# 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.

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

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.

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

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.

Find the total number of diagonal of hexagon.

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 $\quad 15-6=9$.