# Permutations

Let us assume that you have 6 chairs and 4 people. If you randomly choose 4 chairs out of these 6, in how many ways can you arrange the 4 people?

Well, rather than taking each case one by one, permutation helps us solve these types of questions.

Let us have a look at how different choices of arrangements can affect our total number of outcomes.

## Lesson Plan

 1 What do You Mean by Permutation 2 Important Notes on Permutation 3 Solved Examples on Permutation 4 Challenging Questions on Permutation 5 Interactive Questions on Permutation

## Definition of Permutation

Suppose you need to arrange n people in a row. How many distinct arrangements are possible?

We have n options to fill the first chair, $$\left( {n - 1} \right)$$ options to fill the second chair, $$\left( {n - 2} \right)$$ options to fill the third chair, and so on. Thus, using the FPC (Fundamental Principle of counting), the total number of arrangements of the n people in the row of n chairs is

$\left( n \right) \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times ... \times 1 = n!$

Each particular arrangement is termed as a permutation of the n people. Hence, there will be a total of n! different permutations / arrangements.

## Formula of Permutation

Now, we introduce a new notation. The symbol $$^n{P_r}$$ is used to denote the number of permutations of n distinct objects, taken r at a time. Thus, permutation formula can be given as :

 $^n{P_r} = \frac{{n!}}{{\left( {n - r} \right)!}}$

Examples

\begin{align}&^5{P_2} = \frac{{5!}}{{\left( {5 - 2} \right)!}} = \frac{{5!}}{{3!}} = \frac{{120}}{6} = 20\\&^6{P_5} = \frac{{6!}}{{\left( {6 - 5} \right)!}} = \frac{{6!}}{{1!}} = \frac{{120}}{1} = 120\\&^7{P_2} = \frac{{7!}}{{\left( {7 - 2} \right)!}} = \frac{{7!}}{{5!}} = \frac{{7 \times 6 \times 5!}}{{5!}} = 42\end{align}

## Permutation Calculator

Enter the values of n and r in the Permutation and combination calculator shown below to calculate the value of the permutations and combinations

## What are the types of Permutations

There are two types of permutations.

### Permutation with repetition

This formula is used to find the statistics of permutation (number of possible ways in which arrangement can be done) while allowing repetition.

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

### Permutation without repetition

This formula is used to find the statistics of permutation (number of possible ways in which arrangement can be done) while ensuring that there is no repetition.

 $$P\left ( n,r \right )$$ = $$n^{r}$$

Important Notes
1. We can associate the word permutations with the word arrangements. When you read the phrase “number of permutations”, think of the phrase “number of arrangements”.

2. The number of permutations of n distinct objects taken altogether is $$n!$$

3. The number of permutations of n distinct objects, taken r at a time (where r is less than n), is $$^n{P_r} = \frac{{n!}}{{\left( {n - r} \right)!}}$$

## Solved Examples

 Example 1

Emily has 4 chairs and she wants to place 3 dolls on these chairs. In how many possible ways can she do this?

Solution

Given n = 4 and r = 3

Applying the permutation formula

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

We get,

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

\begin{align}P &= \dfrac{n!}{(n-r)!}\\ & = \dfrac{4!}{(4-3)!}\\ &= \dfrac{4!}{1!}\\ &= \dfrac{4\times 3\times 2\times 1}{1}\\ &= 24\end{align}

So, there are 24 ways in which she can place 3 dolls on the 4 chairs.

 $$\therefore$$ There are 24 such possible ways.
 Example 2

Consider a set of 5 alphabets p, q, r, s, t.

In how many ways can 4 alphabets be selected without repetition?

Solution

The set of alphabets are p, q, r, s, t.

4 alphabets are to be selected.

$$\therefore$$  $$p_{r}^{n}$$ = $$p_{4}^{5}$$

= $$\dfrac{n!}{(n-r)!}$$

= $$\dfrac{5!}{(5-4)!}$$

= $$5\times 4\times 3\times 2\times 1$$

= 120

 $$\therefore$$ There are total 120 such ways.
 Example 3

How many 5-digit codes can be created using the digits from 0 to 9 if repetition is allowed?

Solution

Given n = 10 and r = 5

\begin{align}P\left ( n,r \right ) &= n^{r}\\ & =10^{5}\\ &=100,000\end{align}

 $$\therefore$$ 100,000 five-digit codes can be created.
 Example 4

Consider the word EDUCATION. Now, we have to rearrange these letters to form different words, but we want all those permutations in which the string “CAT” occurs.

Solution

Let us treat “CAT” as a single letter/object since this is what we want - we want “CAT” to appear as a single entity. We now have the following objects (letters) which we need to permute:

$\text{E D U “CAT ” I O N}$

Now, there are 7 objects in all.

Thus, they can be permuted in   $${}^7{P_7} = 7!$$  ways. This is the number of permutations that will be having the string “CAT”.

 $$\therefore$$ There are a total of $${}^7{P_7} = 7!$$ ways.
 Example 5

There are 10 students in a class. Two of these students are Jack and Daniel, who don’t get along very well. In how many ways can the teacher arrange the students in a row, so that Jack and Daniel are not together?

### Solution

The total number of arrangements of the 10 students is $$^{10}{P_{10}} = 10!$$.

Now, let us count the number of arrangements in which Jack and Daniel are together. Treating Jack + Daniel as a single unit (call it JD), we have a total of 9 entities which we can permute: the 8 students, and JD. These 9 entities can be permuted in $$^9{P_9} = 9!$$ ways. But for each of 9! permutations, J and D can be permuted among themselves in 2! or 2 ways: JD or DJ. Thus, the total number of permutations in which Jack and Daniel are together, is $$2 \times 9!$$.

We conclude that the number of permutations in which Jack and Daniel are not together, is

$10! - \left( {2 \times 9!} \right)$

 $$\therefore$$ The total arrangements are $10! - \left( {2 \times 9!} \right)$.

Challenging Questions

Find the number of all the possible 4-digit numbers formed using the digits 3, 4, 5, and 6 using each digit once.

Will there be a difference in the final answer if the repetition of the digits is allowed only once?

Using the same logic, can you formulate the number of all the possible 4-digit numbers formed if the repetition of digits is:

1. Not allowed
2. Allowed once
3. Allowed any number of times

## Interactive Questions

Here are a few activities for you to practice.

### 1. What is the difference between  Permutation and Combination?

Ans. Permutation differs with the number of arrangements, while combination differs with the number of selections.

### 2. What are the Signs of  Permutation?

Ans. The sign of permutation is $$^n{P_r}$$.

Combinatorics
grade 9 | Questions Set 1
Combinatorics
Combinatorics
grade 9 | Questions Set 2
Combinatorics