Question

Find the total number of triangles that can be formed by the $5$ points on a single line and $3$ points on the parallel line to the previous line. Choose the correct option below
1) ${}^8{C_3}$
2) ${}^8{C_3} - {}^5{C_3}$
3) ${}^8{C_3} - {}^5{C_3} - 1$
4) None of the above

Answer 1

Hint: We can do this problem by selecting points. We have select points in such a way that the points can form a triangle. We can do this by selecting one point from one line and two points from the line parallel to it. In this way, we can form a triangle.

Complete step-by-step answer:
Given,
There are $$5$$ points on a line and $3$ points on another line that is parallel to the previous line.
That means the points look like

Now as I said in the hint, we need $3$points to form a triangle.
The $3$ points should not be collinear because if those $3$points formed a line then they can not form a triangle.
So, we need to select $2$ points from one line and $1$point from another line.
This can be done in $2$ ways.
That is $2$ points from the first line and one point from the second line, one point from the first line, and $2$ points from the second line.
So,
The total number of triangles are $$ = {}^5{C_2}.{}^3{C_1} + {}^5{C_1}.{}^3{C_2}$$,
By calculating the binomial coefficients, we get
The total number of triangles are $$ = 45$$.
So the number of triangles that can be formed by the $5$ points on a single line and $3$ points on the parallel line to the previous line is $45$.
By calculating the options we get the option is $3$.
So, the correct answer is “Option 3”.

Note: We can solve permutations and combination questions in many ways. This is a natural way of solving this problem without using any formulae. We can solve this problem by the elimination method also that is finding that number of triangles can be formed from $8$ points and removing the number of triangles that can be formed from $5$ points and $3$ points because those are collinear that is ${}^8{C_3} - {}^5{C_3} - {}^3{C_3}$ which is option 3.