Class 9 Advanced Maths Chapter 5 Combinatorics Ex 5.4 Solutions

Students often refer to NCERT Class 9 Advanced Maths Solutions Chapter 5 Combinatorics Ex 5.4 to verify their answers.

Advanced Maths Class 9 Exercise 5.4 Solutions

Class 9 Advanced Maths Ex 5.4 Solutions

Question 1.
In how many ways can 3 students be chosen from a class of 12 to represent the school?
Solution:
Number of selections = ¹²C₃ = \(\frac{12!}{3!9!}\)
= \(\frac{12 \times 11 \times 10}{3 \times 2 \times 1}\) = 220

Question 2.
How many triangles can be formed from 12 points in a plane, of which 5 are collinear?
Solution:
A triangle is formed by choosing any 3 non-collinear points.
Number of triangles = ¹²C₃ – ⁵C₃
= \(\frac{12!}{3!9!}-\frac{5!}{3!2!}\)
= 220 – 10 = 210

Question 3.
An examination paper contains 12 questions divided into two parts, A and B. Part A contains 7 questions and Part B contains 5 questions. A candidate is required to attempt 7 questions, selecting at least 3 from each part. In how many ways can the candidate select the questions?
Solution:
Case 1: 3 questions from A, 4 from B
(⁷C₃) × (⁵C₄) = 35 × 5 = 175
Case 2: 4 questions from A, 3 from B
(⁷C₄) × (⁵C₃) = 35 × 10 = 350
Total number of selections = 175 + 350 = 525

Question 4.
How many diagonals does a polygon with 10 sides have?
Solution:
A diagonal is formed by joining any two vertices of a polygon.
Select any 2 vertices out of 10
Since order does not matter, we use combination:
¹⁰C₂ = \(\frac{10 \times 9}{2}\) = 45
This gives all possible line segments formed by joining two vertices.
Exclude the sides of the polygon

In a 10-sided polygon, there are 10 sides, and each side is formed by joining adjacent vertices.
So, number of diagonals: 45 – 10 = 35

Question 5.
If you invite 15 of your friends to a party and all shake hands exactly once, how many handshakes occur?
Solution:
15 friends shake hands exactly once.
Each handshake is between 2 people.
Number of handshakes = ¹⁵C₂ = 105

Question 6.
A committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of 1 man and 2 women?
Solution:
Total number of ways of selecting 3 persons
= ⁵C₃ = 10
Number of committees with 1 man and 2 women
= ²C₁ × ³C₂ = 2 × 3 = 6