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Class 9 Maths Advanced Chapter 5 Combinatorics Notes
Class 9 Advanced Maths Chapter 5 Notes
Fundamental Principles of Counting and Factorials
The Fundamental Principle of Counting serves as the base for understanding permutations and combinations. It allows us to determine the total number of outcomes for an event by breaking it down into smaller, step-by-step decisions. The Fundamental Principle of Counting provides two core rules to calculate the total number of possible outcomes.
The Rule of Multiplication (The ‘AND’ Rule)
This rule is used when an event requires multiple steps to be completed together or in a specific sequence. We apply this rule, when the logic involves the word ‘AND.’
If one event can happen in m different ways and a second independent event can happen in n different ways then the total number of ways both events can happen together is m × n. This can be extended to any finite number of events.
e.g. In a school with 200 boys and 150 girls, a teacher wants to select a boy and a girl to represent the school in a competition.
Here, the teacher can select a boy in 200 ways and a girl in 150 ways. So, by fundamental principle of multiplication, the teacher can select a boy and a girl in 200 × 150 = 30000 ways.
If there are n events, say E1 E2 ….. En such that E1 can occur in m1 ways, E2 can occur in m2 ways, E3 can occur in m3 ways and so on then the total number of ways in which all the events can occur in the stated order is m1 × m2 × m3 × …., × mn.
The Rule of Addition (The ‘OR ’ Rule)
This rule is used when we have mutually exclusive or disjoint choices. We apply this rule, when the logic involves the word ‘OR.’
If an event can be completed by choosing from one set of m options or a distinct, non-overlapping set of n options then the total number of ways to complete the event is m + n.
e.g. In a school with 200 boys and 150 girls, a teacher wants to select either a boy or a girl to represent the school.
Here, the teacher can select a boy in 200 ways and a girl in 150 ways.
So, by fundamental principle of addition, the teacher can select either a boy or a giri in (200 + 150) = 350 ways.
Example 1.
A room has 7 doors. In how many ways can a man enter the room through one door and come out through a different door?
Solution:
Here, we need to perform two operations.
(i) Selecting a door to enter.
(ii) Selecting a door to come out.
Clearly, the man can enter the room through anyone of the seven doors.
So, there are seven ways of entering into the room. Note that the man can come out through anyone of the remaining six doors. So, he can come out through a different door in 6 ways.
Hence, by fundamental principle of multiplication, required number of ways = 7 × 6 = 42.
Example 2.
In a restaurant, the menu offers 8 different types of sandwiches and 5 different types of fresh juices. A customer wants to order one sandwich and one juice for lunch. In how many ways can the customer make this selection?
Solution:
We know, a customer needs to select one item from each category to form a complete lunch set.
∴ Number of ways to choose a sandwich = 8
and number of ways to choose a fresh juice = 5
Now, we can represent these two independent selections in a table format.
| Selection 1 : Sandwich | Selection 2 : Fresh Juice |
| 8 different types | 5 different types |
| ↓ | ↓ |
| 8 ways | 5 ways |
Hence, by fundamental principle of multiplication, required number of ways = 8 × 5 = 40.
Example 3.
In an examination, there are three multiple choice questions and each question has 4 choices. Find the number of ways in which a student can fail to get all answer correct.
Solution:
Since, each question can be answered in 4 ways.
So, the total number of ways answering 3 questions = 4 × 4 × 4 = 64
Out of these possible ways, only one way will be correct and hence the number of ways in which a student can fail to get all correct answer is 64 – 1 = 63.
Example 4.
How many 4-digit numbers are there, when a digit may be repeated any number of times?
Solution:
The thousand’s place of a 4-digit number cannot be 0. So, thousand’s place can be filled with any digit from 1 to 9. Thus, thousand’s place can be filled in 9 ways.
Since, repetition of digits is allowed, therefore each of the remaining 3 places can be filled in 10 ways using the digits from 0 to 9.
Hence, by the fundamental principle of multiplication, the four digit numbers, when a digit may be repeated any number of times = 9 × 10 × 10 × 10 = 9000.
Example 5.
How many numbers are there between 100 and 1000, which have exactly one of their digits as 7?
Solution:
We have to find numbers between 100 and 1000
i.e. 3-digit numbers which have exactly one of their digits as 7.
Now, consider the following cases.
Case I When 7 is at unit’s place
In this case, ten’s and hundred’s place can be filled using the digits 0, 1, 2, 3, 4, 5, 6, 8 and 9.
Clearly, the number of ways to fill ten’s place = 9 [using anyone of the digits 0, 1, 2, 3, 4, 5, 6, 8, 9] and the number of ways to fill the hundred’s place = 8
[∵ hundred’s place cannot filled by zero]
∴ Number of such numbers = 8 × 9 × 1 = 72
Case II When 7 is at ten’s place
In this case, unit’s and hundred’s place can be filled using the digits 0, 1, 2, 3, 4, 5, 6, 8 and 9.
Clearly, the number of ways to fill unit’s place = 9 [using anyone of digits 0, 1, 2, 3, 4, 5, 6, 8, 9] and the number of ways to fill hundred’s place = 8
[∵ hundred’s place cannot be filled by zero]
∴ Number of such numbers = 8 × 1 × 9 = 72
Case III When 7 is at hundred’s place
In this case, unit’s and ten’s place can be filled using the digits 0, 1, 2, 3, 4, 5, 6, 8 and 9.
Clearly, the number of ways to fill unit’s place = 9 [using anyone of the digits 0, 1, 2, 3, 4, 5, 6, 8, 9] and the number of ways to fill the ten’s place = 9 [using anyone of the digits 0, 1, 2, 3, 4, 5, 6, 8, 9]
∴ Number of such numbers = 1 × 9 × 9 = 81
Hence, the total number of required numbers = 72 + 72 + 81 = 225
Factorial Notation
Let n be a positive integer. Then, the continued product of first n natural numbers is called the n factorial and it is denoted bynl or ∠n.
Thus, n! = n(n – 1) (n – 2)(n – 3) ………… 3.2.1.
Zero Factorial
It does not make any sense to define 0! as product of the integers from 1 to 0. So, we define 0! = 1.
Factorial of proper fractions or negative integers are not defined.
n! is defined only for whole numbers i.e. for non-negative integers.
Example 6.
Evaluate 8!
Solution:
By definition, 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40320
Example 7.
Compute \(\frac{8!}{4!}\). Is \(\frac{8!}{4!}\) = 2!?
Solution:
We have, \(\frac{8!}{4!}\) = \(\frac{8 \times 7 \times 6 \times 5 \times 4!}{4!}\) [∵ n! = n(n – 1)(n – 2) …… 3.2.1]
= 8 × 7 × 6 × 5 = 1680
Also, 2! = 2 × 1 = 2 ≠ 1680
∴ \(\frac{8!}{4!}\) ≠ 2!
Example 8.
Find x, If \(\frac{1}{6!}\) + \(\frac{1}{7!}\) = \(\frac{x}{8!}\)
Solution:
We have \(\frac{1}{6!}\) + \(\frac{1}{7!}\) = \(\frac{x}{8!}\)
⇒ \(\frac{1}{6!}\) + \(\frac{1}{7 \times 6!}\) = \(\frac{x}{8 \times 7 \times 6!}\) ⇒ \(\frac{1}{6!}\)(1 + \(\frac{1}{7}\)) = \(\frac{x}{56 \times 6!}\)
On multiplying both sides by 6!, we get
⇒ 1 + \(\frac{1}{7}\) = \(\frac{x}{56}\) ⇒ \(\frac{7+1}{7}\) = \(\frac{x}{56}\)
⇒ \(\frac{8}{7}\) = \(\frac{x}{56}\) ⇒ x = \(\)
⇒ x = 8 × 8 ⇒ x = 64
Hence, the value of x is 64.
Example 9.
Find the value of \(\sqrt{x-y}\), if 1! + 2 × 2! + 3 × 3! + 4 × 4! + … + 100 × 100! = x! – y!
Solution:
Let S = 1! + 2 × 2! + 3 × 3! + 4 × 4! + … + 100 × 100!
Notice that n × n! = (n + 1 – 1) × n!
∴ n × n! = (n + 1) × n! – 1 × n! = (n + 1)! – n!
On applying this property to each term of the series, we get
S = (2! – 1!) + (3! – 2!) + (4! – 3!) + (5! – 4!) + … + (101! – 100!)
On simplifying, all intermediate terms cancel out.
∴ S = 101! – 1!
Comparing this with the given x! – y!, we get
x = 101, y = 1
Therefore, \(\sqrt{x-y}\) = \(\sqrt{101-1}\)
= \(\sqrt{100}\) = 10
Example 10.
Prove that \(\frac{(2 n)!}{n!}\) = {1.3.5 … (2n – 1)}2n.
Solution:
Permutations
A permutation is an arrangement of distinct objects in a definite order, which can be made by taking some or all objects at a time.
When we arrange objects in a specific sequence, the order matters, meaning that changing the position of even one object creates an entirely different result.
We use permutations in situations, where position or sequence is important such as seating people in chairs, generating passwords or awarding rankings,
e.g. If there are three objects A, B and C, the permutations taking two at a time are AB, BA, AC, CA, BC, CB and the permutations taking all at a time are ABC, ACB, BAC, BCA, CAB, CBA. In both cases, the number of permutations is 6 because, when the order is changed, a different permutation is obtained.
Fundamental Formulas for Permutations
The number of permutations of n distinct objects taken r at a time, where no repetition is allowed, is denoted by nPr or P(n, r) and calculated as
nPr = \(\frac{n!}{(n-r)!}\),
where 0 ≤ r ≤ n
Special Case
I. The number of permutations of n distinct objects, when all n objects are arranged at the same time is given by
nPn = n!
II. The number of permutations of n different objects taken r at a time, when repetition of objects is permitted in the arrangement is nr.
When r = 0, nP0 = \(\frac{n!}{(n!)!}\) = 1
Example 1.
If nP5 = 20 nP3 then find the value of n.
Solution:
We have, nP5 = 20 × nP3
⇒ \(\frac{n!}{(n-5)!}\) = 20 × \(\frac{n!}{(n-3)!}\)
⇒ (n – 3)! = 20 × (n – 5)!
⇒ (n – 3)(n – 4)(n – 5)! = 20 × (n – 5)!
⇒ (n – 3) (n – 4) = 20 = 5 × 4
⇒ (n – 3) (n – 4) = (8 – 3) (8 – 4)
On comparing both sides, we get n = 8
Example 2.
If 9Pr = 3024 then find r.
Solution:
We have, 9Pr = 3024
⇒ \(\frac{9!}{(9-r)!}\) = 3024 ⇒ \(\frac{9!}{(9-r)!}\) = 9 × 336
⇒ \(\frac{9!}{(9-r)!}\) = 9 × 8 × 7 × 6
\(\frac{9!}{(9-r)!}\) = \(\frac{9 \times 8 \times 7 \times 6 \times 5!}{5!}\) [multiplying numerator and denominator by 5!]
⇒ \(\frac{9!}{(9-r)!}\) = \(\frac{9!}{5!}\) ⇒ (9 – r)! = 5!
On comparing both sides, we get
9 – r = 5 ⇒ r = 4
Example 3.
in a class of 10 students, in how many ways can the positions of President, Vice-President and Secretary be filled?
Solution:
We arrange 3 students into specific roles from the 10 students. The order matters here because being elected president is different from being elected Secretary.
The required number of ways is 10P3 = 10 × 9 × 8 = 720 ways.
Example 4.
In how many ways two different rings can be worn in four fingers with atmost one in each finger?
Solution:
Clearly, required number of ways is same as number of permutations (or arrangements) of 4 different things taken 2 at a time.
So, required number of ways = 4P2 = \(\frac{4!}{2!}\) = \(\frac{4 \times 3 \times 2!}{2!}\) = 12
Example 5.
How many words, with or without meaning can be made from the letters of the word ‘MONDAY,’ assuming that no letter is repeated, if
(i) 4 letters are used at a time?
(ii) all letters are used at a time?
Solution:
Here, the word MONDAY contains 6 distinct letters.
(i) Required number of words = Number of permutations of 6 different things taken 4 at a time, 6P4 = \(\frac{6!}{2!}\) = \(\frac{6 \times 5 \times 4 \times 3 \times 2!}{2!}\) = 360
(ii) Required number of words = Number of permutations of 6 different things taken all at a time = 6P6 = 6! = 720
Example 6.
Find the number of 5-digit telephone numbers having atleast one of their digits repeated.
Solution:
Using the digits 0, 1, 2, …, 9, the number of 5-digit telephone numbers, which can be formed is 105 (since, repetition is allowed).
The number of 5-digit telephone numbers, which have none of the digits repeated = 10P5 = \(\frac{10!}{5!}\) = 30240.
Hence, the required number of telephone numbers having atleast one of their digits repeated = 105 – 10P5 = 100000 – 30240 = 69760
Example 7.
Find the number of different words that can be formed from the letters of the word ‘TRIANGLE’ so that
(i) all vowels occur together.
(ii) all vowels do not occur together.
Solution:
There are 8 distinct letters in the word TRIANGLE, out of which 3 are vowels, namely A, E, I and 5 are consonants, namely T, R, N, G, L.
(i) Since, the vowels have to occur together, so let us assume them as a single object (AEI).
A E I T, R, N, G,L
Now, this single object together with 5 remaining letters will be counted as 6 objects and these can be arrange in 6P6 = 6! ways.
Corresponding to each of these permutations, we have 3! = 6
Permutations of the three vowels A, E and I taken all at a time.
Hence, by fundamental principle of multiplication, the required number of words = 6! × 3!
= 4320
(ii) Clearly, required number of words = Number of all possible arrangements of 8 letters taken all at a time – Number of permutation in which the vowels are always together
= 8P8 – 6! × 3! = 8! – 6! × 3!
= 8 × 7 × 6! – 6v × 3! = 6! (56 – 6)
= 720 × 50 = 36000
Combinations
A combination is a selection of some or all of a group of different objects where the order does not matter. Unlike arrangements where position is key, a combination focuses only on which items are picked, not their sequence.
e.g. The different combinations formed from three letters A, B, C taking two at a time are AB, AC,BC.
Permutation vs Combiantion
- A combination is the process of selecting objects while a permutation is the process of arranging them.
- In a combination, only selection is made and the order in which the objects are arranged is immaterial while in a permutation, not only the selection is formed but also an arrangement in a definite order is considered.
Fundamental Formulas for Combinations
The number of combinations of n distinct objects taken r at a time is given by nCr = \(\).
In particular,
(i) when r = 0 then nC0 = 1
(ii) when r = n then nCn = 1
Some Important Results
• nCr = nCr.r!, 0 < r ≤ n
• nCr = nCn-r, 0 ≤ r ≤ n
• nCr + nCr-1 = n+1Cr
• nC1 + nCn-1 = n
Example 1.
If nC9 = nC8, find nC17.
Solution:
We have, nC9 = nC8
⇒ \(\frac{n!}{9!(n-9)!}\) = \(\frac{n!}{(n-8)!8!}\)
⇒ \(\frac{1}{9}\) = \(\frac{1}{n-8}\) ⇒ n – 8 = 9
⇒ n = 17
Therefore, nC17 = 17C17 = 1.
Example 2.
If nPr = 840 and nCr = 35 then find r.
Solution:
We know that nPr = r! × nCr
⇒ 840 = r! × 35
⇒ r! = \(\frac{840}{35}\) ⇒ r! = 24 = 4!
∴ r = 4
Example 3.
If n+2C8 : n-2P4 = 57 : 16 then find the value of n.
Solution:
We have, n+2C8 : n-2P4 = 57 : 16
⇒ (n – 1) n(n + 1) (n + 2) = 19 × 3 × 7 × 6 × 5 × 4 × 3
⇒ (n – 1) n(n + 1) (n + 2) = 19 × (3 × 7) × (6 × 3) × (5 × 4)
⇒ (n – 1) n(n + 1) (n + 2) = 18 × 19 × 20 × 21
On comparing both sides, we get
n – 1 = 18
∴ n = 19
Example 4.
In how many ways, can 5 sportsmen be selected from a group of 10?
Solution:
Since, we have to select 5 sportsmen out of 10.
∴ Required number of ways = 10C5 = \(\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}\)
= 2 × 3 × 7 × 6 = 252
Example 5.
If there are 15 persons in a party and if each two of them shake hands with each other. How many hand-shakes happen in the party?
Solution:
The total number of hand-shakes is same as the number of ways of selecting 2 persons among 15 persons
= 15C2 = \(\frac{15 \times 14}{2 \times 1}\) = 15 × 7 = 105
Example 6.
There are 12 non-collinear points in a plane. How many straight lines can be drawn amongst them?
Solution:
Since, any two points determine exactly one straight line and the 12 points are non-collinear (no three points lie on the same line), each pair of points will form a unique straight line.
The total number of straight lines is the number of ways of selecting 2 points out of 12 points, which is given by
12C2 = \(\frac{12!}{2!\times 10!}\) = \(\frac{12 \times 11}{2 \times 1}\) = 66
Hence, the total number of straight lines that can be drawn is 66.
Example 7.
From 7 boys and 5 girls, In how many ways can a team of 4 boys and 3 gills be formed?
Solution:
The team consists of 4 boys and 3 girls. Since, the order of selection does not matter, we calculate the ways of selecting boys and girls separately and multiply them.
Selection of 4 boys from 7 boys is done in,
7C4 = \(\frac{7!}{4!\times 3!}\) = \(\frac{7 \times 6 \times 5}{3 \times 2 \times 1}\) = 35 ways.
Selection of 3 girls from 5 girls is done in
5C3 \(\frac{5!}{3!\times 2!}\) = \(\frac{5 \times 4}{2 \times 1}\) = 10 ways.
Now, the total number of ways to form the team by applying the fundamental principle of counting = Ways of selecting 4 boys × Ways of selecting 3 girls = 35 × 10 = 350
Hence, the required number of ways is 350.
The post Combinatorics Class 9 Notes Advanced Maths Chapter 5 appeared first on Learn CBSE.
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