Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3

   

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3

Question 1.
If nC12 = nC9 find 21Cn.
Solution:
nCx = nCy ⇒ x = y or x + y = n
Here nC12 = nC9 ⇒ 12 ≠ 9 so 12 + 9 = n (i.e) n = 21
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 1

Question 2.
If 15C2r – 1 = 15C2r + 4, find r.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 2

Question 3.
If nPr = 720 and nCr = 120, find n, r.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 3

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3

Question 4.
Prove that 15C3 + 2 × 15C4 + 15C5 = 17C5
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 4

Question 5.
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 5
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 6

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3

Question 6.
If (n + 1)C8 : (n – 3)P4 = 57 : 16, find the value of n.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 7
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 8

Question 7.
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 101
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 105

Question 8.
Prove that if 1 ≤ r ≤ n then n × (n – 1)Cr – 1 = (n – r + 1)Cr – 1.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 9
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 106
(1) = (2) ⇒ LHS = RHS

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3

Question 9.

(i) A Kabaddi coach has 14 players ready to play. How many different teams of 7 players could the coach put on the court?
Solution:
No. of players in the team = 14
We need 7 players
So selecting 7 from 14 players can be done is 14C7 = 3432 ways

(ii) There are 15 persons in a party and if, each 2 of them shakes hands with each other, how many handshakes happen in the party?
Solution:
Total No. of persons = 15
Each two persons shake hands
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 11

(iii) How many chords can be drawn through 20 points on a circle?
Solution:
A chord is a line join of 2 points
No. of points given = 20
Selecting 2 from 20 can be done in 20C2 ways
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 12

(iv) In a parking lot one hundred, one year old cars are parked. Out of them five are to be chosen at random for to check its pollution devices. How many different set of five cars are possible?
Solution:
Number of cars =100
Select 5 from 100 cars can be done in 100C5 ways

(v) How many ways can a team of 3 boys, 2 girls and 1 transgender be selected from 5 boys, 4 girls and 2 transgenders?
Solution:
We have 5 boys, 4 girls and 2 transgenders. We need 3 boys, 2 girls and 1 transgender The selection can be done as follows Selecting 3 boys from 5 boys can be done is 5C3 ways
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 13
Selecting 2 girls from 4 girls can be done in 4C2 ways
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 14
Selecting 1 transgender from 2 can be done in 2C1 = 2 ways
∴ Selecting 3 boys, 2 girls and 1 transgender can be done in 10 × 6 × 2 = 120 ways

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3

Question 10.
Find the total number of subsets of a set with
(i) 4 elements
(ii) 5 elements
(iii) n elements
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 15
Solution:
If a set has n elements then the number of its subsets = 2n

(i) Here n = 4
So number of subsets = 24 = 16

(ii) n = 5
So number of subsets = 25 = 32

(iii) n = n
So number of subsets = 2

SamacheerKalvi.Guru

Question 11.
A trust has 25 members.

(i) How many ways 3 officers can be selected?
Solution:
Selecting 3 from 25 can be done in 25C3 ways
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 30

(ii) In how many ways can a President, Vice President and a secretary be selected?
Solution:
From the 25 members a president can be selected in 25 ways
After the president is selected, 24 persons are left out.
So a Vice President can be selected in (from 24 persons) 24 ways.
After the selection of Vice President 23 persons are left out
So a secretary can be selected (from the remaining 23 persons) in 23 ways
So a president, Vice president and a secretary can be selected in 25P3 ways 25P3 = 25 × 24 × 23 = 13800 ways

Question 12.
How many ways a committee of six persons from 10 persons can be chosen along with a chair person and a secretary?
Solution:
Selecting a chair person from the 10 persons can be done in 10 ways
After the selection of chair person only 9 persons are left out so selecting a secretary (from the remaining a persons) can be done in 9 ways.
The remaining persons = 8
Totally we need to select 6 persons
We have selected 2 persons.
So we have to select 4 persons
Selecting 4 from 8 can be done in 8C4 ways
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 40

Question 13.
How many different selections of 5 books can be made from 12 different books if,
Solution:
No. of books given = 12
No. of books to be selected = 5

(i) Two particular books are always selected?
Solution:
So we need to select 3 more books from (12 – 2) 10 books which can be done in 10C3 ways
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 41

(ii) Two particular books are never selected?
Solution:
Two particular books never to be selected.
So only 10 books are there and we have to select 5 books which can be done in 10C5 ways
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 42

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3

Question 14.
There are 5 teachers and 20 students. Out of them a committee of 2 teachers and 3 students is to be formed. Find the number of ways in which this can be done. Further find in how many of these committees
(i) a particular teacher is included?
(ii) a particular student is excluded?
Solution:
No. of teachers = 5
No of students = 20
We need to select 2 teachers and 3 students
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 43

(i) A particular teacher should be included. So from the remaining 4 teachers one teacher is to be selected which can be done in 4C1 = 4 ways
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 44
So selecting 2 teachers and 3 students can be done in 4 × 1140 = 4560 ways

(ii) particular student should be excluded.
So we have to select 3 students from 19 students which can be done in 19C3 ways
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 45
∴ 2 teachers and 3 students can be selected in 969 × 10 = 9690 ways

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3

Question 15.
In an examination a student has to answer 5 questions, out of 9 questions in which 2 are compulsory. In how many ways a students can answer the questions?
Solution:
No. of questions given = 9
No. of questions to be answered = 5
But 2 questions are compulsory
So the student has to answer the remaining 3 questions (5 – 2 = 3) from the remaining 7 (9 – 2 = 7) questions which can be done in 7C3 ways
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 46

Question 16.
Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly three aces in each combination.
Solution:
No. of cards = 52 : In that number of aces = 4
No. of cards needed = 5
In that 5 cards number of aces needed = 3
So the 3 aces can be selected from 4 aces in 4C3 = 4C1 = 4 ways
So the remaining = 5 – 3 = 2
This 2 cards can be selected in 48C2 ways
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 47

Question 17.
Find the number of ways of forming a committee of 5 members out of 7 Indians and 5 Americans, so that always Indians will be the majority’ in the committee.
Solution:
We need a majority of Indian’s which is obtained as follows.
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 48
The possible ways are (5I) or (4I and 1A) or (3I and 2A)
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 49

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3

Question 18.
A committee of 7 peoples has to be formed from 8 men and 4 women. In how many ways can this be done when the committee consists of
(i) exactly 3 women?
(ii) at least 3 women?
(iii) at most 3 women?
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 10
We need a committee of 7 people with 3 women and 4 men.
This can be done in (4C3) (8C4) ways
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 50
The number of ways = (70) (4) = 280

(ii) Atleast 3 women
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 51
So the possible ways are (3W and 4M) or (4W and 3M)
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 52
The number of ways (4) (70) + (1) (56) = 280 + 56 = 336

(iii) Atmost 3 women
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 53
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 54

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3

Question 19.
7 relatives of a man comprises 4 ladies and 3 gentlemen, his wife also has 7 relatives; 3 of them are ladies and 4 gentlemen. In how many ways can they invite a dinner party of 3 ladies and 3 gentlemen so that there are 3 of man’s relative and 3 of the wife’s relatives?
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 55
We need 3 ladies and 3 gentlemen for the party which consist of 3 Husbands relative and 3 wifes relative.
This can be done as follows
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 56
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 57

Question 20.
A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw?
Solution:
The box contains 2 white, 3 black and 4 red balls
We have to draw 3 balls in which there should be alteast 1 black ball
The possible draws are as follows
Black balls = 3
Red and White = 2 + 4 = 6
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 555

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3

Question 21.
Find the number of strings of 4 letters that can be formed with the letters of the word EXAMINATION.
Solution:
EXAMINATION
(i.e.) A, I, N are repeated twice. So the number of distinct letters = 8
From the 8 letters we have to select and arrange 4 letters to form a 4 letter word which can
be done in 8P4 = 8 × 7 × 6 × 5 = 1680
From the letters A, A, I, I, N, N when any 2 letters are taken as AA, II or AA, NN or II, NN
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 60
From AA, II, NN we select one of them and from the remaining we select and arrange 3 which can be done in ways
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 600
Total number of 4 letter word = 1680 + 18 + 756 = 2454

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3

Question 22.
How many triangles can be formed by joining 15 points on the plane, in which no line joining any three points?
Solution:
No. of non-collinear points = 15
To draw a Triangle we need 3 points
∴ Selecting 3 from 15 points can be done in 15C3 ways.
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 61

Question 23.
How many triangles can be formed by 15 points, in which 7 of them lie on one line and the remaining 8 on another parallel line?
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 62
7 points lie on one line and the other 8 points parallel on another paraller line.
A triangle is obtained by taking one point from one line and second points from the other parallel line which can be done as follows.
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 63
∴ Number of triangles = (7) (28) + (21) (8) = 196 + 168 = 364

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3

Question 24.
There are 11 points in a plane. No three of these lies in the same straight line except 4 points, which are collinear. Find,

(i) The number of straight lines that can be obtained from the pairs of these points?
Solution:
4 points are collinear
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 64
Total number of points 11.
To get a line we need 2 points
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 69
But in that 4 points are collinear
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 65
From (1) Joining the 4 points we get 1 line
∴ Number of lines = 11C24C2 + 1 = 55 – 6 + 1 = 50

(ii) The number of triangles that can be formed for which the points as their vertices?
A triangle is obtained by joining 3 points.
So selecting 3 from 11 points can be
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 68
But of the 11 points 4 points are collinear. So we have to subtract 4C3 = 4C1 = 4
∴ Number of triangles = 165 – 4 = 161

Question 25.
A polygon has 90 diagonals. Find the number of its sides?
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 699
∴ n = 15

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 Additional Questions Solved

Question 1.
A group consists of 4 girls and 7 boys. In bow many ways can a team of 5 members be selected, if the team has
(i) no girls
(ii) atleast one boy and one girl.
(iii) at least three girls
Solution:
We have 4 girls and 7 boys and a team of 5 members is to be selected.

(i) If no girl in selected, then all the 5 members are to be selected out of 7 boys
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 70

(ii) When at least one boy and one girl are to be selected, then
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 71
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 72
Hence the required number of ways are (i) 21 ways (ii) 441 ways (iii) 91 ways

Question 2.
A committee of 6 is to be choosen from 10 men and 7 women so as to contain atleast 3 men and 2 women. In how many different ways can this be done it two particular women refuse to serve on the same committee?
Solution:
We have 10 men and 7 women out of which a committee of 6 is to be formed which contain atleast 3 men and 2 women
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 73
∴ Total number of committee = 8610 – 810 = 7800
Hence, the value of the filler is 7800

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3

Question 3.
Using the digits 1, 2, 3,4, 5, 6, 7 a number of 4 different digits is formed. Find
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 74
Solution:
(a) Total of 4 digit number formed with 1, 2, 3, 4, 5, 6, 7
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 75
(b) When a number is divisible by 2 = 4 × 5 × 6 × 3 = 360
(c) Total numbers which are divisible by 25 = 40
(d) Total numbers which are divisible by 4 (last two digits is divisble by 4) = 200
Hence, the required matching is (a) ⟷ (z), (b) ⟷ (iii), (c) ⟷ (iv), (d) ⟷ (ii)

Question 4.
If 22Pr + 1 : 20Pr + 2 = 11 : 52, find r.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 76
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 77

Question 5.
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 78
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 79

Question 6.
A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:
(i) exactly 3 girls?
(ii) at least 3 girls?
(iii) almost 3 girls?
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 80

(ii) We have to select at least 3 girls. So the committee consists of 3 girls and 4 boys or 4 girls and 3 boys.
∴ Number of ways of selection = 4C3 × 9C4 + 4C4 × 9C3
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 81

(iii) We have to select at most 3 girls. So the committee consists of no girl and 7 boys or 1 girl and 6 boys or 2 girls and 5 boys or 3 girls and 4 boys.
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 86

Question 7.
Determine n if
(i) 2nC3 : nC2 = 12 : 1
(ii) 2nC3 : nC3 = 11 : 1
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 877
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 88

Question 8.
Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.
Solution:
There are 6 red balls, 5 white balls and 5 blue balls.
We have to select 3 balls of each colour.
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 89

Question 9.
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 90
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 91

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3

Question 10.
If nC4, nC5 and nC4 are in A.P. then find n.
[Hint: 2nC5 = nC6 + nC4]
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 92
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 93