Permutation with Repetition: Definition and Formula

What is Permutation with Repetition?

Permutation with repetition is a counting technique used when arranging objects and the same object is allowed to appear more than once. Unlike ordinary permutations, where each object can be used only once, permutation with repetition allows repeated selections, resulting in a larger number of possible arrangements. This concept is widely used in combinatorics, probability, coding theory, computer science, and competitive examinations.

Permutation with Repetition Meaning in Simple Words

Permutation with repetition means arranging objects where an item can be selected multiple times.

For example, if you are creating a 4-digit PIN using the digits $0$ to $9$, each digit can be used repeatedly. Therefore, numbers like $1111$, $2020$, and $9999$ are all valid arrangements.

Definition of Permutation with Repetition

Permutation with repetition refers to the number of ordered arrangements that can be formed when each object may be selected more than once.

If there are $n$ different objects and $r$ positions to fill, then the total number of arrangements is

$n^r$

where:

• $n$ is the number of available objects.

• $r$ is the number of positions to be filled.

Why Permutation with Repetition is Important

Permutation with repetition is important because many real-life arrangements allow repeated choices. It is widely used in:

• Password generation.

• PIN and OTP creation.

• Computer programming.

• Coding theory.

• Probability problems.

• Artificial intelligence and machine learning.

It is also a frequently tested topic in JEE, CAT, NDA, CUET, and other competitive examinations.

Real-Life Examples of Permutation with Repetition

Some common examples include:

• Creating ATM PINs.

• Generating passwords.

• Forming vehicle registration numbers.

• Creating binary strings.

• DNA sequence analysis.

• Generating computer codes.

Basics of Permutations

Before learning permutation with repetition, it is essential to understand the basic concept of permutations and the role of repetition in counting problems.

What is a Permutation?

A permutation is an arrangement of objects in a specific order.

For example, arranging the letters A, B, and C gives

ABC, ACB, BAC, BCA, CAB, CBA

Since the order changes, each arrangement is considered different.

Understanding Repetition

Repetition means that the same object may be selected more than once.

For example, when forming a 3-letter code from the letters A and B, arrangements such as

AAA, ABB, BAB, BBB

are all allowed because repetition is permitted.

Difference Between Repetition and No Repetition

When repetition is not allowed, each object can appear only once.

For example,

ABC

is valid, but

AAB

is not.

When repetition is allowed, objects can appear multiple times.

For example,

AAA

and

ABA

are both valid arrangements.

Understanding this difference helps students choose the correct counting formula.

Fundamental Principle of Counting

The Fundamental Principle of Counting states that if one task can be completed in $m$ ways and another independent task can be completed in $n$ ways, then both tasks together can be completed in

$m\times n$

ways.

Using this principle, if each of the $r$ positions has $n$ possible choices, then the total number of arrangements becomes

$n\times n\times\cdots\times n=n^r$

This forms the basis of the permutation with repetition formula.

Permutation with Repetition Formula

The formula for permutation with repetition is simple because every position can be filled independently using any of the available objects.

Standard Formula

If there are $n$ different objects and $r$ positions, then

$\text{Number of permutations with repetition}=n^r$

This is one of the most frequently used formulas in combinatorics.

Meaning of Variables

In the formula

$n^r$

• $n$ represents the number of available objects.

• $r$ represents the number of positions to be filled.

For example, if there are $5$ letters and we want to form a $3$-letter code, the number of possible arrangements is

$5^3=125$

Derivation of the Formula

Suppose there are $n$ choices for each position.

The first position can be filled in $n$ ways.

The second position can also be filled in $n$ ways.

Similarly, every position has $n$ possible choices.

Using the Fundamental Principle of Counting,

$n\times n\times\cdots\times n=n^r$

Hence,

$\boxed{n^r}$

is the required formula.

When to Use the Formula

Use the formula $n^r$ whenever:

• Repetition is allowed.

• Order matters.

• Every position has the same number of choices.

Typical examples include passwords, PINs, lock combinations, and coding sequences.

Types of Permutation with Repetition

Permutation with repetition appears in many different forms depending on the nature of the objects being arranged.

Arrangements of Digits

Digit arrangement problems often allow repetition.

For example, the number of 5-digit numbers formed using digits $0$-$9$ with repetition allowed is

$10^5$

provided leading zeros are permitted.

Arrangements of Letters

Letters can also be repeated while forming words or codes.

For example, using the letters A, B, and C to form 4-letter strings gives

$3^4=81$

possible arrangements.

Password and PIN Problems

Passwords and PINs are among the most common applications of permutation with repetition.

If each character can be reused, every position has the same number of possible choices.

Sequence Formation Problems

Permutation with repetition is also used to count binary sequences, DNA sequences, computer instructions, and signal patterns where repeated symbols are allowed.

Difference Between Permutation with and without Repetition

Although both involve arrangements, the counting methods are different depending on whether repetition is permitted.

Formula Comparison

Without repetition,

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

With repetition,

$n^r$

The latter is much simpler because every choice remains available throughout the arrangement.

Order of Arrangement

In both cases, order matters.

However, only permutation with repetition allows the same object to occupy multiple positions.

Number of Possible Arrangements

Since repetition increases the number of available choices, permutation with repetition usually produces more arrangements than ordinary permutation.

Comparison Table

Properties of Permutation with Repetition

Permutation with repetition has several mathematical properties that simplify counting problems.

Repetition of Objects

Each object remains available after every selection.

Choosing an object once does not reduce the number of choices for the remaining positions.

Independence of Choices

Every position is filled independently of the others.

This independence makes the counting process straightforward.

Counting Principle

The total number of arrangements follows directly from the Fundamental Principle of Counting.

If each of the $r$ positions has $n$ choices, then the total number of arrangements is

$n^r$

Important Observations

Some useful observations include:

• Repetition always increases the total number of arrangements.

• The formula does not involve factorials.

• Every position has the same number of choices.

• The method is applicable only when repetition is explicitly allowed.

Applications of Permutation with Repetition

Permutation with repetition has numerous practical applications across mathematics and technology.

Applications in Password Generation

Password systems often allow characters to repeat.

The formula $n^r$ helps determine the total number of possible passwords.

Applications in Coding Theory

Digital communication systems use repeated symbols while generating codes and transmitting information.

Permutation with repetition helps count the total number of valid codewords.

Applications in Computer Science

Computer scientists use permutation with repetition in:

• Password security.

• Cryptography.

• Binary strings.

• Data encoding.

• Algorithm analysis.

Applications in Probability

Many probability problems begin by counting the total number of possible outcomes.

When repetition is allowed, permutation with repetition provides this count directly.

Common Mistakes in Permutation with Repetition

Students often lose marks by confusing different counting techniques.

Confusing with Ordinary Permutations

Many students incorrectly apply

${}^nP_r$

instead of

$n^r$

Always check whether repetition is allowed.

Applying the Wrong Formula

If the question explicitly states that objects may repeat, the correct formula is

$n^r$

Using factorial-based formulas gives incorrect answers.

Ignoring Repetition Conditions

Words such as

• "repetition allowed,"

• "digits may repeat,"

• "with replacement,"

indicate that the permutation with repetition formula should be used.

Counting Duplicate Arrangements

Students sometimes remove duplicate arrangements unnecessarily.

When repetition is allowed and order matters, every distinct arrangement should be counted.

Permutation Formula Chart

The following formulas are commonly used in counting problems involving permutations.

Formula Summary

The most important formulas include:

Permutation with repetition:

$n^r$

Permutation without repetition:

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

Combination:

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

Comparison of Counting Formulas

Permutation of Objects When Few are Identical Formula

If there are $n$ objects of which p objects are of one type, $q$ objects of another type, r objects of yet another type, and all others are distinct, the total number of ways of arranging all the objects is $\frac{\mathrm{n}!}{\mathrm{p}!\mathrm{q}!\mathrm{r}!}$

Derivation of Permutation of Objects When Few are Identical

By doing this we have now all the $n$ things different, therefore number of permutations should be $n!$

Suppose the total number of permutations are $x$, now if we replace all $p$ identical objects with $p$ different objects which are also different from the rest of the objects then we have $x \times p$ ! arrangements.

Again if we replace $q$ identical things with $q$ different things, then no. of permutations will become ( $x$ p! q!). Further, if we replace $r$ identical things with $r$ different things which are different from the rest, then a number of permutations will become ( $x$ p! q!) r!

By doing this we have now all the n things different, therefore number of permutations should be n !

The number of arrangements will be $x \times p!\times q!\times r!$
Now, we have replaced all identical objects and we are left with $n$ different objects which can be arranged in n! Ways.

Hence, $x \times p!\times q!\times r!=n!$

$
x=\frac{n!}{p!q!r!}
$
Example: In how many ways can the letters of the word "MISSISSIPPI" be arranged?

Solution: repeated letters $\mathrm{I}=4$ times, $\mathrm{S}=4$ times and $\mathrm{P}=2$ times
So using the above formula we have $x=\frac{11!}{4!\times 4!\times 2!}$

Circulation Permutations with Repetition

The formula for Circulation Permutations with Repetition for n elements is $=n!=n(n-1)$

Best Books for Permutation with Repetition

Permutation with repetition is an important topic in combinatorics and counting principles. The following books explain its concepts with numerous solved examples and practice questions.

Shortcut Tips and Tricks for Permutation with Repetition

Most permutation with repetition questions can be solved quickly by identifying whether repetition is allowed before choosing the correct formula.

Important Formula Table

The following formulas are frequently used in permutation with repetition problems.

Solved Examples Based on Permutation with Repetition

Example 1: The total number of six-digit numbers formed using the digits $4$, $5$, and $9$ only and divisible by $6$ is: [JEE Main 2023]

Solution:

For a number to be divisible by $6$, it must be divisible by both $2$ and $3$.

Since the available even digit is only $4$, the units digit must be fixed as $4$.

Now consider the remaining five positions.

Case I: Six $4$s

There is only one arrangement.

$\frac{5!}{5!}=1$

Case II: Four $4$s, one $5$, and one $9$

The remaining five positions contain

$4,4,4,5,9$

Number of arrangements

$\frac{5!}{3!}=20$

Case III: Three $4$s and three $5$s

The remaining five positions contain

$4,4,5,5,5$

Number of arrangements

$\frac{5!}{2!3!}=10$

Case IV: Three $4$s and three $9$s

The remaining five positions contain

$4,4,9,9,9$

Number of arrangements

$\frac{5!}{2!3!}=10$

Case V: Two $4$s, two $5$s, and two $9$s

The remaining five positions contain

$4,5,5,9,9$

Number of arrangements

$\frac{5!}{2!2!}=30$

Case VI: One $4$, one $5$, and four $9$s

The remaining five positions contain

$5,9,9,9,9$

Number of arrangements

$\frac{5!}{4!}=5$

Case VII: One $4$, four $5$s, and one $9$

The remaining five positions contain

$5,5,5,5,9$

Number of arrangements

$\frac{5!}{4!}=5$

Therefore,

$\begin{aligned}
\text{Total numbers}
&, =1+20+10+10+30+5+5\
&, =81
\end{aligned}$

Hence, the answer is 81.

Example 2: The number of seven-digit integers with a sum of digits equal to $10$ and formed using the digits $1$, $2$, and $3$ only is: [JEE Main 2021]

Solution:

The possible digit combinations are:

Case I: , $1,1,1,1,1,2,3$

Number of arrangements

$\frac{7!}{5!}=7\times6=42$

Case II: , $1,1,1,1,2,2,2$

Number of arrangements

$\frac{7!}{4!3!}=\frac{7\times6\times5}{6}=35$

Therefore,

$\begin{aligned}
\text{Total numbers}
&, =42+35\
&, =77
\end{aligned}$

Hence, the answer is 77.

Example 3: The number of words (with or without meaning) that can be formed from all the letters of the word LETTER in which the vowels never come together is: [JEE Main 2020]

Solution:

The word is

LETTER

The vowels are

$E,E$

The consonants are

$L,T,T,R$

First, arrange the consonants.

Since the letter $T$ is repeated twice,

the number of arrangements is

$\frac{4!}{2!}=12$

This creates five gaps:

L, T, T, R, _

Choose any two of these five gaps for the two vowels.

This can be done in

${}^5C_2=10$

ways.

Since both vowels are identical,

their arrangement is

$\frac{2!}{2!}=1$

Therefore,

$\begin{aligned}
\text{Total words}
&, =\frac{4!}{2!}\times{}^5C_2\times\frac{2!}{2!}\
&, =12\times10\
&, =120
\end{aligned}$

Hence, the answer is 120.

Example 4: The total number of six-digit numbers in which only the digits $1$, $3$, $5$, $7$, and $9$ appear and each digit appears at least once is: [JEE Main 2020]

Solution:

There are six positions but only five distinct digits.

Therefore, exactly one digit must repeat.

First, choose the digit to be repeated.

This can be done in $5$ ways.

Now arrange the six digits.

Since one digit is repeated twice,

the number of arrangements is

$\frac{6!}{2!}$

Therefore,

$\begin{aligned}
\text{Total numbers}
&, =5\times\frac{6!}{2!}\
&, =\frac{5}{2}(6!)
\end{aligned}$

Hence, the answer is $\frac{5}{2}(6!)$.

Example 5: All possible numbers are formed using the digits $1,1,2,2,2,2,3,4,4$ taken all at a time. The number of such numbers in which the odd digits occupy even places is: [JEE Main 2019]

Solution:

The given digits are

$1,1,2,2,2,2,3,4,4$

There are a total of $9$ digits.

The even positions are

$2,4,6,8$

There are four even positions.

The odd digits are

$1,1,3$

These three digits must occupy any three of the four even positions.

The number of ways to choose these positions is

${}^4C_3$

Since the digit $1$ is repeated twice,

The arrangements of the odd digits are

$\frac{3!}{2!}=3$

Thus, the number of ways to place the odd digits is

${}^4C_3\times\frac{3!}{2!}=4\times3=12$

The remaining six positions are occupied by

$2,2,2,2,4,4$

The number of arrangements is

$\frac{6!}{4!2!}=\frac{6\times5}{2}=15$

Therefore,

$\begin{aligned}
\text{Total numbers}
& =12\times15\
& =180
\end{aligned}$

Hence, the answer is 180.

Related Topics to Permutation with Repetition

Permutation with repetition is closely connected with permutations, combinations, factorials, the Fundamental Principle of Counting, binomial expansion, probability, and combinatorics. Studying these topics strengthens problem-solving skills in counting techniques.