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
