Permutation with Repetition: Definition and Formula

Permutation basically means the arrangement of things. And when we talk about arrangement then the order becomes important if the things to be arranged are different from each other (when things to be arranged are the same then order doesn’t have any role to play). So in permutations order of objects becomes important. In real life, we use permutation for arranging numbers, letters, codes, and alphabets.

In this article, we will cover the Introduction Of Permutation. This topic falls under the broader category of Permutations and combinations, which is a crucial chapter in Class 11 Mathematics. This is very important not only for board exams but also for competitive exams, which even include the Joint Entrance Examination Main and other entrance exams: SRM Joint Engineering Entrance, BITSAT, WBJEE, and BCECE. A total of seven questions have been asked on this topic in JEE Main from 2013 to 2023 including one in 2019, two in 2020, one in 2021, and one in 2023.

What is the Permutation?

A permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement. Arranging $n$ objects in r places (Same as arranging $n$ objects taken $r$ at a time) is equivalent to filling $r$ places from $n$ things.

The number of ways of arranging n objects taken r at a time $={ }^{\mathrm{n}} \mathrm{P}_{\mathrm{r}}$, Where $r \leq n$ and $r \in W$

The number of ways arranging n different objects taken all at a time $=$ ${ }^n P_n=n!$

Permutations Methods

Permutation can be done in two ways,

Permutation with repetition: This method is used when we are asked to make different choices each time and with different objects.

Permutation without Repetition: This method is used when we are asked to reduce 1 from the previous term for each time.

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 $=n!=n(n-1)$

Summary

The permutations represent the number of distinct arrangements of objects where the order matters. The calculation of permutations using factorial notation provides a precise method to quantify and analyze sequential arrangements. Understanding of permutation is necessary for solving complex problems.

Solved Examples Based on Permutation Of Objects When Few Are Identical

Example 1: The total number of six-digit numbers, formed using the digits 4,5, and 9 only and divisible by 6 , is
[JEE MAINS 2023]
Solution: We have, For this, 4 will be fixed as a unit place digit

$
\begin{aligned}
& \text { CaseI }: 4^{\prime} s \rightarrow 6 \text { times } \\
& \text { CaseII }: \quad 4 \text { 's } \rightarrow \text { 4times } \\
& 5^{\prime} s \rightarrow 1 \text { times } \quad \frac{5!}{3!}=20 \\
& 9^{\prime} s \rightarrow 1 \text { times }
\end{aligned}
$

Case III: $\quad 4^{\prime} s \rightarrow 3$ times

$
5^{\prime} s \rightarrow 3 \text { times } \quad \frac{5!}{2!3!}=10
$

Case IV: 4 's $\rightarrow$ 3times

$
9^{\prime} s \rightarrow 3 \text { times } \quad \frac{5!}{2!3!}=10
$
Case $V: \quad 4^{\prime} s \rightarrow 2$ times

$
\begin{aligned}
& 5^{\prime} s \rightarrow 2 \text { times } \quad \frac{5!}{2!2!}=30 \\
& 9^{\prime} s \rightarrow 2 \text { times }
\end{aligned}
$
Case VI: $\quad 4^{\prime} s \rightarrow 1$ times
5 's $\rightarrow 1$ times $\quad \frac{5!}{4!}=5$
9 's $\rightarrow$ 4times
Case VII: 4 's $\rightarrow 1$ times
$5^{\prime} s \rightarrow 4$ times $\quad \frac{5!}{4!}=5$
$9^{\prime} s \rightarrow 1$ times
Total numbers $=81$
Hence, the answer is 81 .

Example 2: The number of seven-digit integers with a sum of the digits equal to 10 and formed by using the digits 1,2 and 3 only is :

[JEE MAINS 2021]
Solution: The first possibility is $1,1,1,1,1,2,3$

$
\text { required number }=\frac{7!}{5!}=7 \times 6=42
$

The second possibility is $1,1,1,1,2,2,2$

$
\begin{aligned}
& \text { required number }=\frac{7!}{4!3!}=\frac{7 \times 6 \times 5}{6}=35 \\
& \text { Total }=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 vowels never come together is $\qquad$
[JEE MAINS 2020]

Solution
LETTER
vowels $=\mathrm{EE}$, consonant $=\mathrm{LTTR}$
First place LTTR
${ }_{-} \mathrm{L}_{-} \mathrm{T}_{-} \mathrm{T}_{-} \mathrm{R}_{-}$
This can be done in $4!$ / 2 ! ways
Now, E, E can be placed in any 2 of the 5 gaps
This can be done in ${ }^5 \mathrm{C}_2$ ways
2 Es can be arranged in these 2 places in 2! / 2! ways
So, total $=\frac{4!}{2!} \times{ }^5 C_2 \times \frac{2!}{2!}=12 \times 10=120$
Hence, the answer is 120

Example 4: Total number of 6-digit numbers in which only the digits $1,3,5,7$ and 9 appear and each of these digits appears at least once, is :

[JEE MAINS 2020]
Solution: If there 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 $\frac{\mathrm{n}!}{\mathrm{p}!\mathrm{q}!\mathrm{r}!}$

Now,
6 places, 5 digits so one digit will repeat
First select the digit to be repeated: 5 options for this
Then arrange these 6 digits in 6 places: $6!$ / 2 !
So, total: $5 \times \frac{6!}{2!}$
Hence, the answer is $\frac{5}{2}(6!)$

Example 5: All possible numbers are formed using the digit $1,1,2,2,2,2,3,4,4$ taken all at a time. The number of such numbers in which the odd digits occupy even places is:
[JEE MAINS 2019]
Solution: $1,1,2,2,2,2,3,4,4=9$ digit
We have a total 4 even places and 3 odd digits $(1,1,3)$ should come at these places

Number of ways placing odd digits at even places $=$ ${ }^4 C_3 \frac{3!}{2!}=4 \times 3=12$

Number of ways to place even digits

$
=\frac{6!}{4!2!}=\frac{6 \times 5}{2}=15
$

So, the total number of ways $=12 \times 15=180$
Hence, the answer is 180