 10 taken 3 at time.
Here,

Different arrangement of for 1, 2, and 3 places are

10P3 = 10! / (7!)
= 10 × 9 × 8 = 720

**Problem 8. Find all the three-letter words beginning and ending with a vowel. Given that repetition of alphabets is not allowed.

**Solution:

Total vowels = 5 (a, e, i, o, u)

First letter (vowel) = 5 choices
Third letter (vowel, no repetition) = 4 choices
Middle letter (any remaining letter) = 26 − 2 = 24 choices

Total words = 5 × 4 × 24 = 480

**Problem 9. An ice-cream shop has 10 flavors of ice cream. Find the number of ways to arrange 3 different flavors in layers on an ice cream cone.

**Solution:

Let us consider n = 10 (total number of flavors) and r = 3 (number of different flavors needed)

For first flavor we have 10 choices

For second flavor we have 10 - 1 choices

For third flavor we have 10 - 2 choices and this is same as (n - r + 1)

The numbers of arrangement would be: 10 × (10 - 1) × (10 - 3 + 1) = 720

From this we can generalize that, the number of ways of arranging r objects out of n different objects is:

n × (n - 1) . . . (n - r + 1) = nPr

**Problem 10. How many even numbers lying between 1000 and 2000 can be formed using the digits 1, 2, 4, 5, and 9?

**Solution:

Since the number is supposed to be even, the digits at units place must either be 2 or 4 leaving us with 2 choices for the digit at units place.

The number is supposed to lie between 1000 and 2000, So the digits at thousand's place must be 1, we thus have

1 choice for the digit at thousands place.

Rest of the 2 digits can be any one from 1, 2, 4, 5 and 9 i.e. 5 choices each

So the total permutations are: 2 × 5 × 5 × 1 = 50

**Problem 11. How many 4-digit numbers divisible by 5 can be formed using 0, 3, 5, 7, and 9 if repetition of digits is not allowed?

**Solution:

For the number to be divisible by 5, the digit at units place must either be 0 or 5, hence we have 2 possibilities.

**Case 1. Digit at units place is 0

Hence the possible arrangements with 0 at units place are

4 × 3 × 2 = 24

**Case 2. Digit at units place is 5

Hence the possible arrangements with 5 at units place are 3 × 3 × 2 = 18
Total Arrangements = Number of arrangements in case 1 + Number of arrangements in case 2
Total Arrangements = 24 + 18 = 42

Practice Problems

**Question 1. In how many ways can 6 prisoners be placed in 4 cells if any number of prisoners can fit in a cell?

**Question 2. Find how many 4-digit numbers divisible by 8 can be formed using 0, 1, 2, 3, 5, 7, and 9 if repetition of digits is not allowed.

**Question 3. Find the number of ways of selecting 8 balls from 10 red, 16 blue, and 15 white given that the selection must have 1 ball of each color.

**Question 4. Find how many even numbers lying between 4000 and 8000 can be formed using the digits 1, 2, 3, 4, 5, and 6 when repetition is allowed.