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.
B) when four points are collinear then the total number of lines= 10C2 but it contains lines using that 4 collinear points make only one line
lines using 4 points = 4C2
No. of ways is 4 points are collinear = 10C2 - 4C2 +1
= 45 - $\frac{4!}{2!2!}$+1
=46- 6
=40
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
Permutation Combination Exercise 3.1
Permutation Combination Exercise 3.2
Permutation Combination Exercise 3.3
Permutation Combination Exercise 3.4
Permutation Combination Exercise 3.5
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| Permutation and Combination Exercise 3.6 11th Class MH Board |
Permutation and Combination Exercise 3.6 11th Class MH Board
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