# Question 5 and 6 Exercise 6.3

Solutions of Question 5 and 6 of Exercise 6.3 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

## Question 5(i)

How many straight lines are determined by $12$ points, no three of which lie on the same straight line?

### Solution

Total points are eight so $n=12$.

Every pair of points determines a line,

so that gives ${ }^{12} C_2=66$ lines.

The question now is whether we have counted any line twice.

the the answer is “No,” because there are no three of the given points on any line.

## Question 5(ii)

How many triangles are determined by $12$ points, no three of which lie on the same straight line?

### Solution

Total points are eight so $n=12$.

Now for triangles, we have ${ }^{12} C_3=220$ ways to choose the vertices.

Again the question is whether we have counted any triangle twice.

Again, the answer is “No.” If there were a fourth point in one of these triangles,

it would lie on a side with two of the points, giving three collinear.

## Question 6

Find the total number of diagonal of hexagon.

### Solution

First we find the total number lines. We know one line can be drawn between each two points, so total number of lines are: $${ }^6 C_2=\dfrac{6 !}{(6-2) ! 2 !}=15 $$ Now $6$ are sides of the hexagon so,

total number of diagonal are $\quad 15-6=9$.

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