Permutation and Combination: Definition and Formulas

'Permutations and Combinations' is concerned with determining the number of different ways of arranging and selecting objects out of a given number of objects, without actually listing them. PnC tests your ability to observe the pattern, your mathematical reasoning, and your creativity. In real life, we use permutations and combinations for arranging people, digits, numbers, and alphabets.

In this article, we will cover the Introduction Of Permutation and Combination. 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 2023.

The chapter 'Permutations and Combinations' is concerned with determining the number of different ways of arranging and selecting objects out of a given number of objects, without actually listing them. These are generally referred to as “PnC”. This chapter is all about logic and “counting”. PnC tests your ability to observe the pattern, your mathematical reasoning, and your creativity. From the exam point of view, PnC is one of the important chapters. Concepts of Permutations and combinations are mostly used while solving problems from probability. The problems of probability are no longer as simple as they used to be in elementary standards. However, if you have a command of Permutations and Combinations, the chapter probability will be a piece of cake for you.

In this chapter, we are going to study the concepts and applications of Permutations and combinations.

Following are the sub-topics that we are going to study:

1. Fundamental Principle of Counting: Addition Rule (OR Rule) and Multiplication Rule (AND Rule)

2. Permutations

3. Combinations

4. Grouping

5. Distribution

6. Derangements

Fundamental Principle of Counting

The principle of Counting includes two rules Addition rule and the Multiplication rule

Addition rule

If work $W$ can be completed by doing task $A$ OR task B, and $A$ can be done in $m$ ways and $B$ can be done in $n$ ways (and both cannot occur simultaneously: in this case, we call tasks A and B as mutually exclusive), hen work $W$ can be done in $(m+n)$ ways.

Multiplication rule

If a certain work W can be completed by doing 2 tasks, first doing task and then doing task $B$. A can be done in $m$ ways and following that $B$ can be done in n ways, then the number of ways of doing the work $W$ is $(m \times n)$ ways.

Permutations

A permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement.

Combinations

The meaning of combination is selection.

Formation of Groups

Formations of the group are used to find the number of ways n distinct objects can be divided into m groups, whose sizes are known.

Distribution

Distribution of things is used to find the number of ways of distributing n different things in r different boxes.

Derangements

If there are n things and n places, one correct place corresponds to each object. Then an arrangement in which none of the objects is in its right place is called a derangement.

Permutations Meaning

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.

For example, suppose we have a set of three letters: A, B, and C. We want to find the number of ways in which 2 letters from this set can be arranged. Each possible arrangement would be an example of a permutation. The complete list of possible permutations would be $A B, A C$, $B A, B C, C A$, and $C B$. Thus there 6 number of permutations. We observe that the order in which letters are occurring is important, i.e., AB and BA are two different arrangements. In mathematics, we use a specific terminology. That is "permutations as $n$ distinct objects taken $r$ at a time". Here n refers to the number of objects from which the permutation is formed, and $r$ refers to the number of objects used to form the permutation. In the above example, the permutation was formed from 3 letters (A, B, and C), so $\mathrm{n}=3$ and the permutation consisted of 2 letters, so $\mathrm{r}=2$.

Permutations Formula

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.

So the number of ways of arranging $n$ objects taken $r$ at a time $=n(n-1)$

$
\begin{aligned}
& (n-2) \ldots(n-r+1) \\
& \frac{n(n-1)(n-2) \ldots(n-r+1)(n-r)!}{(n-r)!}=\frac{n!}{(n-r)!}={ }^n P_r
\end{aligned}
$

Where $r \leq n$ and $r \in W$
So, the number of ways arranging n different objects taken all at a time $=$ ${ }^n P_n=n!$

Combination Meaning

The meaning of combination is selection. Suppose we want to select two objects from four distinct objects a, b, c, and d. This can be stated as a number of combinations of four different objects taken two at a time.

Here we have six different combinations ab, ac, ad, bc, bd, cd. In other words, we can say that there are six ways in which we can select two objects from four distinct objects.

Combination Formula

We can generalize this concept for $r$ object to be selected from given $n$ objects as

$
\begin{aligned}
& { }^n C_r \times r!={ }^n P_r \\
& { }^n C_r=\frac{{ }^n P_r}{r!} \\
& { }^n C_r=\frac{n!}{(n-r)!r!}
\end{aligned}
$

Where $0 \leq r \leq n$, and $r$ is a whole number.

Relation Between Combination And Permutations

The relation between combinations and permutations is given by

$
{ }^n C_r \times r!={ }^n P_r
$

Difference Between Permutation and Combination

The difference between the Permutations and combinations is given below.

Solved Examples Based on Introduction of Permutation and Combination

Example 1: If $(156$ ! $) /(151-r)$ ! : $(154!) /(151-r)!=26020: 1$ find $r_2^P$ ?

Solution: To solve the equation $156_{\mathrm{r}+5}^{\mathrm{P}}: 154_{\mathrm{r}+3}^{\mathrm{p}}=26020: 1$ and find $r_2^P$, we will follow a similar process as before.
Let's start by simplifying the equation:

$
156_{\mathrm{r}+5}^{\mathrm{P}}: 154_{\mathrm{r}+3}^{\mathrm{P}}=26020: 1
$

Rewriting the equation using the permutation formula:

$
(156!) /(156-(\mathrm{r}+5)!):(154!) /(154-(\mathrm{r}+3)!)=26020: 1
$

Simplifying further:

$
(156!) /(151-\mathrm{r})!:(154!) /(151-\mathrm{r})!=26020: 1
$

Next, we can simplify the factorials:

$
(156!) /(151-r)!=26020 \times(154!) /(151-r)!
$

Cancelling out common terms:

$
(156 \times 155 \times \ldots \times(151-r) \times(150-r)!) /(151-r)!=26020
$

Simplifying:

$
156 \times 155 \times \ldots \times(151-r)=26020 \times 154!
$
Now, we can solve for $r_2^{\mathrm{P}}$. The expression $\mathrm{r}_2^{\mathrm{P}}$ represents the number of permutations of 2 objects taken from a set of $r$ objects, which can be calculated as $\mathrm{r}!/(\mathrm{r}-2)!$ :

$
r_2^{\mathrm{p}}=\mathrm{r}!/(\mathrm{r}-2)!
$

Since we have simplified the equation to the form

$
(156 \times 155 \times \ldots \times(151-r) \times(150-r)!)=26020 \times(154!)
$

, we can set $\mathrm{r}=154$ and calculate $\mathrm{r}_2^{\mathrm{p}}$ :

$
\mathrm{r}_2^{\mathrm{P}}=154!/(154-2)!=154!/ 152!=154 \times 153
$

Therefore, $\mathrm{r}_2^{\mathrm{P}}=154 \times 153=23622$.
Hence, the answer is 23622

Example 2: The sum of the series $\sum_{r=1}^n\left(r^2+1\right)(r!)$
Solution: We can write $r^2+1=(r+2)(r+1)-3(r+1)+2$

Thus,

$
\begin{aligned}
& \sum_{r=1}^n\left(r^2+1\right)(r!)=\sum_{r=1}^n[(r+2)(r+1)-(r+1)-2\{(r+1)-1) \\
& =\sum_{r=1}^n[(r+2)!-(r+1)!]-2 \sum_{r=1}^n\{(r+1)!-r!\} \\
& =(n+2)!-2!-2\{(n+1)!-1\}=n(n+1)!
\end{aligned}
$

Hence, the answer is $n(n+1)$ !
Example 3: The remainder when $x=1!+2!+3!+4!+\ldots .+100!$ is divided by 240 , is

Solution: For $r \geq 6 ; r$ is divisible by 240.
Thus, when x is divided by 240,
the remainder is $1!+2!+\ldots+5!=153$.
Hence, the answer is 153

Example 4: In a lottery game, you need to select 4 numbers from a pool of 25 . How many different combinations of numbers are possible?

Solution: In this case, $\mathrm{n}=25$ (the total number of numbers in the pool) and $\mathrm{r}=4$ (the number of numbers to be selected). Plugging these values into the formula, we get:

$
\begin{aligned}
\mathrm{C}(25,4) & =\frac{25!}{(4!(25-4)!)} \\
& =\frac{25!}{(4!21!)} \\
& =\frac{(25 \times 24 \times 23 \times 22)}{(4 \times 3 \times 2 \times 1)} \\
& =12,650
\end{aligned}
$

Therefore, 12,650 different combinations of 4 numbers can be selected from a pool of 25 in the lottery game.

Hence, the answer is 12650

Summary

Permutations focus on arranging objects in a sequence, while combinations focus on selecting groups without considering the order. Both concepts are used in probability, statistics, and computer science and help us solve problems related to possibilities and probabilities. Understanding permutations and combinations not only enhances problem-solving skills but also helps us analyze real-world situations where counting and selection are critical.