unit 4 Elementary descrete mathematics combinatorics.pptx

Welcome to
Permutations & Combinations

Table of contents
𝟎𝟑
04
20
07
Session 01
Fundamental Principle of Counting
Multiplication Principle (Rule of AND) 04
Addition Principle (Rule of OR)
Factorial Notation
𝟐𝟑
38
24
24
36
Session 02
PERMUTATION
FORMULA FOR PERMUTATION
Special Case
Permutation when all the
objects are not distinct
𝟒𝟑
44
61
Session 03
Combination
String Method
𝟔𝟓
72
81
Session 04
Complementary Principle
Including/Excluding Specified
Objects
𝟗𝟎
91
101
102
Session 05
DIVISIBILITY MODEL
Rank
Circular Permutations
𝟏𝟏𝟎
112
116
Session 06
Formation of Groups
𝟏𝟐𝟖
129
143
135
Session 07
All Possible Selections
MODEL BASED ON DIVISORS
Exponent of Prime in 𝑛!
𝟏𝟒𝟕
148
Session 08
GEOMETRY MODELS
𝟏𝟔𝟕
168
176
179
Session 09
Derangements
Principle of Inclusion and Exclusion 173
Distinct Thing → Distinct Boxes
Identical Things → Distinct Boxes
𝟏𝟖𝟕
Session 10
Application of Multinomial Theorem 191

Fundamental Principles
of Counting
Session 01
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Key Takeaways
Total number of ways for two events simultaneously
𝑚 × 𝑛
Event 𝐴
Different ways
𝑚
Event 𝐵
Different ways
𝑛
Fundamental Principle of Counting:
Number of different ways of arranging
• Counting techniques
Selecting different objects
Multiplication Principle (Rule of AND)
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There are 3 ways to travel from 𝐴 to 𝐵 and 4 ways to travel from
𝐵 to 𝐶. In how many ways can a person travel from 𝐴 to 𝐶 via 𝐵.
𝐴
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𝐵 𝐶
3 ways 4 ways
Number of ways to travel from 𝐴 to 𝐵 = 3
Number of ways to travel from 𝐵 to 𝐶 = 4
Number of ways he can travel from 𝐴 to 𝐵 and 𝐵 to 𝐶 = 3 × 4 = 12 Ways

𝐵 𝐶
𝑥
𝑦
1
2
(𝑥, 1) (𝑥, 2) (𝑥, 3) (𝑥, 4)
(𝑦, 1) (𝑦, 2) (𝑦, 3) (𝑦, 4)
𝐴
(𝑧, 1) (𝑧, 2) (𝑧, 3) (𝑧, 4)
𝑧 3
4
• The Multiplication principle can be generalised for any finite number of events.
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Key Takeaways
Addition Principle (Rule of OR):
Total number of ways for an event 𝐴 or 𝐵 to occur
𝑚 + 𝑛
Event 𝐴
Different ways
𝑚
Event 𝐵
Different ways
𝑛
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Observe a road network in the diagram below.
𝐴
𝐵
𝐶
𝐷
In how many ways can a person travel from 𝐴 to 𝐶 via 𝐵 (or) via 𝐷.
Case I:Via 𝐵 𝐴 𝐵 𝐶
3 ways 4 ways
∴ Number of ways he can travel = 3 × 4 = 12 Ways
Case II:Via 𝐷 𝐴 𝐷 𝐶
2 ways 3 ways
∴ Number of ways he can travel = 2 × 3 = 6 Ways
Return To Top ∴ Total number of ways = 12 + 6 = 18 Ways

Key Takeaways
Addition Principle (Rule of OR):
Note:
 𝐴 and 𝐵 are disjoint.
 Can be extended to finite number of disjoint events.
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In how many ways can we pair up one boy with one girl from a group
of 3 boys and 5 girls ?
5 Girls
3 Boys
Total possible ways = 3 × 5 = 15
15
A
3
B
5
C
8
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D

ENTRY
EXIT
= 5 × 5
= 25
There are 5 doors to enter and exit the auditorium, but the same
door cannot be used to enter and exit the auditorium, Then in how
many ways can a student enter and exit the auditorium.
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ENTRY
EXIT
= 5 × 4
= 20
There are 5 doors to enter and exit the auditorium, but the same
door cannot be used to enter and exit the auditorium, Then in how
many ways can a student enter and exit the auditorium.
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4 3 5
Customer has 4 options,
Burgers Fries Beverages Deserts
2
4 × 3 × 5 × 2 = 120
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= 5 × 4 × 3 × 2
= 120
→
→
→
→
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= 12 × 11 × 10
= 1320
Twelve students compete for a race. The number of ways in which
first three places can be taken is ?
2𝑛𝑑
1𝑠𝑡
3𝑟𝑑
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A customer forgets a 3-digit code for an Automated Teller Machine
(ATM) in a bank. Find the largest possible number of unsuccessful
trials necessary to obtain correct code.
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10 × 10 × 10
= 1000
= 1000 − 1
= 999
Number of unsuccessful attempts:

Number of ways to
form 3 digit numbers
Number of ways of filling
3 vacant places in succession
Digits :1, 2, 3, 4, 5, 6
=
For 3 digit even numbers,
1 2 3 4 5 6
2 4 6
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How many 3 digit even numbers can be formed from the digits
1,2, 3, 4,5, 6 if the digits can be repeated.

Unit’s place can be filled in 3 ways.
Ten’s place can be filled in 6 ways. (Repetition is allowed)
Hundred’s place also can be filled in 6 ways. (Repetition is allowed)
6 6 3
∴ Required number of 3-digit even number
= 6 × 6 × 3
= 108
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How many 3 digit even numbers can be formed from the digits
1,2, 3, 4,5, 6 if the digits can be repeated.

i. Any of the 5 persons can leave the lift cabin in 7 different ways.
Thus, by product rule, the total number of ways = 7 × 7 × 7 × 7 × 7
= 75 = 16807.
7𝑡ℎ
6𝑡ℎ
5𝑡ℎ
4𝑡ℎ
3𝑟𝑑
2𝑛𝑑
1𝑠𝑡
𝐺
ii. First person can leave the lift cabin in 7 different ways.
Second person can leave the lift cabin in 6 different ways.
⋯
Thus, by product rule, the total number of ways = 7 × 6 × 5 × 4 × 3
= 2520.
5 persons entered the lift on the ground floor of an 8 −floor building (ground
floor included). Suppose each of them can leave the cabin independently at any
floor beginning with the first. In how many ways can each of the five persons can
leave the lift
i. at any of the 7 floors.
ii. at different floors.
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Key Takeaways
Factorial Notation:
• The product of first ‘𝑛’ natural numbers are denoted by 𝑛!
𝑛! = 1 × 2 × 3 × ⋯ × 𝑛 − 1 × 𝑛
When 𝑛 = 1,1! = 1
When 𝑛 = 2, 2! = 1 × 2 = 2
When 𝑛 = 3,3! = 1 × 2 × 3 = 6
When 𝑛 = 4,4! = 1 × 2 × 3 × 4 = 24
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NOTE:
• We define 0! = 1
• 𝑛! = 𝑛 × 𝑛 − 1 ! = 𝑛 × 𝑛 − 1 × 𝑛 − 2 ! and so on
5! = 5 × 4! = 120
6! = 6 × 5! = 720
7! = 7 × 6! = 5040
• Factorial of a negative integer is not defined.
• The value of 𝑛! ends with zero, if 𝑛 ≥ 5.
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Key Takeaways

Find the remainder when 1! + 2! + 3! + ⋯ + 100! Is divided by 15.
(∵ 5!, 6!, ⋯ are divisible by 15)
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Let 𝑆 = 1! + 2! + 3! + ⋯ + 100!
= 1! + 2! + 3! + 4! + 5! + ⋯ + 100!
= 1! + 2! + 3! + 4! + 15𝑘 𝑘 ∈ 𝐼
= 1 + 2 + 6 + 24 + 15𝑘
= 33 + 15𝑘
= 15 𝑘 + 2 + 3
∴ Remainder is 3.

Session 02
Permutations of distinct
and alike objects
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PERMUTATION
• A permutation is an arrangement in a definite order of a
number of objects taken some or all at a time.
• When all the objects are distinct, the number of permutations of
𝑛 distinct objects taken ′𝑟′ at a time and the objects do not repeat is
denoted by 𝑛 𝑃
𝑟 or 𝑃 𝑛, 𝑟 .
𝑛𝑃𝑟 = 𝑛 × 𝑛 − 1 × 𝑛 − 2 × ⋯ × 𝑛 − 𝑟 + 1
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Key Takeaways

𝑛𝑃𝑟 = 𝑛 × 𝑛 − 1 × 𝑛 − 2 × ⋯ × 𝑛 − 𝑟 + 1
Proof :
By definition, 𝑛𝑃𝑟 = number of permutations of 𝑟 things out of 𝑛 different things
= number of ways of filling up 𝑟 vacant places with 𝑛 different objects
Consider 𝑟 vacant places
1𝑠𝑡 2𝑛𝑑 3𝑟𝑑 𝑟𝑡ℎ
⋯
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𝑛 𝑛 − 1 𝑛 − 2 𝑛 − (𝑟 − 1)
So, we are filling the vacant places in succession without repetition.
∴ 𝑛𝑃𝑟 = 𝑛 × 𝑛 − 1 × 𝑛 − 2 × ⋯ × 𝑛 − 𝑟 + 1
(By Fundamental principle of counting)
Key Takeaways

•
𝑟
𝑛𝑃 =
𝑛 !
𝑛−𝑟 !
Proof :
𝑛𝑃𝑟 = 𝑛 × 𝑛 − 1 × 𝑛 − 2 × ⋯ × 𝑛 − 𝑟 + 1
𝑟
⇒ 𝑛 𝑃 = 𝑛 × 𝑛 − 1 × 𝑛 − 2 × ⋯ × 𝑛 − 𝑟 + 1
𝑛−𝑟 !
𝑛−𝑟 !
𝑟
⇒ 𝑛𝑃 =
𝑛 !
𝑛−𝑟 !
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Key Takeaways

𝑛 𝑃0 = 1 (Arranging nothing)
𝑛 𝑃
𝑛 = 𝑛! (Arranging all the things)
The number of permutations of 𝑛 different objects taken
𝑟 at a time,where repetition is allowed is 𝑛𝑟 .
•
•
•
Note :
Key Takeaways
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Given: 𝑛𝑃5 = 42 ⋅ 𝑛𝑃3.
⇒
𝑛 !
= 42 ⋅
𝑛 !
𝑛−5 ! 𝑛−3 !
𝑛 − 4
⇒ 𝑛2 − 7𝑛 − 30 = 0
⇒ 𝑛 − 10 𝑛 + 3 = 0
⇒ 𝑛 = −3 (rejected)
⇒ 𝑛 = 10
= 42𝑛 𝑛 − 1 (𝑛 − 2)
Find 𝑛 such that 𝑛𝑃5 = 42 ⋅ 𝑛𝑃3.
⇒ 𝑛 𝑛 − 1 𝑛 − 2 𝑛 − 3
⇒ 𝑛 𝑛 − 1 𝑛 − 2 𝑛 − 3
⇒ 𝑛 − 3 𝑛 − 4 = 42
𝑛 − 4 − 42 = 0
(∵ 𝑛 ≥ 5)
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Prove that:
𝑖
𝑛𝑃𝑟
𝑛𝑃𝑟−1
= 𝑛 − 𝑟 + 1
Also,𝑛 𝑃
𝑟 = 𝑛 × 𝑛 − 1 × (𝑛−2)𝑃(𝑟−2) and so on
𝑖𝑖 𝑛𝑃𝑟 = 𝑛 × 𝑛−1 𝑃 𝑟−1
𝑟
𝑖 We know, 𝑛 𝑃 =
𝑛 !
𝑛−𝑟 !
𝑛𝑃𝑟
𝑛𝑃𝑟−1
=
𝑛−𝑟 !
𝑛!
× (𝑛−𝑟+1)!
= (𝑛−𝑟+1)!
= 𝑛 − 𝑟 + 1
𝑛! 𝑛−𝑟 !
𝑛𝑃𝑟 = 𝑛 × (𝑛−1)𝑃(𝑟−1)
𝑟
𝑖𝑖 𝑛𝑃 =
𝑛−𝑟 !
𝑛!
= 𝑛 ⋅ (𝑛−1)!
𝑛−𝑟 !
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The value of (2 ⋅ 1𝑃0 − 3 ⋅ 2𝑃1 + 4 ⋅ 3𝑃2 − ⋯ upto 51st
term) + (1! − 2! + 3! − ⋯ upto 51st term) is equal to : JEE MAINS 2020
(2 ⋅ 1𝑃0 − 3 ⋅ 2𝑃1 + 4 ⋅ 3𝑃2 − ⋯ upto 51st term) + (1! − 2! + 3! − ⋯ upto 51st term)
4 ⋅ 3𝑃2 = 4! and so on.
2 ⋅ 1𝑃0 = 2! 3 ⋅ 2𝑃1 = 3!
= 2! − 3! + 4! − ⋯ − 51! + 52! + 1! − 2! + 3! − ⋯ − 50! + 51!
= 1 + 52!
1 − 5! 5!
A
B 1 + 52!
1 + 51!
D
C
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1

Three-digit numbers are to be formed by using the odd
digits only. The number of such numbers which satisfy the
following conditions :
I. Without repetition.
II. When repetition is allowed.
III. When at-least one digit is repeated.
Odd digits:1, 3, 5, 7, 9
3 2!
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Number of permutations without repetition = 5𝑃 =
5!
= 60
The number of permutations of 𝑛 different objects
taken 𝑟 at a time, where repetition is allowed is 𝑛𝑟 .
Number of permutations with repetition = 53 = 125.

125 − 60 = 65.
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Odd digits:1, 3, 5, 7, 9
Number of permutations with repetition
−
Number of permutations without repetition
Three-digit numbers are to be formed by using the odd
digits only. The number of such numbers which satisfy the
following conditions :

How many words, with or without meaning, can be made from the
letters of the word EQUATION, using each letter exactly once if :
i. 4 letters are used at a time.
ii. All letters are used but first letter is a vowel.
iii. All letters are used but last letter is a consonant.
Given word:E Q U A T I O N
8 different letters
Number of vowels = 5
Number of consonants = 3
Number of words = 8𝑃4 = 8 × 7 × 6 × 5 = 1680.
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_ _ _ _ _ _ _
7!
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O
I
A Number of words = 5 × 7! = 25,200
U
E
_
5

N
T
Q
_ _ _ _ _ _ _ _
3
7!
Number of words = 3 × 7! = 15,120

Key Takeaways
Special Case:
A C T :All the three letters are different.
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• Different arrangements
A C T A T C
C A T C T A
T A C T C A
• Number of arrangements = 6 = 3! (3𝑃3)

Key Takeaways
Special Case:
3!
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2!
Number of arrangements = = 3
•
G E G
G G E G G E
E G G :Out of three, two are identical.
E G G E G G
• Different arrangements G E G

Key Takeaways
•
Permutation when all the objects are not distinct:
The number of permutations of 𝑛 objects of which 𝑝 objects are of same
kind and rest are distinct =
𝑛 !
𝑝 !
The number of permutations of 𝑛 objects where 𝑃1objects are of one
• kind, 𝑃2 objects are of second kind, ⋯,𝑃𝑘 objects are of 𝑘𝑡ℎ kind and rest
(if any)are all distinct =
𝑛 !
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𝑃1!⋅𝑃2!⋯𝑃𝑘!

Find the number of words that can be formed by using all the
letters of the word :
𝑖) MESOPOTAMIA 𝑖𝑖) INDEPENDENCE
Possible words =
11!
2! × 2! ×(2!)
Possible words =
12!
4! × 3! ×(2!)
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Total letters = 11
MM AA OO ESPTI EEEE NNN DD IPC
Total letters = 12
𝑖) MESOPOTAMIA 𝑖𝑖) INDEPENDENCE

Given word :BANANA
6!
2! ×3!
= 60
6 letters
∴ Number of arrangements =
∴ Total number of words = 60
How many words can be formed using the letters of the word
BANANA ?
20
A
B 60
40
D
C
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30

Find the number of permutations of the letters of the word
∴ Number of permutations =
9!
4! × 2!
= 7560
∴ Number of permutations = 1 ×
8!
3!×2!
= 3360
ALLAHABAD. How many of them :
𝑖) start with A. 𝑖𝑖) end with L.
𝑖𝑖𝑖) start with A and end with L.
Given word :A L L A H A B A D
A A A A L L B D H
𝑖) start with A.
A __ _ _ _ _ _ _
A A A L L B D H
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∴ Number of permutations = 1 ×
8!
= 1680
4!
A __ _ _ _ _ _ L
3!
∴ Number of permutations =
7!
= 840
Find the number of permutations of the letters of the word
A A A L B D H
ALLAHABAD. How many of them :
𝑖) start with A. 𝑖𝑖) end with L.
𝑖𝑖) end with L.
_ _ _ _ _ _ _ _ L
A A A A L B D H
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Combination of objects
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Session 03

Key Takeaways
• Selection of 3 objects at a time
{A,B, D}
{A,C, D}
{B,C, D}
4 Different selections
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Combination:
• A combination is a selection of some or all of a number of different objects where
the order of selection is immaterial.
Consider four distinct objects A B
{A,B, C}
C D

Key Takeaways
Combination:
3 elements subset of {
A, B, C, D}
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• Number of combinations of 4 distinct objects
taken 3 at a time
= 4
{A,B, C}
{A,B, D}
{A,C, D}
{B,C, D}
In each selection, we have 3! permutations.
Number of permutations = 4 × 3! = 4𝑃3

Key Takeaways
Combination:
Number of selections of 3 objects out of 4 objects = 4
Number of arrangements of 3 objects taken all at a time = 3!
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•
•
• Number of arrangements of 4 objects taken 3 at a time = 4 × 3! = 4𝑃3

Key Takeaways
Combination:
taken 𝑟 at a time
𝑛 𝑃𝑟
𝑟!
𝑛
= 𝐶𝑟
In general,
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Number of selections of 𝑛 objects
× 𝑟! = 𝑛𝑃𝑟
Number of selections of 𝑛 objects
=
taken 𝑟 at a time

Key Takeaways
Combination:
• The number of combinations (selections) of 𝑛 distinct objects taken 𝑟 at a time,
𝑟
is denoted by 𝑛 𝐶 or 𝐶(𝑛, 𝑟) or
𝑛
𝑟
.
• 𝑛𝐶𝑟 𝑟 ! 𝑟 !(𝑛 −𝑟
)!
𝑛𝑃𝑟 𝑛!
= = (𝑛 ∈ 𝑁, 𝑟 ∈ 𝑊, 0 ≤ 𝑟 ≤ 𝑛)
Note
0
𝑖) 𝑛𝐶 =
𝑛!
0!𝑛 !
= 1 (Selecting nothing)
𝑛
𝑖𝑖) 𝑛𝐶 =
𝑛!
𝑛 !0!
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= 1 (Selecting all the things)

Eight teams participate in a cricket tournament. If each team plays
once against each of the others, find the total number of matches ?
Total :8 teams
Each team plays once against
each of the others
DKR
MSK
RMI
VCB
DRH
SR
KLXI
The total number of matches = 8𝐶2 = 28
DS
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5 boys
6 girls
5𝐶3
6𝐶4
Number of ways of selecting 3 boys = 5𝐶3 = 10
Number of ways of selecting 4 girls = 6𝐶4 = 15
∴ Total number of selections = 10 × 15 = 150
Find the number of ways of selecting 3 boys and 4 girls from
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5 boys and 6 girls.

How many words, with or without meaning, each of 3 vowels and 2
consonants can be formed from the letters of the word INVOLUTE ?
T
L
Given word :INVOLUTE
Vowels : I O U E Consonants : N V
∴ Number of selections of 3 vowels and 2 consonants
= 4𝐶3 × 4𝐶2
For each selection, we have
5! ways
Hence, number of required ways = 4𝐶3 × 4𝐶2 × 5!
= 2880
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Find the number of 6 - digit numbers in which all the odd digits
and only odd digits will appear.
odd digits = 1, 3, 5, 7, 9
_ _ _ _ _ _
6 digit number formed
by using 5 odd digits
⇒ 5 odd digits are to be arranged with one digit repeating in 6 places.
1 , 3 , 5 , 7 , 9 , 1
1 , 3 , 5 , 7 , 9 , 3
1 , 3 , 5 , 7 , 9 , 5
1 , 3 , 5 , 7 , 9 , 7
1 , 3 , 5 , 7 , 9 , 9
Possible selections
of digits
∴ The number of 6-digit numbers = 5𝐶1 2!
× 6!
=
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Total number of cards = 52
13 cards of each type
Total number of black cards = 26 T
otal number of red cards
J
Q
A 2 3 4 5 6 7 8 9 10 K
A 2 3 4 5 6 7 8 9 10 K Q J
A K Q J
2 3 4 5 6 7 8 9 10
A 2 3 4 5 6 7 8 9 10 K Q J
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J
Q
K
K Q J
K Q J
K Q J
Total number of face cards = 12
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A 2 3 4 5 6 7 8 9 10 K Q J
SPADE
A 2 3 4 5 6 7 8 9 10 K Q J
CLUB
A 2 3 4 5 6 7 8 9 10 K Q J
DIAMOND
A 2 3 4 5 6 7 8 9 10 K Q J
HEART
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The number of ways of choosing 4 cards from a pack of 52 playing cards,
such that :
𝑖) All are of the same suit is .
𝑖𝑖) Two are red and two are black cards is .
𝑖𝑖𝑖) All are face cards is .
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Selection
of suit = 4𝐶1
A 2 3 4 5 6 7 8 9 10 K Q J
A 2 3 4 5 6 7 8 9 10 K Q J
A 2 3 4 5 6 7 8 9 10 K Q J
A 2 3 4 5 6 7 8 9 10 K Q J
Selection of 4 cards from same suit = 13𝐶4
𝑖) All are of the same suit = 4𝐶1 × 13𝐶4 = 2860
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The number of ways of choosing 4 cards from a pack of 52 playing cards,
such that :
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𝑖) All are of the same suit is .
𝑖𝑖) Two are red and two are black cards is .
𝑖𝑖𝑖) All are face cards is .
Selection of two black cards = 26𝐶2
Selection of two red cards = 26𝐶2
𝑖𝑖) Two are red and two are black cards = 26𝐶2 × 26𝐶2 = 105625
𝑖𝑖𝑖) All are face cards = 13𝐶4 = 495

Selection of one heart = 13𝐶1
Selection of one spade = 13𝐶1
One is heart and other is spade = 13𝐶1 × 13𝐶1 = 169
Find the number of ways of choosing 2 cards from a pack of 52 playing
cards, such that one is heart and other one is spade.
D
A
B
C
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13𝐶1 × 13𝐶1
26𝐶1 × 26𝐶1
52𝐶2
13𝐶2

Given word :RANDOM
Vowels : A O
Find the number of permutations of the letters of the words RANDOM
such that vowels come together.
Consonants : R N D M
Consider as
a single unit
AO R N D M
5
∴ Number of arrangements
of 5 units = 5!
AO OA In each arrangement, two vowels can
be arranged in 2! Ways.
∴ Total number of arrangements = 5! × 2! = 240
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The number of permutations of 𝑛 distinct things, taken all at
A time when 𝑚 specified things always come together.
𝑚! 𝑛 − 𝑚 + 1 !
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String Method

6 women
The number of ways in which we can choose a committee from 4
men and 6 women so that the committee includes at least two men
and exactly twice as many women as men is .
4 men
Committee: At least 2 men and twice as many women as men.
Case 𝑖 : Two men and Four women
Number of selections = 4𝐶2 × 6𝐶4
Case 𝑖𝑖 : Three men and six women
Number of selections = 4𝐶3 × 6𝐶6
∴ Total number of ways = 4𝐶2 × 6𝐶4 + 4𝐶3 × 6𝐶6 = 94.
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There are 𝑚 men and two women participating in a chess
tournament. Each participant plays two games with every other
participant. If the number of games played by the men between
themselves exceeds the number of games played between the
men and women by 84, then the value of 𝑚 is .
Given: Number of men = 𝑚,
Number of women = 2,
Number of games = 𝑚𝐶2 × 2
vs
vs
Number of games = 𝑚𝐶1 × 2𝐶1 × 2
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2 × 𝑚𝐶2 = 2 × 𝑚𝐶1 × 2𝐶1 + 84
⇒ 2 × 𝑚 𝑚−1
2!
= 4 𝑚 + 84
⇒ 𝑚2 − 𝑚 = 4 𝑚 + 84
⇒ 𝑚2 − 5 𝑚 − 84 = 0
⇒ 𝑚 − 12 𝑚 + 7 = 0
𝑚 = −7
neglected
𝑚 = 12
There are 𝑚 men and two women participating in a chess
tournament. Each participant plays two games with every other
participant. If the number of games played by the men between
themselves exceeds the number of games played between the
men and women by 84, then the value of 𝑚 is .
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Permutations I
ncluding
/Excluding specified Objects
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Session 04

Given: Total number of questions = 12
Number of questions to be attempted = 8
CASE
PART (I) PART (II)
Number of
ways
Total : 5 Total : 7
𝑖 3 5 5𝐶3 × 7 𝐶5
𝑖𝑖 4 4 5𝐶4 × 7 𝐶4
𝑖𝑖𝑖 5 3 5𝐶5 × 7 𝐶3
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In an examination, a question paper consists of 12 questions divided into
two parts ,part (𝑖) and part (𝑖𝑖), containing 5 and 7 questions, respectively.
A student is required to attempt 8 questions in all, selecting at least 3
from each part. In how many ways can a student select the questions?

∴ Total number of ways in which the student can select
= 5𝐶3 × 7𝐶5 + 5𝐶4 × 7𝐶4 + 5𝐶5 × 7𝐶3
= 210 + 175 + 35 = 420
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In an examination, a question paper consists of 12 questions divided into
two parts ,part (𝑖) and part (𝑖𝑖), containing 5 and 7 questions, respectively.
A student is required to attempt 8 questions in all, selecting at least 3
from each part. In how many ways can a student select the questions?

The number of ways in which 5 boys and 3 girls can be arranged in a
row such that :
i) All the boys are together
.
ii) All the girls are not together
.
3 girls
5 boys
i)All the boys are together
Consider as a single unit.
5 boys sitting together,
taken as a single unit.
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i)All the boys are together
4! ways
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5! ways
∴ Total number of arrangements = 4! × 5! = 2880
ii)All the girls are not together
∴ Total number of arrangements in which all the girls are not together
= (Number of ways of arranging 5 boys and 3 girls)
− (Number of ways of arranging in which all the girls are together)

3 girls sitting together,
taken as a single unit
(Number of ways of arranging 5 boys and 3 girls)
= 5 + 3 !
And (Number of ways of arranging in which all the girls are together)
1 + 5 = 6 units
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3! ways
∴ Total number of arrangements = 3! × 6!
∴ Total number of arrangements in which all the girls are not together
= 5 + 3 ! − 6! × 3!
= 8! − 6! × 3! = 6! 8 × 7 − 6 = 36000
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6! ways

Complementary Principle
Note:
In the previous problem, we used complementary principle.
Number of favorable ways = Total number of ways
− Number of unfavourable ways
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4 boys
4 girls
i)No two girls are together
5
4
So, 4 boys can be arranged in 4! ways
and 4 girls can be arranged in five places in 𝐶 × 4! ways
∴ Total number of ways = 4! × 5𝐶4 × 4! = 2880
row such that:
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i) No two girls are together
.
ii)Boys and girls sit alternatively.

4 girls can be arranged in 4! ways.
4 boys can be arranged in 4! ways.
∴ Total number of ways = 4! × 4!
Hence,the total number of ways = 4! × 4! + 4! × 4! = 1152
row such that:
i) No two girls are together
.
ii)Boys and girls sit alternatively.
ii)Boys and girls sit alternatively
Case 1
:
4 boys can be arranged in 4! ways.
4 girls can be arranged in 4! ways
Case 2:
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And consonants can be arranged in
7!
2!
∴ Total number of required arrangements = 2 × 5! ×
7!
2!
= 604800
Find the number of arrangements of the letters of the word
PERMUTATIONS such that :
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i) All the vowels as one unit and all the consonants as one unit.
ii) There are always 4 letters between 𝑃 and 𝑆.
Vowels :E U A I O Consonants: P R M T T N S
There are two units, which can be arranged in = 2! ways
In each arrangement in vowels can be arranged in = 5! ways

Remaining 10 places can be filled with the letters
10!
2!
2!
∴ Total number of ways = 7 × 2 ×
10!
= 25401600
ERMUTATION
𝑃 𝑆
𝑆 𝑃
2 cases
1+1+1+1+1+1+1 = 7
Find the number of arrangements of the letters of the word
PERMUTATIONS such that :
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ii) There are always 4 letters between 𝑃 and 𝑆.
ii)There are always 4 letters between 𝑃 and 𝑆.

= 144
5𝑡ℎ 6𝑡ℎ 7𝑡ℎ
1𝑠𝑡 2𝑛𝑑 3𝑟𝑑 4𝑡ℎ
3! ways
4! ways
The number of permutations of the letters of the word HEXAGON
such that
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i)relative positions of the vowels and consonants are not changed.
ii)Order of the vowels is not changed.
Given word :HEXAGON
i) relative positions of the vowels and consonants are not changed.
Vowels : E A O Consonants : H X G N

4
7𝐶4 × 4!
Given word :HEXAGON
Vowels : E A O Consonants : H X G N
ii)Order of the vowels is not changed
Let us arrange the consonants in 4 out of 7 places.
This can be done in
Now remaining three places can be filled with vowels in
only one way. (∵ order of the vowels not to be
changed)
∴ Total number of ways = 7𝐶4 × 4! × 1 = 840
The number of permutations of the letters of the word HEXAGON
such that
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i)relative positions of the vowels and consonants are not changed.
ii)Order of the vowels is not changed.

Note :
The number of permutations of 𝑛 distinct things, in which the order of 𝑟 things
𝑟 !
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is not to be considered is
𝑛 !
= 𝑛 𝑃
𝑛−
𝑟

Including/Excluding Specified Objects
The number of combination of 𝑛 distinct things taken 𝑟 at a time, such that
𝑖) 𝑝 particular things are always included is
𝑖𝑖) 𝑝 particular things are always excluded is
𝑛−𝑝 𝐶 𝑟−𝑝
𝑛−𝑝 𝐶𝑟
𝑖𝑖𝑖) 𝑝 particular things are included and 𝑞 particular things are excluded is 𝑛−𝑝−𝑞 𝐶 𝑟−𝑝
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Key Takeaways

The number of ways in which a team of 11 players can be selected
from 22 players such that
i) 2 particular players are always included is
ii) 4 particular players are always excluded is
iii)2 particular players are included and 4 particular players are
excluded is
Given :𝑛 = 22, 𝑟 = 11
∴ 𝑛−𝑝 𝐶 𝑟−𝑝 = 20𝐶9
∴ 𝑛−𝑝 𝐶𝑟 = 18𝐶11
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∴ 𝑛−𝑝−𝑞 𝐶 𝑟−𝑝 = 16𝐶9
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The number of ways in which a team of 11 players can be selected
from 22 players such that
iii)2 particular players are included and 4 particular players are
excluded is
iii) 2 particular players are always included and 4 particular players are
excluded is

∴ Number of required subsets =
Given, 𝐴 = 𝐸, 𝑄, 𝑈, 𝐴, 𝑇, 𝐼, 𝑂, 𝑁
∴ 𝑛 = 8, 𝑟 = 5
Condition 𝐴, 𝑇 are always included
8−2 𝐶 5−2
= 6𝐶3
= 20
If set 𝐴 = 𝐸, 𝑄, 𝑈,𝐴, 𝑇, 𝐼, 𝑂, 𝑁 then the number of 5 elements.
Subsets of 𝐴 which always include 𝐴, 𝑇 is
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A guard of 12 men is formed from a group of 𝑛 soldiers. It is found
that two particular soldiers 𝐴, 𝐵 is 3 times as often together on guard
as three particular soldiers 𝐶, 𝐷, 𝐸. Then the value of 𝑛 is .
Number of soldiers = 𝑛, 𝑟 = 12
Number of ways in which 𝐴, 𝐵 are included is 3 times
that of 𝐶, 𝐷, 𝐸 included on guard.
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C
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E
B
A D
Number of ways in which 𝐴, 𝐵 are included is 3 times
that of 𝐶, 𝐷, 𝐸 included on guard.
Number of ways in which A and B are included = 𝑛 −2𝐶12−2
Number of ways in which C, D and E are included = 𝑛−3𝐶12−3

𝑛−2 𝐶 12−2
⇒ = 3 × 𝑛−3 𝐶 12−3
⇒ 𝑛−2 𝐶10 = 3 × 𝑛−3 𝐶9
⇒
𝑛−2 !
10! 𝑛−12 !
= 3 ×
𝑛−3 !
9! 𝑛−12 !
⇒ 𝑛−2
= 3
10
⇒ 𝑛 = 32
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5 girls
7 boys
Consider a class of 5 girls and 7 boys. The number of different teams
consisting of 2 girls and 3 boys that can be formed from this class, if
there are two specific boys 𝐴 and 𝐵 who refuse to be the member of
the same team, is
7𝐶3
Number of ways of
selecting 2 girls = 5𝐶2
Number of ways of
selecting 3 boys =
Team consists of 2 girls and 3 boys
Total number of ways to select 2 girls and 3 boys for the team
= 5𝐶2 × 7𝐶3 = 10 × 35 = 350
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5 girls
7 boys selecting 1 boy = 5𝐶1
the same team, is
Number of ways of
selecting 2 girls = 5𝐶2
Number of ways of
A B
If the two boys are included, only one boy has to be chosen from 5 boys
∴ Number of ways the team can be formed if both the boys are included
= 5𝐶2 × 5𝐶1 = 10 × 5 = 50
∴ Required number of selections = (Total no. of ways) − (No. of ways the team can be
formed if both the boys are included)
= 350 − 50 = 300
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JEE MAINS 2019
A
B
C
D
350
500
200
300
the same team, is
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Session 05
Rank of a word and Circular
Permutations
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0 Multiple of 7
504,3367,5719
DIVISIBILITY MODEL:
• A number is divisible by 3: If the sum of its digits is divisible by 3.
• A number is divisible by 4: If it’s last 2 digits are divisible by 4.
• A number is divisible by 6: If it is divisible by 2 and 3.
• A number is divisible by 7: if [2 × units-place digits − (Number formed by other digits)]
Example:
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DIVISIBILITY MODEL:
• A number is divisible by 8: If it’s last 3 digits are divisible by 8.
• A number is divisible by 9: If the sum of its digits is divisible by 9.
• A number is divisible by 11:
if (sum of the digits in the odd places) − (sum of digits in even places) is divisible by 11.
Example: 209,3564
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Number of 5 digited numbers using the digits 0, 1, 2, 3,4 with
repetition,which are divisible by 4, is:
Given digits:0, 1, 2, 3, 4
A number is divisible by 4 if its last two digits are divisible by 4.
Last two digits can be : 00 04 12 20 24 32 40 44
8 ways
4 × 5 × 5 × 8
∴ Total number of ways = 4 × 5 × 5 × 8 = 800.
800
A
1000
B
400
C
600
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D

216
A 600
B 240
C 3125
D
Given digits:0, 1, 2, 3, 4 and 5
A number is divisible by 3 if the sum of its digit is a multiple of 3.
3 3 + 9 = 12 9
The sum of given six digits = 0 + 1 + 2 + 3 + 4 + 5 = 15
15
⇒ Possible sum of five digits such that
the number formed is divisible by 3
= 12 or 15
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A five-digit number divisible by 3 is to be formed using the numbers
0, 1, 2, 3, 4 and 5 without repetitions. The total number of ways this
can be done is:

Case 𝑖 : Digits are 0, 1, 2, 4, 5 (Sum of digits = 12)
Number of ways = 5𝑃5 − (Number of ways in which 0 comes in first place)
= 5! − 4!
= 120 − 24 = 96
Case 𝑖𝑖 : Digits are 1, 2, 3, 4, 5 (Sum of digits = 15)
Number of ways = 5𝑃5 = 5! = 120
∴ Total number of ways = 96 + 120 = 216
A five-digit number divisible by 3 is to be formed using the numbers
0, 1, 2, 3, 4 and 5 without repetitions. The total number of ways this
can be done is:
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If the letters of the word MOTHER are permutated and all the words
so formed (with or without meaning) be listed as in a dictionary, then
the position of the word ‘MOTHER’ is .
Given word :MOTHER Alphabetic order of letters : E, H, M, O, R,T
From the given letters, the word MOTHER comes after the words :
1) Starting with E
2) Starting with H
3) Starting with M
𝐴) Starting with ME
𝐵) Starting with MH
𝑎) Starting with MOE
𝑏) Starting with MOH
𝐶) Starting with MO
𝑐) Starting with MOR
𝑑) Starting with MOT
𝑖) Starting with MOTE
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Next Word :MOTHER
[JEE MAIN 2020 ]

Starting with E
E
5!
= 120
Starting with H
H
5!
= 120
Starting with ME
M
4!
= 24
E
Starting with MH M
4!
= 24
H
Starting with MOE M O E
Starting with MOH M O H
Starting with MOR M
= 6
3!
= 6
3!
= 6
3!
O R
Starting with MOTE M
2!
= 2
O T E
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Starting with E E = 120
Starting with H
H = 120
Starting with ME M = 24
E
Starting with MH
M = 24
H
Starting with MOE
M
= 6
O E
Starting with MOH M
= 6
O H
Starting with MOR M
= 6
O R
Starting with MOTE M = 2
O E
Next word M O T H E R
T
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∴ Position of the word
MOTHER = 309𝑡ℎ
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Words Number of words Sum
Starting with E 120 −
Starting with H 120 240
Starting with ME 24 264
Starting with MH 24 288
Starting with MOE 6 294
Starting with MOH 6 300
Starting with MOR 6 306
Starting with MOTE 2 308

Given word :AGAIN
Alphabetic order of letters :A, A, G,I,N
Starting with A
A
Starting with G
G
4!
2!
I
= 12
4!
2!
If all the permutations of the letters of the word AGAIN are
arranged in the order as in a dictionary, then find the 49𝑡ℎ word.
Words Number of
words
Sum
Starting with 𝐴 24
Starting with 𝐺 12 36
Starting with 𝐼 12 48
49
= 12
4! = 24
The next word is
N A
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A G I
Starting with I

Rank:
The position of a word when all permutations of that word are written in alphabetical
order is called rank.
Example:
Let the word be CAT.
All possible permutation of the word CAT: CAT, CTA, ATC, TCA, ACT, TAC.
Arranging them in alphabetical order CAT: ACT, ATC, CAT, CTA, TAC, TCA
CAT is the 3𝑟𝑑 in the above list, so the rank of the word CAT is 3.
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Key Takeaways
The number of circular permutations of 𝑛 distinct objects is
𝑛 − 1 !
𝑎1
𝑎2
𝑎3
𝑎𝑛
𝑎𝑛−1
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Key Takeaways
Proof :
Let 𝑥 be the number of circular permutations of
𝑛 distinct objects.
For each circular arrangement, the number
of linear arrangements = 𝑛
∴ The number of linear
arrangements of 𝑛 distinct objects
= 𝑛 ×
Number of circular
arrangements of 𝑛 distinct things
𝑛! = 𝑛 × 𝑥
𝑛
∴ 𝑥 =
𝑛!
= 𝑛 − 1 !
∴ The number of circular permutations of 𝑛 distinct objects is 𝑛 − 1 !
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Key Takeaways
The number of circular permutations of 𝑛 distinct objects
is 𝑛 − 1 !
𝑎1
𝑎2
𝑎3
𝑎𝑛
𝑎𝑛−1
Difference between linear and circular arrangement :
Linear → Recognized starting place
Circular → No starting and ending place
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Note:
If positional places are marked, then circular
arrangements can be treated as linear
arrangements.

i) There is no restriction.
9 children can be arranged in 9 − 1 ! = 8! ways
The number of ways in which 5 boys and 4 girls can be seated at a
round table, if
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i)there is no restriction.
ii) all the girls are together
.
iii) all the girls are not together
.
iv)no two girls are not together
.

ii)All the girls are together.
Consider 4 girls as one unit.
∴ Total 6 units
6 units can be arranged in 6 − 1 ! = 5! ways.
4 girls can be arranged themselves in 4! ways.
Required number of ways = 5! × 4!
iii)All the girls are not together.
Required number of ways
= Total number of ways
− Number of ways in which all girls are
together
= 8! − 5! × 4!
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iv)No two girls are not together.
We can arrange 5 boys in 4! ways.
There will be 5 gaps to arrange 4 girls= 4! × 5C4
Total number of ways = 4! × 5C4 × 4!
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Out of 10 people, 5 are to be selected around a round table and the
remaining to be seated in a row. The number of arrangements is
.
Total :10 people
5! ways
Number of linear arrangements = 5!
Number of circular permutations of 𝑛 distinct
things, taken 𝑟 at a time is 𝑛𝐶𝑟 . 𝑟 − 1 !
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Required number of circular permutations = 10𝐶5 . 5 − 1 ! = 10𝐶5 . 4!
Required number of linear arrangements = 5!
Total number of arrangements = 10𝐶5 . 4! ⋅ 5!
1
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5
= ⋅ 10!
= 2 × 9!
Out of 10 people, 5 are to be selected around a round table and the
remaining to be seated in a row. The number of arrangements is
.

Session 06
Division and
Distribution of objects
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Number of ways =
2
8 − 1 ! 1
2
= × 7!
= 2520
The number of ways in which a garland can be made by using 8
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different flowers is .

Key Takeaways
2
directions are taken as not different is
1
𝑛 − 1 ! .
Circular Permutations :
Note :
Number of circular permutations of 𝑛 distinct things, if clockwise and anticlockwise
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Example :Circular arrangement of flowers in a Garland, Beads in a Necklace.

2 𝑟
when clockwise and anticlockwise directions are same is
1
.𝑛𝐶 . 𝑟 − 1 !
2 𝑟
Required number of ways =
1
.𝑛𝐶 . 𝑟 − 1 !
1
2
18
= × 𝐶12 × 12 − 1 !
1
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18
= × 𝐶12 × 11!
2
How many necklaces of 12 beads each can be made from 18
beads of various colors? .
Number of circular permutations of 𝑛 distinct things, taken 𝑟 at a time

Note :
Number of circular permutations of 𝑛 distinct things, taken 𝑟 at a time is 𝑛𝐶𝑟 . 𝑟 − 1 !.
In the above case, if clockwise and anticlockwise directions are taken as not different,
2 𝑟
then the number of circular permutations is
1
.𝑛𝐶 . 𝑟 − 1 !.
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How many different necklaces be made from 10 different beads such
that three particular beads always come together.
Necklaces be made from 10 different beads =
2
8−1 !
× 3!
=
7! × 3!
2
8 different beads including 3 beads together as 1 bead 9!
2
A
7!
2
B
9!×3!
2
C
7!×3!
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2
D

Key Takeaways
Formation of Groups:
Let us consider 4 people
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Number of ways = 4𝐶1 × 3𝐶3 =
4!
1!3!
= 4
Key Takeaways
Formation of Groups:
𝐼) Let us divide them into two unequal groups of sizes 1 and 3
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and 𝑛 is
𝑚+𝑛 !
𝑚! 𝑛 !
• Number of ways of dividing 𝑚 + 𝑛 + 𝑝 (𝑚 ≠ 𝑛 ≠ 𝑝) things into three unequal groups
of size 𝑚, 𝑛, 𝑝 is
𝑚+𝑛+𝑝 !
𝑚! 𝑛! 𝑝!
Key Takeaways
Results:
• Number of ways of dividing 𝑚 + 𝑛 (𝑚 ≠ 𝑛) things into two unequal groups of size 𝑚
𝑘
𝑛 is
• Number of ways of dividing 𝑛1 + 𝑛2 + ⋯ + 𝑛𝑘 into 𝑘 unequal groups of size 𝑛1, 𝑛2, … ,
(𝑛1+ 𝑛2 +⋯+ 𝑛𝑘)!
𝑛1! 𝑛2! … 𝑛𝑘!
General Results:
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Key Takeaways
𝐼𝐼) Let us divide them into two equal groups
2!
4𝐶2×2𝐶2
Number of ways = =
4!
2!(2!∙2!)
= 3
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• Number of ways of dividing 𝑘𝑛 things into 𝑘 equal groups of size 𝑛 each is
𝑘𝑛 !
𝑘! 𝑛! 𝑘
Key Takeaways
Results:
• Number of ways of dividing 2𝑛 things into two equal groups of size 𝑛 each is
2𝑛 !
2!(𝑛!∙𝑛!)
• Number of ways of dividing 3𝑛 things into three equal groups of size 𝑛 each is
3𝑛 !
3! 𝑛!∙𝑛!∙𝑛!
General Results:
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𝑖)
10
2
3
5
𝑖𝑖)
Number of ways =
10!
2!3!5!
Number of ways =
10!
2! 5! 2
𝑖𝑖𝑖)
𝑖𝑣)
Number of ways =
10!
2!2! 4! 2
10!
3! 3! 31!
Consider division of 10 distinct things into groups, in different cases.
10
2
4
4
10
5 5
10
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3
3 3
Number of ways =
1

10
2
2
2 2
2
10
1 4
10
4 6
10
2 3
Number of ways =
10!
5! 2! 5
2 3
Number of ways =
10!
1!2!3!4!
Number of ways =
10!
4!6!
2 3
Number of ways =
10!
2! 2! 2. 2! 3! 2
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𝑣)
𝑣𝑖)
𝑣𝑖𝑖)
𝑣𝑖𝑖𝑖)

𝑖)
Number of ways =
52!
4! 13! 4
Number of ways =
52!
4! 13! 4
× 4! =
52!
13! 4
Four equal sets
𝑖𝑖) Distributing equally among four people
Four equal sets can be distributed among 4 people in 4! ways
52
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13
13 13
13
The number of ways in which a pack of 52 cards:
𝑖) can be divided into four equal sets is .
𝑖𝑖) can be distributed among four people equally is .

Number of ways =
5!
1!2! 2! 2
= 15 Number of ways =
5!
2! 1! 23!
= 10
Case 1
Total number of ways = 15 + 10 = 25
5
1 2 2
Case 2
5
1
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1 3
Find the number of ways in which 5 books can be
𝑖) tied up in 3 bundles.
𝑖𝑖) distributed among 3 students
such that each student get at-least one book.
𝑖) 5 books → 3 bundles

1!2! 2! 2
Number of ways =
5!
= 15 ⋅ 3! = 90
(Distribution)
Total number of ways = 90 + 60 = 150
𝑖𝑖) 5 books → 3 students
Case 1
5
1 2 2
5!
2! 1! 23!
= 10 ⋅ 3! = 60
Number of ways =
(Distribution)
Case 2
5
1
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1 3

𝑃1 𝑃2 𝑃3 𝑃4
A boat is manned by 8 men, 4 on each side. In how many ways can 8
men be arranged if two particular men can only row on left side and
another one particular man can only row on right side?
8 men → 4 on each side
𝑃5 𝑃6 𝑃7 𝑃8
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5!
Remaining 5 people will be divided into two groups of size 2 and 3 in ways
2!3!
Here we have to arrange the crew on both the sides, which can be done in 4! ways
5!
2!3!
Total number of arrangements = × 4! × 4! = 5760
𝑃8
𝑃7
𝑃6
𝑃5
𝑃1 𝑃2 𝑃3 𝑃4
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Session 07
Selections from distinct and
Identical objects
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Key Takeaways
All Possible Selections:
Case 1:
• The total number of selections out of 𝑛 distinct things taken any number of
things (including nothing)is 2𝑛
• Total number of selections = 𝑛𝐶0 + 𝑛𝐶1 + ⋯ + 𝑛𝐶𝑛 = 2𝑛
• The total number of ways of selecting at least one thing from 𝑛 distinct things
= 2𝑛 − 1
• Total number of selections = 𝑛𝐶1 + 𝑛𝐶2 + ⋯ + 𝑛𝐶𝑛 = 2𝑛 − 𝑛𝐶0 = 2𝑛 − 1
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= 10𝐶1 + 10𝐶2 + ⋯ + 10𝐶10
= 210 − 10𝐶0
Number of ways = 210 − 1 = 1023
There are 10 lamps in a room, each can be switched on independently.
The number of ways in which the room can be illuminated is .
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Key Takeaways
All Possible Selections:
Case 2:
• The total number of selections out of 𝑛 similar things taken any number
of things (including nothing)is 𝑛 + 1.
• The total number of ways of selecting at-least one thing from 𝑛 similar things = 𝑛.
If there are three similar letters,say A,A and A,then
Total number of selections = {no A, one A, two A’s, three A’s}= 4 ways
Case 3:
• If there are 𝑝 similar things of one type, 𝑞 similar things of second type
and 𝑟 similar things of third type, then the number of ways of selecting
atleast one thing is 𝑝 + 1 𝑞 + 1 𝑟 + 1 − 1
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Number of ways of selecting at least one fruit
= 119
4 + 1
= 5 + 1 3 + 1 −1
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Find the number of ways of selecting at least one fruit from 5 mangoes,
4 apples and 3 bananas.

Key Takeaways
Case 4:
• If there are 𝑝, 𝑞, 𝑟 similar things of different kinds and 𝑛 distinct things, then the
number of ways of selecting atleast one thing is 𝑝 + 1 𝑞 + 1 𝑟 + 1 2𝑛 − 1
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2
Case II:2 alike of 1 kind and 2 alike of other kind:
4!
2!2!
Number of ways = 3𝐶 × = 18
Case III:2 alike and 2 distinct
1 2 2!
Number of ways = 3𝐶 × 6𝐶 ×
4!
= 540
Case IV:4 distinct
Number of ways = 7𝐶4 × 4! = 840
∴ Number of 4 letter words = 24 + 18 + 540 + 840 = 1422
Four letters words:
Case I:3 alike and 1 distinct.
1 3!
Number of ways = 1 × 6𝐶 ×
4!
= 24
We can get 4 four letter words, if we permutates
the letters in each of the above cases.
Number of 4 letter words with or without meaning that can be
formed using letters of the word INEFFECTIVE is:
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Key Takeaways
MODEL BASED ON DIVISORS:
• Divisors of 12 are: 1 2 3 4 6 12
• Number of divisors = 12 Sum of divisors = 28
Let us explain these two with ‘combinations’
12 = 22 ∙ 31 (prime factorization)
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2 = 21 ∙ 30
3 = 20 ∙ 31
4 = 22 ∙ 30
6 = 21 ∙ 31
12 = 22 ∙ 31
The number of divisors
= (Number of ways of selecting two 2’s)×
(Number of ways of selecting one 3)
= 2 + 1 1 + 1 = 6
• So, if 12 = 2𝑥 ∙ 3𝑦 then number of divisors = 𝑥 + 1 𝑦 + 1
Key Takeaways
12 = 22 ∙ 31 (prime factorization)
1 = 20 ∙ 30
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Sum of divisors = 20 ∙ 30 + 21 ∙ 30 + 20 ∙ 31 +22 ∙ 30 +21 ∙ 31 + 22 ∙ 31
= (20 + 21 + 22)(30 + 31)
= (7)(4)
= 28
So, if 12 = 2𝑥 ∙ 3𝑦 ,then
Sum of the divisors = (20 + 21 + 22 + ⋯ 2𝑥)(30 + 31 + 32 + ⋯ 3𝑦)
G.P
.
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G.P.
Key Takeaways

Key Takeaways
1
𝑝𝑥1+1
−1
𝑝1−1
2
𝑝𝑥2+1
−1
𝑝2−1
⋯
• → Sum of all the divisors =
Let 𝑁 be a positive integer and 𝑁 = 𝑝𝑥1
⋅ 𝑝𝑥2
⋅ 𝑝𝑥3
⋯
1 2 3
Where 𝑝1, 𝑝2, 𝑝3 ⋯ are distinct primes and 𝑥1, 𝑥2, 𝑥3 ⋯ ∈ 𝑁
• → Number of divisors of 𝑁 = 𝑥1 + 1 𝑥2 + 1 𝑥3 + 1
0
1 1
1
1
2
• → Sum of all the divisors = 𝑝 + 𝑝 + 𝑝 + ⋯ + 𝑝1
𝑥1
2
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0
2
1
2
2
2
𝑥2
𝑝 + 𝑝 + 𝑝 + ⋯ 𝑝 ⋯

𝑥1 + 1 𝑥2 + 1 ⋯ , if 𝑁 is not a perfect square
Let 𝑁 be a positive integer and 𝑁 = 𝑝𝑥1
⋅ 𝑝𝑥2
⋅ 𝑝𝑥3
⋯
1 2 3
Where 𝑝1, 𝑝2, 𝑝3 ⋯ are distinct primes and 𝑥1, 𝑥2, 𝑥3 ⋯ ∈ 𝑁
Number of ways in which 𝑁 can be resolved as a product of two factors is equal to
• 1
2
• 1
2 1
𝑥 + 1 2
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𝑥 + 1 ⋯ + 1 , if 𝑁 is a perfect square

iii. Sum of divisors =
22−1 34−1 53−1
2−1 3−1 5−1
= 3720
iv. Sum of proper divisor = 3720 − 1 + 1350 = 2369
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Consider the number 1350. Find
i. Number of divisors
ii. Number of proper divisors
iii. Sum of divisors
iv. Sum of proper divisors
Given: 1350 = 21 ∙ 33 ∙ 52
i. Number of divisors = 1 + 1 3 + 1 2 + 1 = 24
ii. Number of proper divisor = 24 − 2 = 22 ( Except 1 and 1350 )

In how many ways the number 8100 can be written
as product of two coprime factors?
Given: 8100 = 22 ∙ 34 ⋅ 52
Possible ways to express 8100 as product of two coprime factors are,
→ 1 × 22 ∙ 34 ∙ 52
→ 22 × 34 ∙ 52
→ 34 × (22 ∙ 52)
→ 52 × 22 ∙ 34
∴ Number of ways = 4
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• If ′𝑛′ is the number of different prime factors of 𝑁, then the number of ways in
which 𝑁 can be resolved as product of two coprime factors is equal to 2𝑛−1.
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Prime Factorization of 𝑛!
• Let 𝑛! = 2𝑒1 ⋅ 3𝑒2 ⋅ 5𝑒3 ⋅ 7𝑒4 ⋯
Example:
3! = 21 ⋅ 31
4! = 23 ⋅ 31
5! = 23 ⋅ 31 ⋅ 51
100! =?
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Key Takeaways

1
𝑒 =
𝑛
𝑝1
+
𝑛
𝑝12 +
𝑛
𝑝13 + ⋯
2
𝑒 =
𝑛
𝑝2
+
𝑛
𝑝22 +
𝑛
𝑝23 + ⋯
Exponent of 𝑝2 in 𝑛! Is given as
• Let 𝑝 be a prime number and 𝑛 be any positive integer such that :
𝑛! = 𝑝1
𝑒1 ⋅ 𝑝2
𝑒2 ⋅ 𝑝3
𝑒3 ⋯
Exponent of 𝑝1 in 𝑛! Is given as
Key Takeaways
. denotes greatest integer
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200! = 2𝑒1 ⋅ 3𝑒2 ⋅ 5𝑒3 ⋯
1
𝑒 =
200
2
+
200
22 +
200
23 +
200
24 +
200
25 +
200
26 +
200
27 + ⋯
= 100 + 50 + 25 + 12 + 6 + 3 + 1 + 0 = 197
3
𝑒 =
200
5
+
200
52 +
200
53 +
200
54 + ⋯
= 40 + 8 + 1 + 0 = 49
200! = 2197 ⋅ 3𝑒2 ⋅ 549 ⋯
= 2148 ⋅ 249 ⋅ 3𝑒2 ⋅ 549 ⋯
= 2148 ⋅ 3𝑒2 ⋅ 1049 ⋯
Number of zeros in 200! = 49.
Find the number of zeros in 200!
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Find the number of zeros in 200!
200
5
= 40
40
5
= 8
8
5
= 1
1
5
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= 0
Number of zeros in 200! = 40 + 8 + 1 + 0 = 49
Alternative solution:
Number of zeros in 200! directly equals to multiples of 10 in 200!
Among the prime factors 2 and 5 the highest power of 5 in 200! will be less than
the highest power of 2 in 200!
So, the highest power of 10 in 200! will be equal to highest power of 5 in 200!

Session 08
Applications of selections in
geometry
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HAND SHAKE
PROBLEM
Key Takeaways
GEOMETRY MODELS
• If there are 𝑛 points in a plane (no three are collinear)
then by joining them we can obtain:
i. 𝑛 𝐶2 straight lines
ii.𝑛𝐶3 triangles
𝑃1 𝑃2
𝑃3 𝑃4
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Key Takeaways
GEOMETRY MODELS
• If there are 𝑛 points in a plane of which 𝑚 points are
collinear, then by joining them, we can obtain
i. 𝑛 𝐶2 − 𝑚𝐶2 + 1 straight lines
ii.𝑛𝐶3 − 𝑚𝐶3 triangles
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• 𝑛𝐶2 − 𝑛 = 𝑛 𝑛−3
2
Key Takeaways
GEOMETRY MODELS
The number of diagonals of a polygon of 𝑛 sides (𝑛 vertices) is
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𝑚𝐶2 × 𝑛𝐶2
Key Takeaways
GEOMETRY MODELS
If a set of 𝑚 parallel lines are intersected by another set of 𝑛 parallel
lines, then the number of parallelograms, we can obtain is
•
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𝑛𝐶2
i. 𝑛 straight lines in a plane is
ii. 𝑛 circles in a plane is 2 × 𝑛 𝐶2
Key Takeaways
GEOMETRY MODELS
The maximum number of points of intersection of
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There are 12 points in a place of which 7 are collinear.
By joining them,we can have :
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i. Number of straight lines
ii. Number of triangles
iii.Number of circles
iv.Number of pentagons
i. Number of straight lines = 12𝐶2 − 7𝐶2 + 1 = 46
ii. Number of triangles = 12𝐶3 − 7𝐶3 = 185
iii. Number of circles = 12𝐶3 − 7𝐶3 = 185
iv. Number of pentagons = 12𝐶5 − 7𝐶5 = 771

The maximum number of points of intersection of :
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i)20 lines in a plane is .
ii) 10 circles in a plane is .
iii)20 lines and 10 circles in a plane is .
i)20 line in a plane
The maximum number of points of intersection of 𝑛 straight lines in a plane is 𝑛 𝐶2
⇒ Maximum number of points of intersection of 20 straight lines
= 20𝐶2 = 190
The maximum number of points of intersection of 𝑛 circles in a plane is 2 × 𝑛𝐶2.
⇒ Maximum number of points of intersection of 10 circles
= 2 × 10𝐶2 = 2 × 45 = 90

Case 𝑖
= 2 × 20𝐶1 × 20𝐶1 = 400
Case 𝑖𝑖
= 2 × 10𝐶2 = 90
Case 𝑖𝑖𝑖
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= 2 × 10𝐶2 = 90
Total = 400 + 90 + 190
= 680

A convex polygon has 44 diagonals. Find the number of its sides.
⋮
⋮
⋮
44 diagonals
Let 𝑛 be the number of slides ⇒ 𝑛 points (vertices)
The number of diagonals of a polygon of 𝑛 sides (𝑛
vertices) is 𝑛𝐶2 − 𝑛.
⋮
𝑉1 𝑉2
𝑉𝑛
𝑛𝐶2 − 𝑛 =
2
𝑛 (𝑛−1)
− 𝑛 = 44
⇒ 𝑛 (𝑛−3)
= 44
2
⇒ 𝑛 𝑛 − 3 = 88
⇒ 𝑛2 − 3𝑛 − 88 = 0
𝑉3
⇒ 𝑛 − 11 𝑛 + 8 = 0
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⇒ 𝑛 = 11 (∵ 𝑛 = −8 is not possible)

𝐵 𝐶
Triangle can be formed by
𝐴
𝐴𝐵 → 3 points 𝐵𝐶 → 4 points 𝐴𝐶 → 5 points
Selecting one point each on one side
Selecting two points on one side and
third point on any other side
OR
In a triangle 𝐴𝐵𝐶, the sides 𝐴𝐵, 𝐵𝐶 and 𝐴𝐶 have 3, 4 and 5 points
respectively on them.The number of triangles that can be
constructed using these points as vertices is .
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𝐵 𝐶
Case (i)
Selecting one point each on = 3𝐶1 × 4𝐶1 × 5𝐶1 = 60
one side
Case (ii)
Selecting two points on one side
and third point on any other side
= 3𝐶2 × 9𝐶1 + 4𝐶2 × 8𝐶1 + 5𝐶2 × 7𝐶1
= 27 + 48 + 70 = 145
∴ Number of triangles = 60 + 145 = 205
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𝑛 points
Given : 𝑇𝑛 +1 − 𝑇𝑛 = 21
Number of triangles = 𝑇𝑛 = 𝑛 𝐶3, 𝑇𝑛 +1 = 𝑛 +1𝐶3
𝑇𝑛+1 − 𝑇𝑛 = 21
𝑛𝐶2 + 𝑛𝐶3 = 𝑛+1𝐶3
⇒ 𝑛+1𝐶3 − 𝑛𝐶3 = 21
⇒ 𝑛𝐶2 = 21
⇒ 𝑛 𝑛 − 1 = 42
⇒ 𝑛 = 7
Let 𝑇𝑛 denotes the number of triangles which can be formed
by joining the 𝑛 points which lie on a circle. If 𝑇𝑛+1 − 𝑇𝑛= 21,
then the value of 𝑛 is .
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Given : A regular polygon of 10 sides
two vertices are consecutive
Number of ways in which no =
Number of ways of selecting any 3 vertices.
−
Number of ways of selecting 3 consecutive vertices.
−
Number of ways of selecting 3 vertices such
that two are consecutive.
A regular polygon of 10 sides is constructed. In how many ways can 3
vertices be selected so that no two vertices are consecutive ?
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Let the vertices be :𝑉1, 𝑉2, 𝑉3, ⋯ , 𝑉10
Number of ways of selecting any 3 vertices = 10𝐶3
𝑉1 𝑉2 𝑉3, 𝑉2 𝑉3 𝑉4 , ⋯ , 𝑉10 𝑉1 𝑉2
𝑉2
𝑉1
𝑉3
𝑖) Starting with 𝑉 𝑉 = 6
𝑉1 𝑉2 𝑉4
𝑉7
𝑉8
𝑉9
6 ways
𝑉6
𝑉5
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𝑉4
𝑉5
𝑉6
𝑉7
𝑉8
𝑉9
𝑉10
1 2
Similarly,we have starting with
𝑉2 𝑉3, 𝑉3 𝑉4, 𝑉4𝑉5, ⋯ , 𝑉10 𝑉1
Number of ways of selecting 3 consecutive vertices = 10
Number of ways of selecting 3 consecutive vertices
such that two are consecutive = 10 × 6 = 60

Number of ways of selecting any 3 vertices = 10𝐶3
Number of ways of selecting 3 consecutive vertices = 10
Number of ways of selecting 3 vertices
such that two are consecutive = 60
Number of ways in which no two vertices are consecutive
= 10𝐶3 − 10 − 60
= 120 − 70
= 50
𝑉
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2
𝑉1
𝑉3
𝑉4
𝑉5
𝑉6
𝑉7
𝑉8
𝑉9
𝑉10

How many squares are there in a chess board?
A chessboard has 64 small squares.
There are many more different sized
squares on the chessboard.
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1. No.of 8 × 8 𝑠𝑞. = 1
2. No.of 7 × 7 𝑠𝑞. = 4
3. No.of 6 × 6 𝑠𝑞. = 9
4. No.of 5 × 5 𝑠𝑞. = 16
5. No.of 4 × 4 𝑠𝑞. = 25
6. No.of 3 × 3 𝑠𝑞. = 36
7. No.of 2 × 2 𝑠𝑞. = 49
8. No.of 1 × 1 𝑠𝑞. = 64
∴ Total number of squares on chessboard
⇒ 64 + 49 + 36 + 25 + 16 + 9 + 4 + 1 = 204

Given word : ROORKEE
Let A ≡ {words begin with R }
B ≡ {words end with E }
A’ ∩ B’ ≡ {words neither begin with R nor end with E }
ROORKEE 𝑛 𝑈 =
7!
2! . 2! . 2!
𝑛 𝐴 =
6!
2! . 2!
R
Find the number of words that can be made by permutating all
the letters of the word ROORKEE, which neither begins with R nor
end with E.
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Let A ≡ {words begin with R } B ≡ {words end with E }
A’ ∩ B’ ≡ {words neither begin with R nor end with E }
ROORKEE 𝑛 𝑈 =
7!
2! . 2! . 2!
𝑛 𝐴 =
6!
2! . 2!
R
E
𝑛 𝐵 =
6!
2! . 2!
𝑛 𝐴 ∩ 𝐵
5!
2!
R E
=
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ROORKEE
𝑛 𝐴 =
6!
2! . 2!
𝑛 𝐵 =
6!
2! . 2!
𝑛 𝐴 ∩ 𝐵 =
5!
2!
𝑛 𝐴′ ∩ 𝐵′
= 𝑛 𝑈
= 𝑛 𝑈 − 𝑛 𝐴 𝖴 𝐵
− 𝑛 𝐴 − 𝑛 𝐵 + 𝑛 𝐴 ∩ 𝐵
=
7! 6!
− −
6!
+ 5!
2! .2! .2! 2! . 2! 2! . 2! 2!
𝑛 𝑈 =
7!
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2! . 2! . 2!
= 630 − 180 − 180 + 60 = 330

Session 09
Distribution of objects to
distinct boxes
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Key Takeaways
𝑛
𝐷 = 𝑛! 1 −
1
+
1
−
1
+ ⋯ + −1 𝑛 1
1! 2! 3! 𝑛 !
where 𝑛 ≥ 2
Derangements
If 𝑛 things are arranged in a row, then the number of rearrangements such that none of
them occupy their original positions are called Dearangements.
The number of Derangements of 𝑛 distinct things can be denoted by 𝐷𝑛 .
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Given word : BAG
All permutations = 3! = 6 ways =
ABG
AGB
BAG
BGA
GAB
GBA
Number of derangements = 𝐷3 = 2
Find the number of derangements of the letters of the word BAG.
𝑛
𝐷 = 𝑛! 1 −
1
+
1
−
1
+ ⋯ + −1 𝑛 1
1! 2! 3! 𝑛 !
where 𝑛 ≥ 2
3 1! 2! 3!
𝐷 = 3! 1 −
1
+
1
−
1
= 2
Using formula:
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Six letters with numbers 1,2,3,4,5,6written on them are put in six
envelopes with numbers 1,2,3,4,5,6written on them too.
𝑖) No letter is in its correct envelope is
𝑖𝑖) At least one letter is in its correct envelope is
𝑖𝑖𝑖) At most two letters are not in their correct envelopes is
𝑖𝑣) Exactly three letters are not in their correct envelopes is
Given:
.
.
Lette
r 1
.
Lette
r 2 Lette
r 3 Lette
r 4 Lette
r 5
. . . . . .
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. . . . .
. . . .
Lette
r 6
1 2 3 4 5 6

𝑖) No letter is in its correct envelope is
.
.
Lette
r 1
.
Lette
r 2 Lette
r 3 Lette
r 4 Lette
r 5
. . . . . .
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. . . . .
. . . .
Lette
r 6
1 2 3 4 5 6
Number of ways = 𝐷6= 265
𝑖𝑖) At least one letter is in its correct envelope is
Number of ways =Total number of permutations
− No letter is in its correct envelope
= 6! − 265
= 455

𝑖𝑖𝑖) At most two letters are not in their correct envelopes is
+ +
All letters are
in its correct
envelope
One letter isn’t
in its correct
envelope
Two letters are
not in their
correct envelope
= 1 + 0 + 6𝐶2 × 𝐷2
= 16
𝑖𝑣) Exactly three letters are not in their correct envelopes is
= 6𝐶3 × 𝐷3
= 40
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Principle of Inclusion and Exclusion
𝑛 𝐴 𝖴 𝐵
= + -
𝑛 𝐴 𝑛 𝐵 𝑛 𝐴 ∩ 𝐵
include exclude
𝑛 𝐴 𝖴 𝐵 = 𝑛 𝐴 + 𝑛 𝐵 − 𝑛 𝐴 ∩ 𝐵
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𝐴 𝐵
𝐶
𝑛 (𝐵 ∩ 𝐶)
𝑛 (𝐴 ∩ 𝐵 ∩ 𝐶)
𝑛 𝐴 𝖴 𝐵 𝖴 𝐶
= 𝑛 (𝐴) + 𝑛 (𝐵) + 𝑛 (𝐶) − 𝑛 (𝐴 ∩ 𝐵) − 𝑛 (𝐴 ∩ 𝐶) −
+
include
include
exclude
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In general, 𝑛 𝐴1 𝖴 𝐴2 𝖴 𝐴3 𝖴 ⋯ 𝖴 𝐴𝑛
= ෍ 𝑛 𝐴𝑖 − ෍ 𝑛 𝐴𝑖 ∩ 𝐴𝑗 + ෍ 𝑛 𝐴𝑖 ∩ 𝐴𝑗 ∩ 𝐴𝑘 − ⋯ + −1 𝑛 𝑛 𝐴1 ∩ 𝐴2 ∩ 𝐴3 ∩ ⋯ ∩ 𝐴𝑛
⇒ 𝑛 𝐴1′ 𝖴 𝐴2′ 𝖴 𝐴3′ 𝖴 ⋯ 𝖴 𝐴𝑛′ = 𝑛 𝑈 − 𝑛 𝐴1 𝖴 𝐴2 𝖴 𝐴3 𝖴 ⋯ 𝖴 𝐴𝑛
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Key Takeaways
Distinct Thing → Distinct Boxes:
The number of ways of distributing 𝑛 distinct things in 𝑟 distinct boxes such
that each box is filled with
Case 1
0 or more things (i.e. empty boxes are allowed) = 𝑟𝑛
Case 2
Atleast one thing (i.e. empty boxes are not allowed)
= 𝑟𝑛 − 𝑟 𝐶1 𝑟 − 1 𝑛 + 𝑟 𝐶2 𝑟 − 2 𝑛 − ⋯
+
−1 𝑟−1 ∙ 𝑟𝐶𝑟−1 1 𝑛
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In how many ways 5 different balls can be distributed into 3 boxes so
that no box remains empty.
= 35 − 3𝐶1 3 − 1 5 + 3𝐶2 3 − 2 5 − 3𝐶3 3 − 3 5
= 150
The required number of ways
Atleast one ball (i.e. empty boxes are not allowed)
−1 𝑟−1 ∙ 𝑟𝐶𝑟−1 1 𝑛
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= 𝑟𝑛 − 𝑟𝐶1 𝑟 − 1 𝑛 + 𝑟𝐶2 𝑟 − 2 𝑛 − ⋯ +
where 𝑛 = 5, 𝑟 = 3

𝑖𝑖) Number of functions from 𝐴 to 𝐵 such that
every element in 𝐵 has atleast one pre-image
If 𝐴 = 𝑎1, 𝑎2, 𝑎3, 𝑎4, 𝑎5 and 𝐵 = 𝑏1, 𝑏2, 𝑏3 then find
𝑖) Number of functions from 𝐴 to 𝐵
= 𝑟𝑛 = 35 = 243
𝑩
𝑨
𝒃𝟏
𝒃𝟐
𝒃𝟑
= 𝑟𝑛 − 𝑟𝐶1 𝑟 − 1 𝑛 + 𝑟𝐶2 𝑟 − 2 𝑛 − ⋯ +
= 35 − 3𝐶1 3 − 1 5 + 3𝐶2 3 − 2 5
= 150
−1 𝑟−1 ∙ 𝑟𝐶𝑟−1 1 𝑛
𝒂𝟏
𝒂𝟐
𝒂𝟑
𝒂𝟒
𝒂𝟓
If 𝐴 = 𝑎1,𝑎2, 𝑎3, 𝑎4, 𝑎5 and 𝐵 = 𝑏1,𝑏2, 𝑏3 then find
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𝑖) Number of functions from 𝐴 to 𝐵
𝑖𝑖) Number of functions from 𝐴 to 𝐵 such that every element in
𝐵 has atleast one pre-image

Key Takeaways
Proof:
Let 𝑛 identical objects put on floor
1 2 3 4 𝑛 − 1
⋯ ⋯ 𝑛
Identical Things → Distinct Boxes:
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Case 1
0 or more things (i.e. empty boxes are allowed) = 𝑛+𝑟−1𝐶𝑟−1

𝑟 − 1 identical partitions
𝑛 identical objects 𝑟 − 1 identical partitions
𝑛! 𝑟−1 !
𝑛+𝑟−1 !
= 𝑛+𝑟−1𝐶𝑟 −1
Proof (Cont.⋯):
To form 𝑟 −groups we need
1 2 3 4 ⋯ ⋯ 𝑛 − 1
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𝑛
Key Takeaways

𝑛−1𝐶𝑟−1
0 or more things (i.e.empty boxes are allowed) = 𝑛+𝑟−1𝐶𝑟−1
Now where 𝑛 − 𝑟 objects to be distributed in 𝑟 boxes.
= 𝑛−𝑟+𝑟−1𝐶𝑟−1 = 𝑛−1𝐶𝑟−1
Key Takeaways
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Case 2
Atleast one thing (i.e. empty boxes are not allowed) =
Proof:
Let us give each box 1object to satisfy atleast one condition.

Key Takeaways
For the equation 𝑥1 + 𝑥2 + 𝑥3 + ⋯ + 𝑥𝑟 = 𝑛
Where 𝑛 is 𝑛 identical units of 1.
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𝑖) Number of non-negative integral solution 𝑛 +𝑟 −1𝐶𝑟 −1
𝑖𝑖) Number of positive integral solution 𝑛−1𝐶𝑟−1

In how many ways 5 identical balls can be distributed into 3 distinct
boxes so that no box remains empty.
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Given:
𝑛 = 5, 𝑟 = 3
= 𝑛−1𝐶𝑟−1
= 4𝐶2
= 6

For the equation 𝑎 + 𝑏 + 𝑐 + 𝑑 = 12.
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𝑖) Number of non-negative integral solutions are
𝑖𝑖) Number of positive integral solutions are
𝑖) Number of non-negative integral solutions are
Given:
𝑛 = 12, 𝑟 = 4
𝑛+𝑟−1𝐶𝑟−1 = 12+4−1𝐶4−1
= 15𝐶3 = 455
𝑖𝑖) Number of positive integral solutions are
𝑛−1𝐶𝑟−1
= 12−1𝐶4−1
= 11𝐶3 = 165

In how many ways 10 apples, 5 mangoes, 4 oranges can be distributed
among 4 persons, when each person may receive any number of
fruits. (Fruits of each type are identical)
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Given:
Number of persons = 4
Let 𝑥1, 𝑥2, 𝑥3, 𝑥4 be number of apples receive by 𝑃1,𝑃2,𝑃3,𝑃4respectively
∴ 𝑥1 + 𝑥2 + 𝑥3 + 𝑥4 = 10
10+4−1𝐶4−1
= 13𝐶3
Let 𝑦1, 𝑦2, 𝑦3, 𝑦4 be number of mangoes receive by 𝑃1,𝑃2,𝑃3,𝑃4respectively
∴ 𝑦1 + 𝑦2 + 𝑦3 + 𝑦4 = 5
5+4−1𝐶4−1
= 8𝐶3

Let 𝑧1, 𝑧2, 𝑧3, 𝑧4 be number of oranges receive by 𝑃1,𝑃2,𝑃3,𝑃4 respectively
∴ 𝑧1 + 𝑧2 + 𝑧3 + 𝑧4 = 4
4+4−1𝐶4−1
= 7𝐶3
= 13𝐶3 × 8𝐶3 × 7𝐶3
= 560560
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In how many ways 10 apples, 5 mangoes, 4 oranges can be distributed
among 4 persons, when each person may receive any number of
fruits. (Fruits of each type are identical)

Multinomial theorem and
its applications
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Session 10

16
3 3 3 3
4
𝑎 𝑏 𝑐 𝑑
𝑎 + 𝑏 + 𝑐 + 𝑑 = 4
4+4−1𝐶4−1
= 7𝐶3
= 35
Given:
Remaining 4 to be distributed among
4 Persons so that any one receive any
amount,which is equal to the number of
non-negative integral solutions of
In how many ways can Rs. 16 be divided among 4 persons such that
none of them gets less than Rs. 3.
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Find the number of non-negative integral solutions of
𝑎 + 𝑏 + 𝑐 + 𝑑 ≤ 20.
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Given: 𝑎 + 𝑏 + 𝑐 + 𝑑 ≤ 20 ⋯ (𝑖)
Let 𝑒 ≥ 0 be a dummy variable such that
𝑎 + 𝑏 + 𝑐 + 𝑑 + 𝑒 = 20 ⋯ (𝑖𝑖)
Now, number of non-negative integral solutions of (𝑖)
= number of non-negative integral solutions of (𝑖𝑖)
= 𝑛+𝑟−1𝐶𝑟−1
𝑛 = 20, 𝑟 = 5
= 20+5−1𝐶5−1

𝑥 + 𝑦 + 𝑧 = 20
𝑢 − 4 + 𝑣 + 1 + 𝑧 = 20
𝑢 + 𝑣 + 𝑧 = 23⋯ (𝑖𝑖)
where 𝑢 ≥ 0, 𝑣 ≥ 0 and 𝑧 ≥ 0
Given: 𝑥 + 𝑦 + 𝑧 = 20 ⋯ (𝑖)
𝑥 ≥ −4 ⇒ 𝑥 + 4 ≥ 0
𝑦 ≥ 1 ⇒ 𝑦 − 1 ≥ 0
Let,𝑥 + 4 = 𝑢 and 𝑦 − 1 = 𝑣
Number of required solutions of (𝑖)
= Number of non- negative integral solutions of (𝑖𝑖)
= 23+3−1𝐶3−1 = 25𝐶2 = 300
Find the number of integral solutions of
𝑥 + 𝑦 + 𝑧 = 20 where 𝑥 ≥ −4, 𝑦 ≥ 1, ≥ 𝑧 ≥ 0.
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Key Takeaways
Application of Multinomial Theorem
In this theorem we try to write all possible outcomes in powers of a random
variable 𝑥 and then calculate coefficient of required outcomes.
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Find the number of ways of distributing 10 identical apples
among 3 children without any restrictions
Let 𝑎, 𝑏, 𝑐 be number of apples distributed to three children
𝑎 + 𝑏 + 𝑐 = 10; 𝑎 ≥ 0, 𝑏 ≥ 0. 𝑐 ≥ 0
Possible outcomes = 𝑎
𝑥0 + 𝑥1 + 𝑥2 + ⋯
𝑏
𝑥0 + 𝑥1 + 𝑥2 + ⋯
𝑐
𝑥0 + 𝑥1 + 𝑥2 + ⋯
Number of solution will be = Coefficient of 𝑥10 in
𝑥0 + 𝑥1 + 𝑥2 + ⋯ 𝑥0 + 𝑥1 + 𝑥2 + ⋯ 𝑥0 + 𝑥1 + 𝑥2 + ⋯
Infinite G.P
.
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Coefficient of 𝑥𝑟 in 1 − 𝑥 −𝑚 is 𝑚+𝑟−1𝐶𝑟
Coefficient of 𝑥10 in
1 1 1
1−𝑥 1−𝑥 1−𝑥
= Coefficient of 𝑥10 in 1 − 𝑥 −3
𝑎
𝑆∞ = 1−𝑟
= 66
= 10+3−1𝐶10
= 12𝐶10
𝑚 = 3, 𝑟 = 10
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Find the number of non-negative integral solutions of
𝑥1 + 𝑥2 + 𝑥3 + 4𝑥4 = 20.
Given: 𝑥1+𝑥2 + 𝑥3 + 4𝑥4 = 20
Number of non-negative integral solutions
= the coefficient of 𝑥20 in the product 1 + 𝑥 + 𝑥2 + ⋯ 3
1 + 𝑥4 + 𝑥8 + ⋯
𝑚+1 𝐶2 𝑥2 − 𝑚+2 𝐶3 𝑥3 + ⋯
= coefficient of 𝑥20 in 1 − 𝑥 −3 1 − 𝑥4 −1 1 + 𝑥 −𝑚
= 1 − 𝑚𝐶1 𝑥 +
= Coefficient of 𝑥20 in
(1 + 3𝐶1𝑥 + 4𝐶2𝑥2 + ⋯ + 𝑟+2𝐶𝑟𝑥𝑟) 1 + 𝑥4 + 𝑥8 + ⋯
= 1 + 6𝐶4 + 10𝐶8 + 14𝐶12 + 18𝐶16 + 22𝐶20
= 536
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Given :
15days → to visit for 4 days,
No two visits are consecutive.
Let 𝑉1, 𝑉2, 𝑉3, 𝑉4 be the 4 visiting days
An engineer is required to visit a factory for exactly 4 days during the
first 15 days of every month and it is mandatory that no two visits
take place on consecutive days. Then the number of all possible ways
in which such visits to the factory can be made by the engineer
during 1-15June 2021 is .
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JEE (ADVANCED)2020

= 8+5−1𝐶5−1 = 12𝐶4
= 495
𝑎 𝑏 𝑐 𝑑
𝑉1 𝑉2 𝑉3 𝑉4
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𝑒
where, 𝑎, 𝑏, 𝑐, 𝑑, 𝑒 be the remaining days
So, 𝑎 + 𝑏 + 𝑐 + 𝑑 + 𝑒 = 11 and 𝑎 ≥ 0, 𝑏 ≥ 1, 𝑐 ≥ 1, 𝑑 ≥ 1, 𝑒 ≥ 0
Put 𝑏 = 1 + 𝑥, 𝑐 = 1 + 𝑦, 𝑑 = 1 + 𝑡.
then, 𝑎 + 𝑥 + 𝑦 + 𝑡 + 𝑒 = 8 and 𝑎 ≥ 0, 𝑥 ≥ 0, 𝑦 ≥ 0, 𝑡 ≥ 0, 𝑒 ≥ 0
Number of non-negative integral solution
= 𝑛+𝑟−1𝐶𝑟−1
Let 𝑉1, 𝑉2, 𝑉3, 𝑉4 be the 4 visiting days
Consider,

Key Takeaways
Result :
If 𝑑1, 𝑑2, ⋯ , 𝑑𝑛 are the given non-zero digits, then the sum of all 𝑛 digit numbers
(without repetition) is equal to
𝑛 − 1 ! (𝑑1 + 𝑑2 + ⋯ + 𝑑𝑛 ) (100 + 101 + ⋯ + 10𝑛 −1)
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Find the sum of all four-digit numbers formed by using the
digits 2, 3, 4,5 without repetition.
_
Given digits:2, 3, 4, 5
_ _ _ _
Numbers having 2 in unit’s place = 3! 5
4
3
2
3!
Similarly,Numbers having 3, 4, 5 in
units place = 3!
_ _ _ _ 5
4
3
2
3!
Sum of all digits in unit’s place = 3! 2 + 3 + 4 + 5
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Number of 4-digit numbers = 4! = 24 _ _ _
Click to add tex4
t!

_ _ _ _
2
Sum of all digits in unit’s place = 3!
Number having 2 in ten’s place= 3!
2 + 3 + 4 + 5
3
4
Similarly,Numbers having 3, 4, 5 in
5
ten’s place = 3! _ _ _ _
Sum of digits in ten’s place =3! 2 + 3 + 4 + 5 × 10
Sum of all digits = Sum of all digits of unit’s ,ten’s, hundred’s and
thousand’s places.
Hence, the sum of all 4 digited number
= 3! 2 + 3 + 4 + 5 1111
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= 3! 2 + 3 + 4 + 5 100 + 101 + 102 + 103
= 6 × 14 × 1111 = 93324

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THANK
YOU
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unit 4 Elementary descrete mathematics combinatorics.pptx

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