1. PERMUTATIONS AND COMBINATIONS

2. INTRODUCTION
The concept of permutation and combination is used to determine the ways and pattern in which objects can be arranged without actually listing them.
1,2,3,4?
1234
1243
1324
1423
1432……

3. CONTENTS Greetings!! I am a math Professor.
I am going to introduce you to the topic of permutations and combinations.
1. Fundamental Principle of Counting
2. Permutations
Distinct objects
Factorial Notation
Non-distinct objects
Theorems
3. Combinations
4. Summary
FUN FACT:
Do you know in how many distinct sequences can we study these 4 topics?
4 x 3 x 2 x 1 = 24 different sequences!

4. MULTIPLICATION PRINCIPLE study the multiplication principle
Welcome to my
class! Here we will
study the multiplication principle
Also called the FUNDAMENTAL PRINCIPLE OF COUNTING.
“If an event can occur in m different ways, following which another event can occur in n different ways, then the total number of occurrence of the events in the given order in m x n”

5. 6 POSSIBILITIES EXAMPLE: when repetition is not allowed. (2 X 3 = 6)
If a person has 2 T-shirts, 2 Bags and 2 pairs of Shoes, we can find the different combinations of these by the multiplication principle.
6 POSSIBILITIES
(2 X 3 = 6)

6. EXAMPLE: when repetition is allowed.
If we have to form three letter words, with or without meaning, by the letters in the word ‘CAT’ and the letters can be repeated, we can find the number of words by the formula n
Thus, we can make 3 X 3 X 3=27 words.
PROOF: n
CCC
CTA
ACT
ATA
TTC
TCC
CCA
CTT
ACC
ATT
TAT
TCA
CCT
CTC
AAT
ATC
TCT
CAA
CAT
AAC
TTT
TAC
CAC
AAA
ACA
TTA
TAA

7. you the concept of permutations.
Now, I will explain
you the concept of permutations.
Are you ready?
PERMUTATIONS are arrangements in a definite order of some objects taken few or all at a time.
Represented as P n r
n! represents n(n-1)(n-2)…..3 X 2 X 1
EXAMPLE: 5! = 5 X 4 X 3 X 2 X 1= 120

8. WHEN OBJECTS ARE DISTINCT
The number of permutations of n different objects taken r at a time, where 0 < r < n and the objects do not repeat is n(n-1)(n-2)…(n-r+1).
FORMULA: n! , 0<r<n
(n-r)! n r
P =

9. NOTE: When objects are distinct and repetition is allowed, then the permutation of n objects, taken r at a time is n .
EXAMPLE: If we have to make three letter words from the word ‘EIGHTY’, there can be two cases to solve:
Repetition not allowed
Repetition allowed. r

10. are clearly understanding this.
CASE 1: Repetition not allowed
n= 6
r= 3
I think that you
are clearly understanding this.
So, let’s continue…. n r
P = n! / (n-r)! = 6! / (6-3)!
=6 X 5 X 4 X 3 X 2 X 1 / 3 X 2 X 1
= 120 words.
CASE 2: Repetition allowed
n= 6
r= 3
n r r
P = n = 6 X 6 X 6 = 216

11. WHEN OBJECTS ARE NOT DISTINCT
This means that some objects are repeated more than one time.
EXAMPLE : If we have to find the number of words we can form with the words ‘BOOK’ taking all 4 letters at a time, we can’t simply do it by the previous methods.

12. NOTE: In this case the letter ‘O’ is being repeated
NOTE: In this case the letter ‘O’ is being repeated. So the formula will be n! = 4!
p! 2!
Where n= number of total objects
and p= number of times an object is repeated.
EXAMPLE: If we have to make words from the word ‘ASSASSIN’, we can use this formula.

13. ‘A S S A S S I N’ n= 8 p1= 2 (A is repeated twice)
p2= 4(S is repeated four times)
thus, number of words = 8! = 8X7X6X5X4!
2! X4! X1 X 4!
= 4 X 5 X 6 X 7
= 840 words 4

14. COMBINATIONS
If we are given n different objects, then taking r objects at a time, we can find the number of combinations by the formula:
C = n!
(n-r)! X r! n r

15. EXAMPLE: If we have to find the number of combinations of three letter words formed by the letters A,B,C,D & E, we can easily find it.
C = n!
(n-r)! X r!
= !
2! X 3!
= 10 n r

16. THANK
YOU!!!
Tanishi Gola
11th A