Suppose you have a suitcase with a number lock. The number lock has $5$ wheels each labeled with digits from $0$ to $9$. You forget the sequence of the digits to open the lock but remember only the first digit. In order to open the lock, how many sequences of $4$ digits you may have to check with?

To answer this question, you may, immediately, start listing all possible arrangements of $9$ remaining digits taken $4$ at a time. But, this method will be tedious, because the number of possible sequences may be large. In such situations, using permutations is an easy and time-saving way of finding the total number of possible arrangements.

Let’s understand what is permutation and their types, and the formulas used to calculate with examples.

## What is Permutation?

A permutation is an arrangement of objects in a particular way or order. While using permutation one should be concerned about the selection as well as the arrangement. Ordering is very important in permutations. In other words, the permutation is considered an ordered combination.

### Examples of Permutation

**Example 1: **Find the number of $4$-letter words, with or without meaning, which can be formed out of the letters of the word $\text{YOUR}$, where the repetition of the letters is not allowed.

To find the different arrangements, let first consider a sequence of $4$ blank spaces – _____ _____ _____ _____.

Starting with the first leftmost blank. The number of choices available for this blank space is $4$, (either of ‘Y’, ‘O’, ‘U’, and ‘R’).

The number of choices available for the second blank space is $3$. (As one letter is already used in the first blank space and repetition of letters is not allowed).

Similarly, the number of choices available for the third blank space is $2$.

And finally, the number of choices available for the fourth blank space is $1$.

Thus by principle of counting the total number of ways is $4 \times 3 \times 2 \times 1 = 4! = 24$

**Example 2:** How many $2$-digit numbers can be formed from the digits $1$, $2$, $3$, $4$, $5$ if

a) the digits can be repeated

b) the digits cannot be repeated

The total number of digits is $5$

Let first consider a sequence of $2$ blank spaces – _____ _____. (Representing a $2$-digit number)

a) Starting with the first leftmost blank. The number of choices available for this blank space is $5$, (either of $1$, $2$, $3$, $4$, or $5$).

The number of choices available for the second blank space is $5$. (As digits can be repeated).

Thus the total number of $2$-digit numbers that can be formed using the digits $1$, $2$, $3$, $4$, $5$ if the digits can be repeated is $5 \times 5 = 25$.

b) Starting with the first leftmost blank. The number of choices available for this blank space is $5$, (either of $1$, $2$, $3$, $4$, or $5$).

The number of choices available for the second blank space is $4$. (As digits cannot be repeated).

Thus the total number of $2$-digit numbers that can be formed using the digits $1$, $2$, $3$, $4$, $5$ if the digits cannot be repeated is $5 \times 4 = 20$.

**Example 3:** How many $2$-digit even numbers can be formed from the digits $1$, $2$, $3$, $4$, $5$ if

a) the digits can be repeated

b) the digits cannot be repeated

The total number of digits is $5$

Let first consider a sequence of $2$ blank spaces – _____ _____. (Representing a $2$-digit number)

a) Starting with the rightmost blank(units place). Since an even number is required, we have only $2$ choices from the $5$ given digits(viz., $2$ and $4$)

The number of choices available for the second blank space is $5$. (As digits can be repeated and any digit can come at tens place).

Thus the total number of $2$-digit even numbers that can be formed using the digits $1$, $2$, $3$, $4$, $5$ if the digits can be repeated is $2 \times 5 = 10$.

b) Starting with the rightmost blank(units place). Since an even number is required, we have only $2$ choices from the $5$ given digits(viz., $2$ and $4$)

The number of choices available for the second blank space is $4$. (As digits cannot be repeated and any digit can come at tens place).

Thus the total number of $2$-digit even numbers that can be formed using the digits $1$, $2$, $3$, $4$, $5$ if the digits cannot be repeated is $2 \times 4 = 8$.

## Permutation Formula

The formula for permutation of $n$ objects selected $r$ at a time is given by $^{n} \text{P}_{r} = \frac{n!}{(n – r)!}$.

**Note:** Other ways of representing permutation:

- $\text{P}(n, r)$
- $\text{P}_{r}^{n}$
- $_{n} \text{P}_{r}$
- $^{n} \text{P}_{r}$

### Permutation Formula Examples

**Example 1:** Evaluate $^{6} \text{P}_{2}$

$^{6} \text{P}_{2} = \frac{6!}{(6 – 2)!} = \frac{6!}{4!} = \frac{6 \times 5 \times 4!}{4!} = 30$

**Example 2:** Find $r$, if $^{5} \text{P}_{r} = 2 \times ^{6} \text{P}_{r – 1}$

$^{5} \text{P}_{r} = 2 \times ^{6} \text{P}_{r – 1} => \frac{5!}{(5 – r)!} = 2 \times \frac{6!}{(6 – (r – 1))!}$

$=> \frac{5!}{(5 – r)!} = 2 \times \frac{6!}{(6 – r + 1)!}$

$=> \frac{5!}{(5 – r)!} = 2 \times \frac{6!}{(7 – r)!}$

$=> \frac{(7 – r)!}{(5 – r)!} = 2 \times \frac{6!}{5!}$

$=> \frac{(7 – r) \times (6 – r) \times (5 – r)!}{(5 – r)!} = 2 \times \frac{6 \times 5!}{5!}$

$=> (7 – r) \times (6 – r) = 12 => (7 – r) \times (6 – r) = 4 \times 3$

$=> 7 – r = 4$ or $6 – r = 3$

$=> r = 3$

## Types of Permutations

Permutation can be classified into three different categories:

- Permutation of $n$ different objects (when repetition is not allowed)
- Repetition, where repetition is allowed
- Permutation when the objects are not distinct (Permutation of multisets)

### Permutation of $n$ Different Objects

If $n$ is a positive integer and $r$ is a whole number, such that $r \lt n$, then $^{n} \text{P}_{r}$ represents the number of all possible arrangements or permutations of $n$ distinct objects taken $r$ at a time. In the case of permutation without repetition, the number of available choices will be reduced each time. It can also be represented as $^{n} \text{P}_{r}$.

$^{n} \text{P}_{r} = n(n-1)(n-2)(n-3)… \text{ upto } r \text{ factors}$

$^{n} \text{P}_{r} = n(n-1)(n-2)(n-3)…(n – r +1)$

$=> ^{n} \text{P}_{r} = \frac{n!}{(n – r)!}$.

Here, $^{n} \text{P}_{r}$ represents the $n$ objects to be selected from $r$ objects without repetition, in which the order matters.

#### Examples of Permutation of $n$ Different Objects

**Example 1: **How many $4$ letters words can be formed out of the letters of the word “CABLES” when repetition is not allowed?

The word “CABLES” has six letters so $n = 6$ and we have to form all possible four-letter words without repetition hence $r = 6$. Therefore, possible permutations can be calculated as $^{n} \text{P}_{r} = \frac{n!}{(n – r)!} = \frac{6!}{(6 – 4)!} = \frac{6!}{2!} = 6 \times 5 \times 4 \times 3 = 360$.

**Example 2:** How many signals can be given by $6$ lights of different colours while using $3$ lights at the same time?

The total number of lights $n = 6$

The number of used lights at a time $r = 3$

$^{6} \text{P}_{3} = \frac{6!}{(6 – 3)!} = \frac{6!}{(3)!} = \frac{6 \times 5 \times 4 \times 3! }{3!} = 6 \times 5 \times 4 = 120$.

Therefore, the total number of signals formed using $6$ lights of different colours while using $3$ colours at a time is $120$.

**Example 3:** How many $4$-digit numbers can be formed by using the digits $3$, $4$, $5$, $6$, $7$, $8$, and $9$ only once?

The total number of digits = $7$

The number of digits used at a time = $4$

$^{7} \text{P}_{4} = \frac{7!}{(7 – 4)!} = \frac{7!}{3!} = \frac{7 \times 6 \times 5 \times 4 \times 3!}{3!} = 7 \times 6 \times 5 \times 4 = 840$.

Therefore, the number of $4$-digit numbers can be formed by using the digits $3$, $4$, $5$, $6$, $7$, $8$, and $9$ only once is $840$.

**Example 4:** How many words can be formed by using all the letters of the word ‘PENCIL’? (Repetition of words is not allowed).

The word “PENCILS” has six letters so $n = 7$ and we have to form all possible words using all the letters, therefore, $r = 7$.

Therefore, possible permutations can be calculated as $^{n} \text{P}_{r} = \frac{n!}{(n – r)!} = \frac{7!}{(7 – 7)!} = \frac{7!}{0!} = \frac{7!}{1} = 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$.

### Permutation When Repetition is Allowed

The permutation with repetition of objects can be written using the exponent form.

When the number of objects is $n$, and we have $r$ to be the selection of an object, then choosing an object can be done in $n$ different ways (each time).

Thus, the permutation of objects when repetition is allowed will be equal to $n \times n \times n \times ……(r \text{ times}) = n^r$

This is the permutation formula to compute the number of permutations feasible for the choice of “r” items from the “n” objects when repetition is allowed.

#### Examples of Permutation When Repetition is Allowed

**Example 1: **How many 4-letter words can be formed out of the letters of the word “CABLES” when repetition is allowed?

Permutation in the case of repetition is calculated in exponential form. If the number of objects is $n$ and the number $r$

is to be selected from $n$ objects, and if repetition is allowed, then permutation is expressed as $n^r$.

So, in this case, the number of letters is $6$ so $n = 6$ while $r = 4$.

Permutation = $6^4 = 1296$

Thus, $1296$ 4-letter words can be formed out of the letters of the word “CABLES” when repetition is allowed

### Permutation of Multi-Sets

Permutation of $n$ different objects when $m_1$ objects among $n$ objects are similar, $m_2$ objects of the second kind are similar, $m_3$ objects of the third kind are similar and so on, $m_k$ objects of the $k^{th}$ kind are similar and the remaining of all are of a different kind. Thus it forms a multiset, where the permutation is given as $\frac{n!}{m_1 ! \times m_2 ! \times m_3 ! \times … m_k !}$.

#### Examples of Permutation of Multi-Sets

**Example 1:** Find the number of permutations of the letters of the word ‘ALLAHABAD’.

Here, there are $9$ objects (letters) of which there are $4$ A’s, $2$ L’s, and the rest are different, therefore, $n = 9$, $m_1 = 4$, $m_1 = 2$

Therefore, the required number of arrangements = $\frac{n!}{m_1 ! \times m_2 ! } = \frac{9!}{4! \times 2! }$

$ = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4! \times 2 \times 1} = \frac{9 \times 8 \times 7 \times 6 \times 5}{2 \times 1} = 9 \times 4 \times 7 \times 6 \times 5 = 7560$.

**Example 2:** In how many of these arrangements, do the words start with $\text{P}$ in the arrangements of the letters of the word $\text{INDEPENDENCE}$?

There are $12$ letters, of which $\text{N}$ appears $3$ times, $\text{E}$ appears $4$ times, and $\text{D}$ appears $2$ times and the rest all different. Therefore, the total number of arrangements = $\frac{12!}{3! \times 4! \times 2!} = 1663200$.

Let us fix $\text{P}$ at the extreme left position, we, then, count the arrangements of the remaining $11$ letters. Therefore, the required number of words starting with $\text{P}$ is $\frac{11!}{3! \times 4! \times 2!} = 138600$.

## Difference Between Permutation and Combination

The following are the major differences between permutation and combination

Permutation | Combination |

The permutation means the selection of objects, where the order of selection matters. | The combination means the selection of objects, in which the order of selection does not matter. |

It is the arrangement of $r$ objects taken out of $n$ objects | It is the selection of $r$ objects taken out of $n$ objects irrespective of the object arrangement. |

The formula for permutation is $^{n} \text{P}_{r} = \frac{n!}{(n – r)!}$ | The formula for combination is $^{n} \text{C}_{r} = \frac{n!}{r!(n – r)!}$ |

## Practice Problems

- How many numbers are there between 99 and 1000 having 7 in the unit’s place?
- How many numbers are there between 99 and 1000 having at least one of their digits 7?
- In how many ways can 5 children be arranged in a line such that
- two particular children of them are always together
- two particular children of them are never together.

- If all permutations of the letters of the word AGAIN are arranged in the order as in a dictionary. What is the 49th word?
- In how many ways 3 mathematics books, 4 history books, 3 chemistry books, and 2 biology books can be arranged on a shelf so that all books of the same subjects are together?
- All the letters of the word ‘EAMCOT’ are arranged in different possible ways. Find the number of such arrangements in which no two vowels are adjacent to each other.
- Ten different letters of the alphabet are given. Words with five letters are formed from these given letters. Find the number of words that have at least one letter repeated.
- Find the number of signals that can be sent by 6 flags of different colours taking one or more at a time.

## FAQs

### What is the mathematical definition of permutations?

The number of ways of selecting and arranging $r$ things out of $n$ things is called the number of permutations. Wherever arrangement has importance, we have to use the permutation.

### What is the formula for permutation?

The formula for permutation is $^{n} \text{P}_{r}$, where $n$ is the total number of objects and $r$ is the number of objects taken at a time.

### What are the different types of permutations?

There are three types of permutations.

a) Arrangement of $n$ different objects where repetition is not allowed

b) Arrangement of $n$ different objects where repetition is allowed

c) Permutation of multisets

### What is the permutations of 5?

Permutation of $n$ objects taken $r$ at a time is given by $^{n} \text{P}_{r} = \frac{n!}{(n – r)!}$. Here $n = 5$ and $r = 5$, therefore, permutation of 5 is $^{5} \text{P}_{5} = \frac{5!}{(5 – 5)!} = 5! = 120$.

### What is the permutation without repetition?

The formula used for permutation without repetition is $^{n} \text{P}_{r} = \frac{n!}{(n – r)!}$, where $r$ different things are taken from $n$ different things and repetition is not allowed.

### What is the permutation with repetition?

The formula used for permutation with repetition is $n^r$, where $r$ different things are taken from $n$ different things, and repetition is allowed.

### Is $^{n} \text{P}_{r}$ and $^{n} \text{C}_{r}$ the same?

$^{n} \text{P}_{r}$ is calculating the permutations as arrangements where the order matters, whereas, $^{n} \text{C}_{r}$ is calculating the combinations, where the order doesn’t matter.

## Conclusion

A permutation is an arrangement of objects in a particular way or order. While using permutation one should be concerned about the selection as well as the arrangement. Ordering is very important in permutations. There are three types of permutations. These are the arrangement of $n$ different objects where repetition is not allowed, the arrangement of $n$ different objects where repetition is allowed, and the permutation of multisets.