Class 9 Advanced Maths Chapter 5 Combinatorics Ex 5.4 Solutions - #NCSOLVE 📚

Accurate and updated NCERT Class 9 Advanced Maths Solutions Chapter 5 Combinatorics Ex 5.4 help students build a strong mathematical foundation.

Ex 5.4 Class 9 Advanced Maths Solutions

Advanced Maths Class 9 Exercise 5.4 Solutions

Exercise 5.4 Class 9 Advanced Maths Solutions

Question 1.
In how many ways can 3 students be chosen from a class of 12 to represent the school?
Solution:
Clearly, required number of ways is same as number of combinations (or selections) of 12 different things taken 3 at a time.
So, required number of ways
= ¹²C₃ = \(\frac{12!}{3!(12-3)!}\) = \(\frac{12!}{3!\times 9!}\)
= \(\frac{12 \times 11 \times 10 \times 9!}{3 \times 2 \times 1 \times 9!}\) = \(\frac{12 \times 11 \times 10}{6}\)
= 2 × 11 × 10 = 220

Question 2.
How many triangles can be formed from 12 points in a *plane of which 5 are collinear?
Solution:
Clearly, a triangle is formed by selecting any 3 non-collinear points.
Total number of ways of selecting 3 points out of 12 points
= ¹²C₃ = \(\frac{12 \times 11 \times 10}{3 \times 2 \times 1}\)
= 220
Since, 5 points are collinear any selection of 3 points from these 5 will not form a triangle.
Number of such selections
= ⁵C₃ = ⁵C₂
= \(\frac{5 \times 4}{2 \times 1}\) = 10
Required number of triangles = Total selections of 3 points – Selections from collinear points = 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 atleast 3 from each part. In how many ways can the candidate select the questions?
Solution:
There are 12 questions in total with 7 in Part A and 5 in Part B. The candidate must attemp 7 questions selecting atleast 3 from each part.
The possible combinations for selecting 7 questions are 3 questions from Part A and 4 questions from Part B.
Number of ways = ⁷C₃ × ⁵C₄ = 35 × 5 = 175
4 questions from Part A and 3 questions from Part B.
Number of ways = ⁷C₄ × ⁵C₃ = 35 × 10 = 350
∴ Required number of ways = 175 + 350 = 525.

Question 4.
How many diagonals does a polygon with 10 sides have?
Solution:
A polygon with 10 sides has 10 vertices. Aline segment is formed by joining any two vertices.
Total number of line segments formed by joining 10 vertices taken 2 at a time
= ¹⁰C₂ = \(\frac{10 \times 9}{2 \times 1}\) = 45
Out of these 45 line segments, 10 are the sides of the polygon. The remaining line segments are the diagonals.
∴ Required number of diagonals = Total line segments – Number of sides
= 45 – 10 = 35

Question 5.
If you invite 15 of your friends to a party and all shake hands exactly once, how many hand-shakes occur?
Solution:
Total number of persons in the party (including you and your 15 friends) is 16.

Clearly, a hand-shake occurs between 2 people. When 16 persons shake hands with each other exactly once, the total number of hand-shakes is the same as the number of combinations (or selections) of 16 different people taken 2 at a time.
So, required number of hand-shakes
= ¹⁶C₂ = \(\frac{16!}{2!(16-2)!}\)
= \(\frac{16!}{2!\times 14!}\) = \(\frac{16 \times 15 \times 14!}{2 \times 1 \times 14!}\)
= 8 × 15 = 120

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:
There is a group of 5 persons consisting of 2 men and 3 women. A committee of 3 persons is to be constituted.
The total number of ways to constitute the committee is same as the number of combinations of 5 different persons taken 3 at a time.
∴ Required number of ways = ⁵C₃ = ⁵C₂ = \(\frac{5 \times 4}{2 \times 1}\) = 10.
For a committee consisting of 1 man and 2 women.
Number of ways to select 1 man out of 2 = ²C₁ = 2
Number of ways to select 2 women out of 3 = ³C₂ = ³C₁ = 3
∴ Required number of committees = 2 × 3
= 6

The post Class 9 Advanced Maths Chapter 5 Combinatorics Ex 5.4 Solutions appeared first on Learn CBSE.

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