Permutation and Combinations - Topics, Formula, Books, FAQs

'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. Permutations and Combinations 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 Permutation 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.

What is factorial?

A factorial is the product of all positive integers up to a given number.

It is denoted by n!.

n!=n ×(n-1)×(n-2)×(n-3)×………×2 ×1

It is calculated as follows.

6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

Note that:

1! = 1

0! = 1

What is the meaning of n!?

n! is read as “n factorial”. It represents the products all positive integers from 1 to n.

n! = n × (n -1) × (n - 2) × (n - 3) × . . . . . . . × 1

It is used in various mathematical and real-world contexts like:

• Permutations and combinations

• In calculations involving probability distributions.

• Algebra and calculus

• Computer science algorithms involving combinatorial logic.

What are the Fundamental Principles of Counting?

Counting principles are foundational rules used to count the number of ways certain arrangements can occur.

It has two types.

• Principle of Addition

• Principle of Multiplication

Principle of Addition(OR 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.

Principle of Multiplication(AND 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.

What is 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.

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 $\mathrm{r}=2$.

What is the formula for Permutation?

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 $=$ ${ }^n P_n=n!$

Types of Permutations

There are different types of permutations.

• 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 (when repetition is not allowed)

In this case, each object is used only once in the arrangement.

The formula to find the number of permutations of n distinct objects is given by:

p(n) = n!

Where n! (n factorial) is the product of all positive integers up to n

Permutation of n different objects, where repetition is allowed

When repetition is allowed, each object can be used more than once in the arrangement. The formula to find the number of permutations when repetition is allowed is given by:

p(n,r) = nr

Where n is the total number of objects to choose from

And r is the number of positions to be filled

Permutation when the objects are not distinct (Permutation of multisets)

When some objects are identical, the formula to find the number of distinct permutations is adjusted to account for the repetitions. The formula is given by:

p(n; n1, n2, n3,......., nk) = $\frac{n!}{n_1! × n_2! × n_3! ×.........×n_k!}$

Where n is the number of total objects, n1, n2, n3,......., nk are frequencies of different objects.

What is Circular Permutation

When n objects are to be arranged circularly, then the formula to find the number of distinct permutations is given by (n-1)!.

What is the Combination?

A combination is a selection of items from a larger pool, where the order of the items does not matter. In other words, it is a way of selecting items where the arrangement or sequence of the selection is not considered.

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

Here we have three different combinations ab, bc, ca, bc. In other words, we can say that there are three ways in which we can select two objects from three 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.

When is the Combination to be used?

Combinations are used when the order of selection does not matter. Here are some examples of when combinations are appropriate:

1. Lottery Selection Method: In the Lottery selection method, we use the combination method. Order does not matter in selecting lotteries to choose prize winners.

2. Forming Committees: When forming a committee from a group of people, the order in which members are chosen does not matter.

3. Choosing Subsets: When choosing a subset of items from a larger set, where the order in which items are chosen is irrelevant.

4. Hand of Cards: When selecting a hand of cards from a deck, the order of the cards in the hand does not matter.

Relation between Permutation and Combination

Theorem 1: $^np_r=^nc_r × r!$

Corresponding to each combination of $^nc_r$ we have r! permutations because r objects in every combination can be rearranged in r! ways.

Proof:

We have to prove that $^np_r=^nc_r × r!$

Taking the left side of the equation:

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

Taking the right side of the equation

$^nc_r × r! = \frac{n!}{(n-r!) r!} × r! = \frac{n!}{(n-r)!}=^np_r$

Hence, $^np_r=^nc_r × r!$ proved.

Theorem 2: ${ }^n\mathrm{C}_r +{ }^n\mathrm{C}_{r-1} = { }^{n+1}\mathrm{C}_r$

Proof:

We have to prove that ${ }^n\mathrm{C}_r +{ }^n\mathrm{C}_{r-1} = { }^{n+1}\mathrm{C}_r$

Taking the left side of the equation, we get,

${ }^n\mathrm{C}_r +{ }^n\mathrm{C}_{r-1}$

= $\frac{n!}{r!(n-r)!}+\frac{n!}{(n-r+1)!(r-1)!}$

= $\frac{n!}{(n-r)!r(r-1)!}+\frac{n!}{(n-r+1)!(r-1)!}$

= $\frac{n!}{(r-1)!}[\frac{1}{(n-r)!r}+\frac{1}{(n-r+1)!}]$

= $\frac{n!}{(r-1)!}[\frac{1}{r(n-r)!}+\frac{1}{(n-r+1)(n-r)!}]$

= $\frac{n!}{(r-1)!(n-r)!}[\frac{1}{r}+\frac{1}{(n-r+1)}]$

= $\frac{n!}{(r-1)!(n-r)!}[\frac{n-r+1+r}{r(n-r+1)}]$

= $\frac{n!}{(r-1)!(n-r)!}[\frac{n+1}{r(n-r+1)}]$

= $\frac{(n+1)n!}{r(r-1)!(n-r+1)(n-r)!}$

= $\frac{(n+1)!}{r!(n-r+1)!}$

= ${ }^{n+1}\mathrm{C}_r$

Hence, proved.

What are the differences between Permutations and Combinations?

Important Points

Recommended Video on Permutation and Combination

Solved Examples Based on Permutation and Combination

Example 1: A bookstore has 5 different mystery novels, 3 different romance novels, and 2 different fantasy novels. How many ways can you choose three mystery novels, two romance novels, and one fantasy novel to create a set?

1. 40 ways

2. 50 ways

3. 60 ways

4. 70 ways

Solution:

For the mystery novels, you have 5 choices, and you want to choose 3.
${ }^5 \mathrm{C}_3=\frac{5!}{3!(5-3)!}$ = 10 ways

For the romance novels, you have 3 choices, and you want to choose 2.
${ }^3 \mathrm{C}_2=\frac{3!}{2!(3-2)!}$ = 3 ways

For the fantasy novels, you have 2 choices, and you want to choose 1.
${ }^2 \mathrm{C}_1=\frac{2!}{1!(2-1)!}$ = 2 ways

$\therefore$ Total combination = 10 × 3 × 2 = 60 ways

So, there are 60 different ways to choose three mystery novels, two romance novels, and one fantasy novel to create a set.

Hence, the correct answer is 60 ways.

Example 2: You have a collection of 10 different candies. If you want to create a bag of 3 candies, how many different combinations of candies can you put in the bag?

1. 30 combinations

2. 120 combinations

3. 220 combinations

4. 720 combinations

Solution:

To find the number of different combinations of 3 candies you can put in the bag from a collection of 10 different candies, you can use combinations. Specifically, you want to calculate the number of ways to choose 3 candies out of 10.

${ }^{10} \mathrm{C}_3=\frac{10!}{3!(10-3)!}=120$

So, you can create 120 different combinations of candies in the bag.

Hence, the correct answer is 120 combinations.

Example 3: In how many different ways can the letters of the word 'PHENOMENAL' be arranged so that the vowels always come together?

1. 30,240 ways

2. 80,640 ways

3. 120,960 ways

4. 161,280 ways

Solution:

To find the number of different ways the letters of the word 'PHENOMENAL' can be arranged so that the vowels always come together, treat the vowels (E, O, E, A) as a single entity, which we can call "V."

Now, you have the letters "V," P, H, N, M, N, and L. These can be arranged in 7! ways.
However, the letter 'N' is repeated twice, so we must divide by 2! to account for the overcounting:

$= \frac{7!}{2!}$
$=\frac{7 × 6 × 5 × 4 × 3 × 2 × 1 }{2}$
$= \frac{5040}{2}$
$= 2520$

But the vowels (2 Es, O and A) can be permuted themselves.

$= \frac{4!}{2!}=\frac{4×3×2×1}{2×1}=12$

$\therefore$ The total number of arrangements will be = 2,520 × 12 = 30,240

Hence, the correct answer is 30,240 ways.

Example 4: 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 5: 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 6: 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 7: 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

List of Topics According to NCERT/JEE MAIN

How to Prepare Permutation and Combination

Follow these tips to prepare the chapter permutation and combination created by the subject matter experts:

Understand the use of $nPr$ and $nCr$:-

When you first prepare for permutation and combination, the two-term you’ll learn are $nPr$ and $nCr$ and initially, you will find that you are using these two terms wrongly to solve problems. So in the beginning, solve the problems without using them. Slowly by practice, you’ll understand the pattern and the need for these two terms in solving certain questions. And only after that, you’ll learn the usage of these terms perfectly.

Practice Different types of Questions:-
In Permutation and Combination, you can form an infinite number and different type of questions and still cannot be confident that you will be able to solve all questions of this topic. So to be able to solve the maximum number of questions you need to practice a lot, that’s the key. Solve the variety of questions and first try to understand the question. Think about the concept that can be used to solve this question. And don’t jump directly to the solution after just reading the question. In JEE Main questions from this topic won’t be too exaggerated and will test your basics. Make sure you practice the ball-box and other related problems, as many questions previously asked previously in JEE Main revolve around this concept.

Best Books For Preparation of Permutation and Combination

First, finish all the concept, example and question given in NCERT Maths Book. You must be thorough with the theory of NCERT. Then you can refer to the book Cengage Mathematics Algebra. PnC is explained very well in this book and there are ample amount of questions with crystal clear concepts. You can also refer to the book Arihant Algebra by SK Goyal or RD Sharma. But again the choice of reference book depends on person to person, find the book that best suits you the best depending on how well you are clear with the concepts and the difficulty of the questions you require.

NCERT Solutions Subject-wise link:

• NCERT solutions for class 12 Physics.
• NCERT solutions for class 12 Chemistry.
• NCERT solutions for class 12 Mathematics.
• NCERT solutions for class 12 Biology.