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Last updated date: 26th Mar 2023

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**Permutation and combination are methods **for representing a collection of things by picking them from a set and forming them into subsets. It specifies the numerous methods for organising a set of data. Permutations are used for selecting data or objects from a group, whereas combinations are used to represent the order in which they are represented. In mathematics, both concepts are crucial.

Here we will go through the definition, explanation and formulas of the permutation and combination. Then we will solve some of the examples and **previous year's permutations and combinations questions** for better understanding.

Fundamental principle of counting

Permutations as an Arrangement

Combinations as Selections

Permutation and Combination of n items taken r at a time

Application of Permutation and Combination

Many important concepts are discussed in the chapter Permutation and Combination. These concepts are discussed in detail below.

Permutation

A permutation is an act of placing all the members of a set into a sequence or order in mathematics. To put it another way, permuting is the process of rearranging the components of a previously sorted set. Permutations can be found in practically every branch of mathematics, with different degrees of importance. When different orderings on certain finite sets are explored, permutations commonly occur.

Combination

The combination is a method of selecting elements from a collection in which the order of selection is not important (unlike permutations). In smaller instances, the number of possible combinations can be counted. Combination refers to the combination of n number of things taken k at a time without repetition. Also, there are some combinations in which repetition is allowed, the terms k-selection or k-combination with repetition are often used.

Permutation and combination concepts involve a lot of formulas. The two key formulas are as follows:

Permutation Formula

A permutation is the selection of r items from a collection of n items without replacement, with the order of the items being important.

nPr = (n!) / (n-r)!

Combination Formula

A combination is a selection of r items from a set of n items with no replacements and where order doesn't matter.

\[^{n}C_{r} = \frac {n!}{r!(n-r)!} = \frac {^{n}P_r}{r!}\]

Since a permutation involves selecting r distinct items without replacement from n items and order is important, by the fundamental counting principle, we have

P (n, r) = n. (n-1) . (n-2) . (n-3)…… (n-(r-1)) ways.

This can be written as:

P (n, r) = n.(n-1).(n-2). (n-3) …. (n-r+1)---------------> (1)

Multiplying and Dividing (1) by (n-r) (n-r-1) (n-r-2)........... 3. 2. 1, we get

P (n, r) = $\dfrac{n.(n-1).(n-2).…. (n-r+1)[(n-r) (n-r-1) (n-r-2)... 3. 2. 1]}{[(n-r) (n-r-1) (n-r-2)....3. 2. 1]}$

P (n, r) = $\dfrac{n!}{(n-r)!}$

Since combinations involve choosing r objects out of n objects where the order doesn't matter, we can determine that:

C(n,r) = the number of permutations /number of ways to arrange ‘r’ objects. $[$Since by the fundamental counting principle, we know that the number of ways to arrange ‘r’ objects in r ways = r!$]$

C(n,r) = P (n, r)/ r!

C(n,r) = $\dfrac{\dfrac{n!}{(n-r)!}}{r!}$

Thus we derive C(n,r) =$\dfrac{n!}{r!.(n - r)!}$

Example 1: Find the number of permutations and combinations if the data given is n = 12 and r = 2.

Solution: Given,

n = 12

r = 2

Using the formula given above:

Permutation:

nPr = (n!) / (n-r)! =(12!) / (12-2)! = 12! / 10! = (12 x 11 x 10! )/ 10! = 132

Combination:

$\begin{array}{l}^{n}C_{r} = \frac{n1}{r!(n-r)!}\end{array} $

$\begin{array}{l}\frac{12!}{2!(12-2)!} = \frac{12!}{2!(10)!} = \frac{12\times 11\times 10!}{2!(10)!} = 66\end{array} $

Example 2: Determine how many ways a committee of 5 men and 3 women may be chosen from a group of 9 men and 12 women?

Solution:

Choose 5 men from the group of 9 men = 9C5 ways = 126 ways

Choose 3 women from the group of 12 women = 12C3 ways = 220 ways

The committee was formed in 27720 ways.

Solved Previous year Questions

1. Prove that if each of the ‘m’ points in one straight line is joined to each of the n points on the other straight line, excluding the points on the given two lines. The number of points of intersection of these lines is $\begin{array}{l}\frac{1}{4}\end{array} $ mn(m-1)(n-1).

Solution:

Two points on the first line and two points on the second line are required to obtain one point of intersection. These can be chosen from a set of n points in nC2 ways or from a set of m points in mC2 ways.

Therefore, the required number = mC2 × nC2 = (m(m-1))/2! x (n(n-1))/2! =$\begin{array}{l}\frac{1}{4}\end{array}$

mn(m – 1)(n – 1)

Question 2. Number of divisors of n = 38808 (except 1 and n) is _____.

Solution:

Since, 38808 = 8 × 4851

= 8 × 9 × 539

= 8 × 9 × 7 × 7 × 11

= 23 × 32 × 72 × 11

So, the number of divisors = (3 + 1) (2 + 1) (2 + 1) (1 + 1) = 72.

This includes two divisors 1 and 38808.

Hence, the number of divisors required is = 72 – 2 = 70.

Question 3. A five-digit number divisible by 3 has to be formed using the numerals 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways in which this can be done is ________.

Solution:

We know that a five-digit number is divisible by 3 if and only if the sum of its digits (= 15), hence we should avoid using 0 or 3 while creating five-digit numbers.

Now,

(i) So, in the first case, we do not use 0; the five-digit number can be formed (from the digit 1, 2, 3, 4, 5) in 5P5 ways.

(ii) And, in the second case, we do not use 3; the five-digit number can be formed (from the digit 0, 1, 2, 4, 5) in 6 ! −5! × 2 = 480 ways.

The total number of such 5 digit number = 5P5 + (5P5 − 4P4)

= 120 + 96

= 216.

1. In how many ways can a committee of 5 be made out of 6 men and 4 women containing at least one woman?

(A) 246

(B) 222

(C) 186

(D) None of these

2. How many numbers greater than 10 lakhs can be formed from 2, 3,0, 3,4,2,3?

(A) 420

(B) 360

(C) 400

(D) 300

**Answer:** 1-(A), 2-(B)

Permutation and combination are explained elaborately in this article, along with the difference between them. We have discussed both the topics here with their formulas, examples and solved questions. With this students will be able when to use permutation and when to use combination formula in a question. Students can also work on** Permutations and Combinations **of different questions to enhance their knowledge and understanding of this chapter.

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FAQ

**1. What is the contribution of the chapter permutation and combination?**

At least one or two problems from this chapter will appear in JEE Main and other entrance exams every year. You will learn some fundamental counting strategies from the concept of permutation and combinations, which will allow you to answer questions without having to list 3-digit arrangements. In fact, similar methods can be used to determine the number of distinct ways to arrange and pick things without having to list them.

**2. How difficult is the chapter Permutation and Combination?**

You can make an endless number of different types of Permutation and Combination questions, but you still can't be sure you'll be able to answer all of them. So, in order to be able to answer the maximum amount of questions, you must practise a lot. Solve a variety of questions and first try to understand the question. Think about the concept that can be used to solve this question.

**3. What are the real-life examples of permutations and combinations?**

Permutations include arranging individuals, numerals, numbers, alphabets, letters, and colours. Combinations include menu selection, cuisine, clothing, subjects, and the team.

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