Applications of Permutations

Before knowing about the Application of permutation and combination, let's revise Permutations. 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. In real life, we use permutation for arranging numbers, letters, codes, and alphabets.

In this article, we will cover the Applications 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 fifty-five questions have been asked on this topic in JEE Main from 2013 to 2023 including one in 2013, one in 2014, one in 2018, one in 2019, two in 2020, thirteen in 2021, sixteen in 2022, and nineteen 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.

Application of Permutations

Some Applications of Permutations are:

Arrangement of Digits

To make 5-digit numbers from 1,2,3,4,5,6,7
1. If the repetition of digits is not allowed

In this case,
The first place of the digit (from left) can be filled in 7 ways
Second place can be filled in 6 ways (as one number is already used at the first place)

Third place can be filled in 5 ways
Fourth in 4 ways
Fith in 3 ways
Hence the total number of ways is $7 \times 6 \times 5 \times 4 \times 3=2520$.

2. If the repetition of digits is allowed

In this case,
First place can be filled in 7 ways
Second place can be filled in 7 ways (as repetition is allowed)
Similarly, all other places can be filled in 7 ways
Hence the total number of ways is $7 \times 7 \times 7 \times 7 \times 7=7^5$.

3. If the number is EVEN

In this case,

One's place can be filled by three numbers $(2,4,6)$ as even numbers can only end with $(0,2,4,6,8)$. Hence we have three different possibilities.

If one's place is occupied by 2 then the remaining 6 digits can be arranged in the remaining 4 places in ${ }^6 P_4$ ways

OR,
If one's place is occupied by 4 hence remaining can be arranged in ${ }^6 P_4$ ways

OR,
If one's place is occupied by 6 hence remaining can be arranged in ${ }^6 P_4$ ways

Hence the total possibilities are ${ }^6 P_4+{ }^6 P_4+{ }^6 P_4=1080$

Let us take another example
Using $0,1,2,3,4,5$ how many 4 -digit numbers can be formed if the repetition of digits is not allowed?

Here 0 cannot be at a leading place as in such a case the number will become a 3-digit number.
Hence the first place can be filled in 5 ways and the second place is left with 5 possibilities (now we have one lesser digit available as it is used up at a leading place, but we have one more digit '0' that can be filled at second place) that's why it can also be filled with 5 ways and the thirdplace left with 4 possibilities and the fourth place left with 3 possibilities.

So the total number of 4 digit numbers $=5 \times 5 \times 4 \times 3=300$
Solving the same question if repetition of digits was allowed
In that case, the first place (from left) has 5 choices (excluding 0 otherwise it will not be 4 digit number). 2nd, 3rd, and 4th place will have 6 choices (all choices available, repetition allowed), so the total number, in this case, $=5 \times 6 \times 6 \times 6$

Find the number of Distinct 4-digit numbers that can be formed using 1,2,3,4,5,6,7 for each of the following conditions:

1. If the number is greater than 3000 .

The thousand's digit can be filled in 5 ways (any one of $3,4,5,6$ or 7 ). Now one is left with 6 digits and has to arrange 3 of them. This can be done in ${ }^6 \mathrm{P}_3=6 \times 5 \times 4$ ways. Thus a total of $5 \times(6 \times 5 \times 4)=600$ such numbers can be formed.
2. If the number is EVEN.

The unit's digit can be filled in three ways $(2,4$, or 6 ). Now we are left with 6 digits and have to arrange 3 of them. This can be done in ${ }^6 \mathrm{P}_3=6 \times 5 \times$ 4 ways. Thus a total of $(6 \times 5 \times 4) \times 3=360$ such numbers can be formed.
3. If the number is EVEN and greater than 3000 .

The unit digit could be 2,4 , OR 6

Considering the case that the unit digit is 2 :
The unit digit can be filled in only 1 way, with a 2 . Having filled this, all of $3,4,5,6$, and 7 are available (so that the number is greater than 3000 ) for the thousands place and thus it can be filled in 5 ways $(3,4,5,6,7)$. Next, we are left with 5 digits two of which have to be arranged i.e. can be done in $5 \times 4$ ways. Thus total possible numbers with unit digit being 2 will be 5 $\times(5 \times 4) \times 1=100$

Considering the case that the unit digit is 4 : The unit digit can be filled in 1 way. Having filled this, the digits available for the thousands place are 3 , $5,6,7$ i.e. there are further 4 possibilities. Next, we are left with 5 digits two of which have to be arranged i.e. can be done in $5 \times 4$ ways. Thus total possible numbers with the unit's digit being 4 or 6 will be $4 \times(5 \times 4) \times$

$
1=80
$

Considering the case that the unit digit is 6 : The unit digit can be filled in 1 way. Having filled this, the digits available for the thousands place are 3 , $4,5,7$ i.e. there are further 4 possibilities. Next, we are left with 5 digits and two of which have to be arranged i.e. can be done in $5 \times 4$ ways.
Thus total possible numbers with the unit's digit being 4 or 6 will be $4 \times(5$ $\times 4) \times 1=80$.

Thus the total number of even numbers greater than 3000 that can be formed will be $100+80+80=260$

ARRANGEMENT OF PEOPLE IN A ROW

Arranging 4 boys and 4 girls such that 4 girls have to be together:
Considering all four girls as one unit, we have to arrange 5 units in 5 places and this can be done in 5 ! ways. Now one of the units is of 4 girls, who can be arranged amongst themselves in $4!$ ways. Thus the total number of ways of arranging is $5!\times 4!$.

Arranging 4 boys and 4 girls such that no two girls should stand together:
The four boys can first be arranged in 4 ! ways. Now there will be 5 places (Three between 4 boys and one at each end) for the 4 girls and they could be arranged in $5 \times 4 \times 3 \times 2$ ways. Thus the total number of arrangements possible is $4!\times 5 \times 4 \times 3 \times 2$.

Arranging 4 boys and 4 girls such that girls and boys have to be alternate:
When girls and boys have to be alternate, it would just be either G B G B G B G B or B G B G B G B G. In each of these ways, there are 4 places for the boys and 4 places for the girls and thus they can be arranged in 4 ! $\times 4!$ in each of these. Thus, the total number of arrangements possible is $2 \times 4!\times 4!$

QUESTION-BASED ON FACTORS

Factors of a number N refer to all the numbers that divide N completely.

These are also called divisors of a number.
Basic formula related to factors of a number:
These are certain basic formulas pertaining to factors of a number N , such that,

$
\mathrm{N}=\mathrm{p}^{\mathrm{a}} \mathrm{q}^{\mathrm{b}} \mathrm{r}^{\mathrm{c}}
$

Where, $p, q$, and r are prime factors of the number n .
$\mathrm{a}, \mathrm{b}$, and c are non-negative powers/ exponents
- Number of factors of $\mathrm{N}=(\mathrm{a}+1)(\mathrm{b}+1)(\mathrm{c}+1)$
- Sum of factors: $\left(p^{a+1}-1\right)\left(q^{b+1}-1\right)\left(r^{c+1}-1\right) /(p-1)(q-1)(r-1)$

Let us take an example:
Find the Number and Sum of the factor of 18.
Factors of 18 are 1,2,3,6,9,18
Number of factors $=6$
And their Sum is $39(=1+2+3+6+9+18)$
Using the Formula

$
18=2^1 \times 3^2
$

Number of factors of $N=(1+1)(2+1)=6$
Sum of factors: $\left(2^2-1\right)\left(3^3-1\right) /(2-1)(3-1)=39$

Number of odd and even factors:

Let us understand this concept with an example:
Find the number of Even and Odd factors of 72
Prime factorization of $72=2^3 \times 3^2$
We know the fact that all the factors of 72 will be in the form of $2^a \times 3^b$.
For ODD factors, the exponent of 2 i.e., a has to be 0 always. Or, the number of ways using 2 for making the combination is 1 , i.e., $2^0$. Also, the number of values that the exponent of 3 i.e., b can take is $3(0,1$ or 2$)$

Hence the number of odd factors of $72=1 \times 3=3$.
Extending the logic for EVEN factors, we can say that for a factor to be Even, it must contain 2 at least once.

So, a can take values 3 values ( 1,2 , or 3 , note: 0 is not possible for the number to be even)
b can take 3 values $(0,1$ or 2$)$
So, the number of values $a$, and b can take are 3 and 3 respectively.
Therefore, the total number of even factors of $72=3 \times 3=9$.

A number of factors that are perfect squares:

Find the number of factors that are perfect squares of 72.
Prime factorization of $72=2^3 \times 3^2$
If we prime factorize any number which is a perfect square, we would observe that in all cases the exponent of all the prime factors of the number is even only.

For example, 36 is a perfect square $36=2^2 \times 3^2$, here we can see that the exponent of both 2 and 3 are even.

Again, any factor 72 will be in the form of $2^{\mathrm{a}} \times 3^{\mathrm{b}}$. For the factors to be perfect squares, all the values a , and b have to be even only.

Or, the possible values which a can take $=0,2$ i.e. 2 values only. Similarly, b can take 0,2 i.e. 2 values.

Therefore, the different combinations we can have $=2 \times 2=4$.
Hence, 72 has 4 factors which are perfect squares.

A number of factors that are perfect cube:

Find the number of factors that are perfect cubes of 72 .
Prime factorization of $72=2^3 \times 3^2$
If we prime factorize any number which is a perfect cube, we would observe that in all cases the exponent of all the prime factors of the number is a multiple of 3 .

For example, 27 is a perfect cube $27=3^3$, here we can see that the exponent of 3 is a multiple of three.

Again, any factor 72 will be in the form of $2^{\mathrm{a}} \times 3^{\mathrm{b}}$. For the factors to be perfect cubes, all the values a , and b have to be divisible by 3 .

Or, the possible values which a can take $=0,3$ i.e. 2 values only. Similarly, b can take 0 i.e. 1 value.

Therefore, the different combinations we can have $=2 \times 1=2$.
Hence, 72 has 2 factors which are perfect cubes.

DIVISIBILITY OF A FACTOR BY A NUMBER

Let us understand with an example:
Find the number of factors of 72 which are divisible by 8
Prime factorization of $72=2^3 \times 3^2$
Prime factorization of $8=2^3$
For the factors to be divisible by 8 the factor should be multiple of 8 or $2^3$.
Or we can say that the values a should be equal to 3 or greater than 3 , and b can take any value.

Hence, the possible values which a can take $=3$, i.e. 1 value only. Similarly, b can take 0,1 , and 2 i.e. 3 values.

Therefore, the different combinations we can have $=1 \times 3=3$.
Hence, 72 has 3 factors which are divisible by 8 i.e. $8,24,72$.

Product of factors:

Let a number be $N$
Its Prime factorization is $N=2^{\mathrm{a}} \times 3^{\mathrm{b}} \times 5^{\mathrm{c}}$

Product of factors $=\mathrm{N}^{\frac{\text { na of fantars }}{2}}$

For example: Find the Product of the factors of 72
Product of factors of $72=72^{\frac{(3+1)(2+1)}{2}}$
Product of factors of $72=72^6$

Exponent of Prime P in n!

Where $[\mathrm{x}]$ stands for the greatest integer value of $x \in R$.
If $m$ is the index of the highest power of a prime $p$ that divides $n$ ! then

$
m=\left[\frac{n}{p}\right]+\left[\frac{n}{p^2}\right]+\left[\frac{n}{p^3}\right]+\ldots \ldots \ldots \ldots
$

Example: Find the number of trailing zeros in 20 !
Solution: $\square$
10 can be written as $2 \times 5$
If you want to figure out the exact number of zeroes, you would have to check how many times the number N is divisible by 10 .

When dividing N by 10 , factors of 10 will be the smaller of the power of 2 or 5 .

Number of trailing zeros is going to be the power of 2 or 5 , whichever is lesser.

$
\begin{aligned}
& m=\left[\frac{20}{2}\right]+\left[\frac{20}{2^2}\right]+\left[\frac{20}{2^3}\right]+\left[\frac{20}{2^4}\right]+\left[\frac{20}{2^4}\right] \ldots \ldots \ldots \ldots=18 \\
& m=\left[\frac{20}{5}\right]+\left[\frac{20}{5^2}\right]+\left[\frac{20}{5^3}\right]+\ldots \ldots \ldots \ldots=4
\end{aligned}
$

Hence the number of trailing zeros is 4

Solved Examples
Example 1: All words, with or without meaning, are made using all the letters of the word MONDAY. These words are written in a dictionary with serial numbers. The serial number of the word MONDAY is [JEE MAINS 2023]

Solution

Hence, the answer is 327

Example 2 : A total number of 3 -digit numbers that are divisible by 6 and can be formed by using the digits $1,2,3,4$, and 5 with repetition, is $\qquad$
[JEE MAINS 2023]
Solution

Hence, the answer is 16

Example 3: A boy needs to select five courses from 12 available courses, out of which 5 courses are language courses. If he can choose at most two language courses, then the number of ways he can choose five courses is
[JEE MAINS 2023]
Solution

\begin{equation}
\begin{aligned}
&\begin{aligned}
& =5_{C_0} \cdot 7_{C 5}+5_{C_1} \cdot 7_{C 4}+5_{C_2} \cdot 7_{C 5} \\
& =21+175+350 \\
& =546
\end{aligned}\\
&\text { Hence, the answer is } 546
\end{aligned}
\end{equation}

Example 4: The number of five-digit numbers, greater than 40000 and divisible by 5 , which can be formed using the digits $0,1,3,5,7$, and 9 without repetition, is equal to
[JEE MAINS 2023]

Solution

Hence, the answer is 120
Example 5: The number of permutations of the digits $1,2,3, \ldots ., 7$ without repetition, which neither contain the string 153 nor the string 2467 is $\qquad$ [JEE MAINS 2023]

Solution: Numbers are 1, 2, 3, 4, 5, 6, 7
Numbers having string $(154)=(154), 2,3,6,7=5$ !
Numbers having string $(2467)=(2467), 1,3,5=4$ !
Number having string (154) and (2467)

$
=(154),(2467)=2!
$
Now n $(154 \cup 2467)=5!+4!-2!$

$
=120+24-2=142
$
Again total numbers $=7!=5040$
Now required numbers $=\mathrm{n}$ (neither 154 nor 2467)

$
\begin{aligned}
& =5040-142 \\
& =4898
\end{aligned}
$

Hence, the answer is 4898.

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 permutation is necessary for solving complex problems.