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Permutations & Combinations

Permutations And Combinations

MODULES

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Basics of Permutations
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Basics of Combinations
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Letters Technique
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Using Blanks
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Blanks with Repetition
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Slotting Technique
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Conditional Permutations
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Special Types
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Circular Permutations
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Derangement
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Binomial Expansion
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Forming Groups
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Identical Elements
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Identical Groups
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Selection in Geometry
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Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

SPEED CONCEPTS

Permutations & Combinations 1
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Permutations & Combinations 2
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Permutations & Combinations 3
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Permutations & Combinations 4
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Permutations & Combinations 5
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Permutations & Combinations 6
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PRACTICE

Permutations & Combinations : Level 1
Permutations & Combinations : Level 2
Permutations & Combinations : Level 3
ALL MODULES

CAT 2025 Lesson : Permutations & Combinations - Basics of Permutations

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1. Introduction

Generally 222 to 333 questions are asked from Permutations & Combinations and Probability in CAT and other MBA entrance tests. These are perceived to be difficult chapters and are skipped very often. However, once you have a thorough understanding of the concepts, these questions take very little time to answer. Therefore, this is a scoring area which should not be skipped.

In this lesson, we will learn the basics of factorials (!)(!)(!), permutation and combination, and look at the different possible types of questions. In simple terms,
Permutations is the number of ways in which items can be selected and arranged.
Combinations is the number of ways in which items can be selected only.

1.1 Factorial and its properties

Factorial is represented by an exclamation mark ('!!!'). 'n!\bm{n}!n!' is read as 'n\bm{n}n factorial\bold{factorial}factorial'.

n!\bm{n!}n! is the product of all natural numbers less than or equal to n\bm{n}n.
⇒ n!=1×2...×nn! = 1 \times 2... \times n n!=1×2...×n
For example, 10!=1×2×3×4×5×6×7×8×9×1010! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 1010!=1×2×3×4×5×6×7×8×9×10

Exception: The only exception to this definition is 0!0!0!.
000 is not a natural number, however, 0!0!0! is defined. 0!\bold{0!}0! = 1.\bold{1.}1.

To enhance your speed, you should memorise the values of the following commonly used factorials.

Factorial Value
0!0!0! 111
1!1!1! 111
2!2!2! 222
3!3!3! 666
4!4!4! 242424
5!5!5! 120120120
6!6!6! 720720720
7!7!7! 5,0405,0405,040
8!8!8! 40,32040,32040,320


Note, n!=n×(n−1)!=n×(n−1)×(n−2)!=...n! = n \times (n - 1)! = n \times (n - 1) \times (n - 2)! = ...n!=n×(n−1)!=n×(n−1)×(n−2)!=...

For instance, 13!=13×12!=13×12×11!=...13! = 13 \times 12! = 13 \times 12 \times 11! = ...13!=13×12!=13×12×11!=...

1.2 Basics of Permutation

Permutation refers to the arrangement of elements or things. The order of arrangement matters.

Note that n\bm{n}n elements can be arranged in n!\bm{n!}n! ways. However, if r\bm{r}r elements are to be selected from a set of n\bm{n}n distinct elements and then arranged, this can be done in nPr=n!(n−r)!^{n}P_r = \dfrac{n!}{(n - r)!}nPr​=(n−r)!n!​ ways.

So, as the definition suggests, permutation includes selection and arrangement of elements. The following example explains the logic behind this formula.

Example 1

In how many ways can 444 students be seated on 222 chairs?

Solution

Let the 444 students be A, B, C and D. Let's begin by seating them in 444 chairs (instead of 222).

In the 1st1^{\mathrm{st}}1st chair, any of the 444 students can sit. In the 2nd2^{\mathrm{nd}}2nd chair, any of the 333 remaining students can sit (as 111 student is already seated in the 1st1^{\mathrm{st}}1st chair). Likewise, 222 students can sit on the 3rd3^{\mathrm{rd}}3rd chair and 111 student in the 4th4^{\mathrm{th}}4th chair. This concept is detailed in section 2.4 Using Blanks.

∴\therefore∴ the number of ways 444 students can sit in 444 chairs =4!=4×3×2×1=24= 4! = 4 \times 3\times 2 \times 1 = 24=4!=4×3×2×1=24

Below are the 242424 different ways in which 4\bm{4}4 students can be seated in 4\bm{4}4 chairs.

A B     C D B A     C D C A     B D D A     B C
A B     D C B A     D C C A     D B D A     C B
A C     B D B C     A D C B     A D D B     A C
A C     D B B C     D A C B     D A D B     C A
A D     B C B D     A C C D     A B D C     A B
A D     C B B D     C A C D     B A D C     B A


The question, however, requires us to arrange only 2\bold{2}2 of the 4\bold{4}4 students. So, the students seated in the 3rd3^{\mathrm{rd}}3rd and 4th4^{\mathrm{th}}4th chairs need not be arranged.

Let us take the case where A and B sit on the 1st1^{\mathrm{st}}1st and 2nd2^{\mathrm{nd}}2nd chairs respectively. This occurs twice in the above arrangements - A B C D and A B D C.

In this case, C and D, the unselected students, have been arranged as well. In every such case, 2\bold{2}2 unselected students are arranged in 2!\bold{2!}2! ways. Therefore, this has to be divided from the total possibilities.

∴\therefore∴ Number of ways 444 students can sit in 222 chairs =4!2!=242=12= \dfrac{4!}{2!} = \dfrac{24}{2} = 12=2!4!​=224​=12

A B B A C A D A
A C B C C B D B
A D B D C D D C


Alternatively (Recommended Method)

If there are n\bm{n}n elements, then number of ways they can be arranged =n!= \bm{n}!=n!

From these n\bm{n}n elements, if r\bm{r}r elements need to be selected and arranged, then there are n−r\bm{n - r}n−r elements that are not selected. The arrangement of these unselected elements in n!\bm{n}!n! does not matter. The unselected elements can be arranged in n−r!\bm{n - r}!n−r! ways.

∴ nPr=n!n−r!\therefore \space ^{n}{P}_r = \dfrac{\bm{n}!}{\bm{n - r}!}∴ nPr​=n−r!n!​

In this question, 222 students are to be selected from 444 students and then seated.

Number of ways =∴ 4P2=4!4−2!=242=12= \therefore \space ^{4}{P}_2 = \dfrac{\bm{4}!}{\bm{4 - 2}!} = \dfrac{24}{2} = 12=∴ 4P2​=4−2!4!​=224​=12

Answer: 12

So, the nPr^{n}{P}_rnPr​ formula is the number of arrangements of n\bm{n}n elements, i.e. n!\bm{n!}n!, divided by the number of arrangements of the unselected elements (n−r)(n - r)(n−r), i.e. (n−r)!\bm{(n - r)!}(n−r)!.

Example 2

In how many ways can 444 students be seated on 444 chairs?

Solution

As all 444 students are to be arranged, and there are no unselected items, the total number of ways = 4!=24\bm{4!} = \bm{24}4!=24

Alternatively

Total number of students =n=4= \bm{n} = 4=n=4

Number of students to be arranged =r=4= \bm{r} = 4=r=4

Total number of permutations = nPr= 4P4−4!(4−4)!=4!0!=24= \space ^{n}{P}_r = \space ^{4}{P}_4 - \dfrac{4!}{(4 - 4)!} = \dfrac{4!}{0!} = 24= nPr​= 4P4​−(4−4)!4!​=0!4!​=24

Answer: 24

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