Combination Practical Problems :
In this section, we will learn, how to solve practical problems on combinations.
Problem 1 :
How many triangles can be formed by 15 points, in which 7 of them lie on one line and the remaining 8 on another parallel line?
Solution :
Line 1 has 7 points and Line 2 has 8. We cannot select all 3 points from the same line to form a triangle.
Case 1 :
Select 2 points from Line 1 and 1 from Line 2
= ^{7}C_{2 }⋅ ^{8}C_{1}
= 21 ⋅ 8
= 168
Case 2 :
Select 1 point from Line 1 and 2 from Line 2.
= ^{7}C_{1 }⋅ ^{8}C_{2}
= 7 ⋅ 28
= 196
Total number of ways = 168 + 196
= 364 ways
Problem 2 :
There are 11 points in a plane. No three of these lies in the same straight line except 4 points, which are collinear. Find,
(i) the number of straight lines that can be obtained from the pairs of these points?
(ii) the number of triangles that can be formed for which the points are their vertices?
Solution :
Total points = 11
Collinear points = 4
No of lines formed from 11 points = ^{ 11}C_{2}
= (11 ⋅ 10) / (2 ⋅ 1)
= 55
(Since we require two points to form one line.)
Five points are collinear = ^{4}C_{2}
= (4 ⋅ 3)/(2 ⋅ 1)
= 6
Required no of straight line =^{ 11}C_{2} - ^{4}C_{2} + 1
= 55 - 6 + 1
= 50
(ii) Number of triangles to be formed:
= ^{ 11}C_{3 }- 4
Because we cannot draw a triangle by joining 4 collinear points.
= (11 ⋅ 10 ⋅ 9)/(3 ⋅ 2 ⋅ 1)
= 165
Total number of ways = 165 - 4 = 161
Problem 3 :
A polygon has 90 diagonals. Find the number of its sides?
Solution :
An n sided polygon can have n ⋅(n - 3)/2 diagonals
n ⋅ (n - 3)/2 = 90
n ⋅ (n - 3) = 180
n^{2} - 3n - 180 = 0
(n - 15) (n + 12) = 0
n = 15 and n = -12
Hence the number of sides of the polygon is 15.
After having gone through the stuff given above, we hope that the students would have understood, how to solve practical problems on combinations.
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