2nd PUC Basic Maths Question Bank Chapter 2 Permutations and Combinations Ex 2.3

Students can Download Basic Maths Exercise 2.3 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 2 Permutations and Combinations Ex 2.3

Part – A

2nd PUC Basic Maths Permutations and Combinations Ex 2.3 One Mark Questions and Answers

Question 1.
In how many ways a committee of 5 can be chosen from 10 students.
Answer:
¹⁰C₅ ways = 252 ways,

Question 2.
Find n if  ⁿC₁₀ = ⁿC₁₅
Answer:
Given ⁿC₁₀ = ⁿC₁₅ ⇒ n = 10 + 15 = 25

Question 3.
Find x if  ⁹C_x + ⁹C₇ = ¹⁰C₇
Answer:
Given ⁹C_x + ⁹C₇ = ¹⁰C₇
w.k.t ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_r
∴ x = 8

Question 4.
How many straight lines can be formed from 10 points if no three of them are collinear.
Answer:
¹⁰C₂ ways = 45 ways

Question 5.
How many triangles can be formed from 8 non collinear points.
Answer:
⁸C₃ ways = 56 ways.

Part – B

2nd PUC Basic Maths Permutations and Combinations Ex 2.3 Two Marks Questions and Answers

Question 1.
If ⁿC₈ = ⁿC₁₂ find ⁿC₅.
Answer:
Given ⁿC₈ = ⁿC₁₂
⇒ n = 8 + 12 = 20
ⁿC₅ = ²⁰C₅ = 15,504.

Question 2.
If ⁿP_r = 30240 ⁿC_r = 252 find r
Answer:
If ⁿP_r = 30, 240, ⁿC_r = 252
w.k.t. \(^{n} C_{r}=\frac{^{n} P_{r}}{r !} \Rightarrow r !=\frac{^{n} P_{r}}{^{n} C_{r}}=\frac{30,240}{252}\)
r! = 120
r! = 5!
⇒ r = 5

Question 3.
Find the number of ways in which a committee of 4 Students and 2 Lecturers can be chosen out of 10 students and 8 lecturers.
Answer:
4 students out of 10 and 2 lecturers out of 8 can be selected in ¹⁰C₄ × ⁸C₂, ways = 210 × 28 = 5,880

Question 4.
Find r if  ²⁵C_r+3 = ²⁵C_r+3
Answer:
²⁵C_r+3 =  ²⁵C_r+3 ⇒ r + 3 = 2r – 3 ⇒ r = 6.

Question 5.
Find the number of straight lines and triangles that can be formed out of 20 points of which 8 are collinear.
Answer:
Number of straight lines =
²⁰C₂ – ⁸C₂ + 1 = 190 – 28 + 1 = 163 straight lines.
Number of triangles = ²⁰C₃ – ⁸C₃ = 1140 – 56 = 1084.

Question 6.
Find the number of parallelograms that can be formed from a set of 6 parallel lines intersecting another set of 4 parallel lines.
Answer:
This can be selected in ⁶C₂ × ⁴C₂ = 15 × 6 ways = 90 ways.

Question 7.
Find the number of diagonals of a polygon of 20 sides.
Answer:
Number of diagonals = ⁿC₂ – n = ²⁰C₂ – 20 = 190 – 20 = 170 ways.

Question 8.
In a party each person shakes hands with everyone else. If there are 25 members in the party, Calculate the number of handshakes.
Answer:
Number of hand shakes = ²⁵C₂ = \(\frac{25 \times 24}{2 \times 1}\) = 300 ways

Question 9.
In how many ways can 6 red and 4 white marbles be chosen from a bag containing 10 Red and 6 White marbles.
Answer:
We have to select 6 marbles out of 10 and 4 white out of 6 white marbles = ¹⁰C₆ × ⁶C₄ = 210 × 15 = 3,150.

Question 10.
In how many ways can 6 people be chosen out of 10 people if one particular person is always included.
Answer:
One particular person is included, we have to choose from 9, this can be done in ⁹C₃ ways.

Part – C

2nd PUC Basic Maths Permutations and Combinations Ex 2.3 Three marks Questions and Answers

Question 1.
If a convex polygon has 170 diagonals. Find the number of sides of the polygon.
Answer:
Number of diagonals = ⁿC₂ – n = 170.
\(\frac{n(n-1)}{2}-n=170\)
n² – n – 2n = 340 ⇒ x² – 3x – 340
(n + 17) = 0
n = 20 or -17
∵ n can’t be negative n = 20

Question 2.
A team of conveners consisting of 4 lecturers, 4 boys and 2 girls has to be selected from 15 lecturers, 10 boys and 12 girls. In how many ways the selection can be made if the 2 lecturers who were conveners last year have to be included and 3 boys who failed in the examination cannot be chosen.
Answer:

∴ Number of selections = ¹³C₂, × ⁷C₄ × ¹²C₂
= 78 × 35 × 66
= 1,80,180.

Question 3.
A committee of 4 has to be chosen from 10 boys and 8 girls. In how many ways can this be done if girls are in a majority.
Answer:

Question 4.
If ⁿP₅ = 24 ×ⁿC₄ find n
Answer:
Given ⁿP₅ = 24 × ⁿC₄

n – 4 = 1
⇒ n = 4 + 1 = 5

Question 5.
If  ⁿp_r = 240, ⁿC_r = 120 find n and r.
Answer:
Given ⁿp_r = 240, ⁿC_r = 120
w.k.t ⁿc_r = \(\frac{^{n} P_{r}}{r !}\)
∴ \(r !=\frac{240}{120}=2\)
r! = 2! ⇒ r= 2.

Question 6.
A team of 8 players has to be selected from 14 players. In how many ways the selections can be made if
(a) Two particular players are always included.
(b) Two particular players are always excluded.
Answer:
(a) Two players are always included: We have to choose 6 from 12 players. This can be done in ¹²C₆ ways = 924 ways.
(b) Two players are always excluded: We have to choose 8 from 12 players. This can be done in ¹²C₈ ways = 495 ways.

Part – D

2nd PUC Basic Maths Permutations and Combinations Ex 2.3 Five marks Questions and Answers

Question 1.
From a class of 9 boys and 7 girls, 12 students are to be chosen for a competition which includes atleast 6 boys and atleast 4 girls. In how many ways can this be done if a particular boy is always chosen.
Answer:
There are 9 boys and 7 girls, 12 students are to be chosen which includes atleast 6 boys & 4 girls one particular boy is always chosen ∴ we have to choose atleast 5 boys and 4 girls from 8 boys and 7 girls.

Question 2.
A man has 10 relatives, 4 of them are ladies 3 gentlemen and 3 children. In how many ways can he invite 7 relatives to a dinner party so that
(1) there are exactly 2 ladies, 3 gentlemen and 2 children.
(2) there are exactly 2 gentlemen and atleast 3 ladies. (1)
Answer:
1.

Number of selections = ⁴C₂ × ³C₃ × ³C₂ = 6 × 1 × 3 = 18 ways

(2)

Question 3.
From a class of 12 boys and 10 girls, 10 students are to be chosen for a competition, including atleast 4 boys and atleast 4 girls. The 2 girls who won the prize last year should be included. In how many ways can the selections be made.
Answer:

Question 4.
Abox contains 5 red, 4 black and 3 white balls. How many selections of 8 balls can be made if the selection contains.
(1) Exactly 4 red, 2 black and 2 white balls.
(2) Atleast 3 red, atleast 3 black and atleast 1 white ball.
Answer:

Question 5.
An examination paper consists of 12 questions divided into parts A and B contains 7 questions in part A and part B contains 5 questions. A candidate is required to answer 8 questions selecting atleast 3 from each part. In how many ways can the candidate select the questions.
Answer:

Question 6.
A man has 10 relatives, 4 of them all ladies, 3 gentlemen and 3 children. In how many ways can he invite 17 relatives to dinner party so that
(i) there are exactly 2 ladies, 3 gentlemen and 2 children
(ii) there are exactly 2 gentlemen and atleast 3 ladies
(iii) there are exactly 3 children, atleast 1 lady and atleast 2 gentlemen.
Answer:

Question 7.
A team of eleven is to be chosen out of 16 cricket players of whom 4 are bowlers and 2 wicket keepers. In how many ways can the team be chosen so that
(i) there are exactly 3 bowlers and one wicket keeper
(ii) there are atleast 3 bowlers and atleast one wicket keeper.
Answer:

Question 8.
Out of 4 officers and 10 clerks in an office a committee consisting of 2 officers and 3 clerks is to be formed. In how many ways can this be done if
(í) any officer and any clerk can be included
(ii) one particular clerk must be on committee
(iii) one particular officer cannot be on the committee.
Answer:
(i)

(ii)
One particular clerk is already on the committee.
∴ There are 2 clerks to be chosen from the remaining 9 clerks can be done in ⁴C₂ × ⁹C₂ = 6 × 36 = 216 ways.

(iii) One particular officer can not be on the committee. We have to choose 2 officers from 3 officers, can be done in ³C₂ × ¹⁰C₃ = 3 × 120 = 360 ways.

Question 9.
A committee of 5 are to be formed from 8 Americans and 5 Anglo Indians. In how many ways can this be done when the committee contains
(i) exactly two Anglo – Indians
(ii) atleast two anglo Indians.
Answer:

Question 10.
A candidate is required to answer 6 out of 12 questions which are divided into two groups containing 6 questions in each group. Find the number of choices he has if he cannot attempt more than five questions from any group.
Answer: