COMBINATION PRACTICAL PROBLEMS

Problem 1 :

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 :

Line 1 has 7 points and Line 2 has 8. We cannot select all 3 points from the same line to form a triangle.

Case 1 :

Select 2 points from Line 1 and 1 from Line 2 

  =  ⁷C₂ ⋅ ⁸C₁

  =  21 ⋅ 8

  =  168 

Case 2 :

Select 1 point from Line 1 and 2 from Line 2.

 =  ⁷C₁ ⋅ ⁸C₂

  =  7 ⋅ 28

  =  196

Total number of ways  =  168 + 196

  =  364 ways

Problem 2 :

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?

(ii) the number of triangles that can be formed for which the points are their vertices?

Solution :

Total points  =  11

Collinear points  =  4

No of lines formed from 11 points  =  ¹¹C₂

  =  (11 ⋅ 10) / (2 ⋅ 1)

  =  55

(Since we require two points to form one line.)

Five points are collinear  =  ⁴C₂

  =  (4 ⋅ 3)/(2 ⋅ 1)

  =  6

Required no of straight line = ¹¹C₂ - ⁴C₂ + 1

  =  55 - 6 + 1

  =  50

(ii) Number of triangles to be formed:

=    ¹¹C₃ - 4

Because we cannot draw a triangle by joining 4 collinear points.

  =  (11 ⋅ 10 ⋅ 9)/(3 ⋅ 2 ⋅ 1)

  =  165

Total number of ways  =  165 - 4  =  161

Problem 3 :

A polygon has 90 diagonals. Find the number of its sides?

Solution :

An n sided polygon can have n ⋅(n - 3)/2 diagonals 

n ⋅ (n - 3)/2 = 90

n ⋅ (n - 3) = 180

n² - 3n - 180  =  0

(n - 15) (n + 12)  =  0

n  =  15 and n = -12

Hence the number of sides of the polygon is 15.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.