====== 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 $\quad 15-6=9$.



====Go To====
<text align="left"><btn type="primary">[[math-11-kpk:sol:unit06:ex6-3-p4 |< Question 4 ]]</btn></text>
<text align="right"><btn type="success">[[math-11-kpk:sol:unit06:ex6-3-p6|Question 7 >]]</btn></text>