# COMBINATION PRACTICAL PROBLEMS

## About "Combination Practical Problems"

**Combination Practical Problems :**

Here we are going to see some practice questions base d on the concept combination.

## Combination Practical Problems - Questions

**Question 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

= ^{7}C_{2 }⋅ ^{8}C_{1}

= 21 ⋅ 8

= 168

**Case 2 :**

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

= ^{7}C_{1 }⋅ ^{8}C_{2}

= 7 ⋅ 28

= 196

Total number of ways = 168 + 196

= 364 ways

**Question 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 = ^{ 11}C_{2}

= (11 ⋅ 10) / (2 ⋅ 1)

= 55

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

Five points are collinear = ^{4}C_{2}

= (4 ⋅ 3)/(2 ⋅ 1)

= 6

Required no of straight line =^{ 11}C_{2} - ^{4}C_{2} + 1

= 55 - 6 + 1

= 50

(ii) Number of triangles to be formed:

= ^{ 11}C_{3 }- 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

**Question 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

n^{2} - 3n - 180 = 0

(n - 15) (n + 12) = 0

n = 15 and n = -12

Hence the number of sides of the polygon is 15.

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