Exercise 3.6 12) 13) 14)

 12.There are 20 straight lines in a plane so that no two lines are parallel and no three lines are concurrent. Determine the number of points of intersection.

Answer-

There are 20 lines such that no two of them are parallel and no three of them are concurrent.    

 Since no two lines are parallel 

∴ they intersect at a point        

∴ Number of points of intersection if no two lines are parallel and no three lines are concurrent.                                                                                                                                                                                       = 20C2                                                    

 $=\frac{20!}{2!18!}$

$=\frac{20×19×18!}{2×1×18!}$

= 190

13.10 points are plotted on a plane. Find the number of straight lines obtained by joining these points if                                                 

 A)no three points are collinear.

 B) four points are collinear.

Answer-

A line is drawn by joining 2 points from the given 10 points.       

  ∴ number of straight lines= 10C2                                           

=$\frac{10!}{2!8!}$

$=\frac{10×9×8!}{2×8!}$

= 45, if no three points are collinear. 

14.Find the number of triangles formed by joining 12 points if         

a) no three points are collinear

b) 4 points are collinear.

Answer-

a)A triangle can be formed by selecting 3 non-collinear points

3 points be selected from 12 points

$^{12}C_3=\frac{12!}{9!3!}$

$=\frac{12.11.10.9!}{9!3!}$

=220 ways

b)There are 12 points on the plane

when 4 of these points are collinear

$\therefore$ number of triangles that can be from these points=$^{12}C_3-^{4}C_3$

=220-$\frac{4!}{3!1!}$

=220-4 

=216

15. A word has three consonant 8 vowels. How many distinct words can be formed if 4 consonants and 2 vowels are use and?

Answer-

4 consonants can be selected from 8 consonants in $^8C_4$ ways 2 vowels can be selected from 3 vowels in $^3C_2$

$\therefore$ number of words with 4 consonants and 2 vowels

=$^8C_4\times ^3C_2$

=$\frac{8!}{4!4!}\times \frac{3!}{2!1!}$

=70 $\times$ 3

=210

now, each of these 6 letters can be arranged in 6P6 ways=6!

$\therefore$ total numbers can be formed 4 consonants and 2 vowels

$=210 \times 6!$

=151200