Permutation

In Mathematics, Permutation is defined as a mathematical concept that 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.

For example, take a set {1, 2, 3}:
All Permutations taking all three objects are {1, 2, 3}, {1, 3, 2}, {2, 1, 3}, {2, 3, 1}, {3, 1, 2}, {3, 2, 1}.
All Permutations taking two objects at a time are, {1, 2}, {1, 3}, {2, 3}, {3, 2}, {3, 1}, {2, 1}.
Note: In Permutations, order matters, for example (2, 1) and (1, 2) are counted as different.

How to Calculate Permutations

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,

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

• P(n, r)
• _nP_r
• ⁿP_r
• P_{n, k}

What is Factorial?

The factorial of a natural number n is denoted by the notation n! which represents the product of all natural numbers starting from 1 till n, including 1 and n. i.e. n! = n × (n-1) × (n-2) × (n-3) . . . × 1.

For example, If n = 3, then 3! = 3 × 2 × 1 = 6, and If n = 5 then 5! = 5 × 4 × 3 × 2 × 1 = 120.

Note: The factorial of 0 is defined as 1 by convention, i.e.

0! = 1

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

ⁿP_r = 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

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

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

⇒ ⁿP_r = n! / (n - r)! where  0 ≤ r ≤ n

Permutation Examples in Real Life

Some real-life examples of Permutation include password generation, seating arrangements, shuffling cards, phone numbers, lock combinations, etc.

Properties of Permutations

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

• ⁿP_n = n (n-1) (n-2) . . . 1 = n!
• ⁿP₀ = n! / n! = 1
• ⁿP₁ = n
• ⁿP_{n - 1} = n!/1
• ⁿP_r / ⁿP_r-1 = n - r + 1
• ⁿP_r =n × ^n-1P_r-1 = n × (n-1) × ^n-2P_r-2 = n × (n-1) × (n-2) × ^n-3P_r-3 = and so on.
• ^n-1P_r + r × ^n-1P_r-1 = ⁿP_r

Solved Examples of Permutation

Let's consider some problems based on the derived formula to better understand its use.

Example 1: Find ⁶P₃

As per the formula,

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

Example 2: Find n if ⁿP₂ = 12

ⁿP_r = 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

Types of Permutation

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

• Permutation with Repetition
• Permutation without Repetition
• Permutation of Multi-Sets

Let's explain these cases in detail with solved examples as follows:

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

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

Permutation of Multi-Sets

A permutation is when the objects are not distinct

This can be thought of as the distribution of n objects into r boxes where the repetition of objects is allowed and any box can hold any number of objects.

1^st  box can hold n objects

2^nd box can hold n objects

3^rd box can hold n objects
.                                          . 
.                                          . 
.                                          . 

r^th box can hold n objects

Hence the total number of arrangements is,

n × n × n . . . (r times) = n^r

Examples: A police officer visits the crime scene 3 times a week for investigation. Find the number of ways to schedule his visit if there is no restriction on the number of visits per day?

Number of ways to schedule first visit is 7 (any of the 7 days)

Number of ways to schedule second visit is 7 (any of the 7 days)

Number of ways to schedule third visit is 7 (any of the 7 days)

Hence, the number of ways to schedule first and second and third visit is

7 × 7 × 7 = 7³ = 343

Permutation and Combination

Permutation and Combination are the two most fundamental concepts in combinatorics, as these lay the foundation of this branch of mathematics. Combinatorics deals with all the mathematics which we need to solve problems based on selection, arrangement, and operation within a finite or discrete system.

Relation Between ⁿP_r and ⁿC_r

We can understand ⁿC_r 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 ⁿP_r by r!

nCr = nPr/r! = n!/{(n - r)! × r!}

Hence, the relation between ⁿP_r and ⁿC_r is,

nCr = nPr/r!

Permutation vs Combination

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

Learn more: Permutation vs Combination

Fundamental Counting Principle

Fundamental principle of the counting states that, "To perform a operation if we have m ways and to perform second operation if we have n different ways then to perform both the operation together there are m × n different ways."

We can also extend the principle of counting for more operations such that, if for 1 operation we have m, 2 operation we have n, for third operation we have o, different ways respectively than, for performing all the operation together we have m × n × o ways.

People Also Read:

• Combinations
• Binomial Theorem
• Probability

Permutation and Combinations Class 11

By understanding these above concepts and practicing various problems, you'll be well-prepared to tackle questions on permutations and combinations in your Class 11 math exams.

Resources related to Permutations Class 11:

• Permutations and Combinations Class 11 Notes
• Permutations and Combinations Class 11 NCERT Solutions

Solved Problems of Permutation

Problem 1. 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 2. 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 3. 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 4. 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 5. Find all the three-letter words beginning and ending with a vowel. Given that repetition of alphabets is not allowed.

Solution:

Total vowels in english = 7 ( a, e, i, o, u, y, w)

Total consonants in english = 26 - 7 = 19

• Choices for the first letter are 7
• Choices for the third letter are 6 (since 1 vowel was placed as first letter)
• Choices for the middle letter are 19 + (7 - 2) = 24 (19 consonants + the vowels which were not placed)

Hence, total permutations are 7 × 6 × 24 = 1008

Do observe that here we first satisfied the vowel condition for first and third letter and then placed the middle letter.

Problem 6. An ice-cream shop has 10 flavors of ice cream. Find the number of ways of preparing an ice cream cone with any 3 different flavors.

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) = ⁿP_r

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

Solution:

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

• n = 10
• r = 3

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

¹⁰P₃ = 10! / (7!)
= 10 × 9 × 8 = 720

Problem 8. 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 9. 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

• There are 4 choices for 10³ place (all numbers except 0)
• There are 3 choices for the 10² place (1 got used up at 10³ place)
• There are 2 choices for the 10¹ place (1 got used up at 10³ place and 1 at 10² place)

Hence the possible arrangements with 0 at units place are

4 × 3 × 2 = 24

Case 2. Digit at units place is 5

• There are 3 choices for 10³ place (all except 0 and 5)
• There are 2 choices for 10² place (1 got used up at 10³ place)
• There is 1 choice for 10¹ place (1 got used up at 10³ place and 1 at 10² place)

Hence the possible arrangements with 5 at units place are 3 × 2 × 1 = 6
Total Arrangements = Number of arrangements in case 1 + Number of arrangements in case 2
Total Arrangements = 24 + 6 = 30

Practice Problems on Permutation

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.

Conclusion

Permutations and Combinations are extremely important in many problems in mathematics where arrangement or selection is involved. Mastering permutation formulae and being able to solve permutation questions builds a person with all the tools needed to break down analysis and predict the possibility of outcomes in many practical situations, from computer science to event planning. Problems on permutations and combinations and mastering their differences allow one to deal efficiently with complex problems of ordering or selection of elements of exact order or selection.