Question

There are two parallel line segments (pq and ab) in a plane. Pq contains 14 points whereas ab contains 10 points. How many triangles can be formed by using these points as vertices ?

Answer 1

Answer:

the shape and I am happy for

Answer 2

Given:

Points on pq=14

Points on ab=10

To find:

The number of triangles that can be formed using these points as vertices

Solution:

The required number of triangles that can be formed using these points as vertices is 1,540.

We know that any triangle can be made by joining three vertices and these at least one and at most two of the three vertices can lie on pq and ab.

So, we will use the combination to determine the ways of choosing three vertices on the two lines.

The number of triangles that can be formed=14C2×10C1 + 14C1×10C2

On solving, we get

=14!/2!(12!)×10!/1!(9!) + 14!/1!(13!)×10!/2!(8!)

=91×10+14×45

=910+630

=1,540

Therefore, the number of triangles that can be formed using these points as vertices is 1,540.