Question

There are $10$ points in a plane, out of these $6$ are collinear. If $N$ is the number of triangles formed by joining these points, then
A. $N > 190$
B. $N \leqslant 100$
C. $100 < N \leqslant 140$
D. $140 \leqslant N \leqslant 190$

Answer 1

Hint:In order to find the number of triangles that can be formed according to the situation given above, we need to know some important information about collinear points, that collinear points are the points that are situated in a straight line, and we know that points in a straight line cannot form a triangle.

Formula used:
$^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$

Complete step by step answer:
We are given a situation that there are $10$ points in a plane, out of them $6$ are collinear. N number of triangles is formed by joining these points. Since, we know a triangle is a polygon of three sides, so out of ten points, we need three points to form a triangle and the number of ways of selecting the three points from the $10$ points to form a triangle, in a plane can be done by combination, numerically written as:
$^{10}{C_3}$ ……………..(1)
Now, it's given that out of $10$ points $6$ are collinear. Collinear means that the lines are in a single straight line, that means points in a straight line can never form a triangle.
So, the number of ways of selecting the three points from $6$ points that are collinear are:
$^6{C_3}$ ……………..(2)

So, in equation 1, we obtained the total number of ways in which we can form a triangle, and in equation 2 we got the number of ways in which we cannot form a triangle as they are collinear. So, the remaining number of ways in which the triangles can be formed is:
$^{10}{C_3}{ - ^6}{C_3}$ ……(3)
From the combination formula, we know that $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.
Substituting this in equation 3, we get:
$^{10}{C_3}{ - ^6}{C_3} \\
\Rightarrow \dfrac{{10!}}{{3!\left( {10 - 3} \right)!}} - \dfrac{{6!}}{{3!\left( {6 - 3} \right)!}} \\ $
Solving the equation:
$ \Rightarrow \dfrac{{10!}}{{3!7!}} - \dfrac{{6!}}{{3!3!}}$
Expanding the factorial:
$ \Rightarrow \dfrac{{10 \times 9 \times 8 \times 7!}}{{3!7!}} - \dfrac{{6 \times 5 \times 4 \times 3!}}{{3!3!}}$
Cancelling the similar values:
\[ \Rightarrow \dfrac{{10 \times 9 \times 8}}{{3 \times 2 \times 1}} - \dfrac{{6 \times 5 \times 4}}{{3 \times 2 \times 1}}\]
\[ \Rightarrow 120 - 20\]
\[ \Rightarrow 100\]
Therefore, $^{10}{C_3}{ - ^6}{C_3} = 100$
Since, there was N number of triangles that can be formed, so by this $N = 100$. This value is fulfilled by Option 2, that is $N \leqslant 100$.

Hence, option B is correct.

Note:Combination is a way of defining the maximum number of possible ways of selecting an object at random, when the order is not important. We used $N \leqslant 100$ instead of $100 < N \leqslant 140$ as both of them had $100$ in it, because we obtained a result $N = 100$, and in $100 < N \leqslant 140$ , $100$ is not included or $N \ne 100$, that’s why the other option is used in which $N = 100{\text{ or N < 100}}$.