Question

Out of 18 points in a plane, no three are in the same straight line except five points which are collinear. How many straight lines fill triangles can be formed by joining them?

Answer 1

Hint: First remember the conditions given in the question. According to that, apply the concept of combinations here to get the solution of the question. There is only one line possible by the collinear points as they all are on the same line. Therefore the number of lines can be formed is \[{}^{18}{C_2} - {}^5{C_2} + 1\] and number of triangles can be formed is \[{}^{18}{C_3} - {}^5{C_3}\] .Then solve it further by using the formula of combination to get the answer.

Complete step-by-step solution:
First we need to find the number of different straight lines that can be drawn from the \[18\] points in which \[5\] points are collinear. We know that to draw a line we need at least two points and the collinear points will lie on the same line and only one line can be drawn by joining any two points of these collinear points.
Let N be the number of lines formed that is
N \[ = \] (total number of lines formed by all \[18\] points) – (number of lines formed by collinear points) \[ + {\text{ 1}}\]
Here we add one because the collinear points can form only one line. Therefore,
\[ \Rightarrow N = {}^{18}{C_2} - {}^5{C_2} + 1\] ------------ (i)
We know that \[{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}\] .Therefore by applying this formula in equation (i) we get
\[ \Rightarrow N = \left( {\dfrac{{18!}}{{\left( {18 - 2} \right)!2!}}} \right) - \left( {\dfrac{{5!}}{{\left( {5 - 2} \right)!2!}}} \right) + 1\]
On further solving the above expression becomes
\[ \Rightarrow N = \left( {\dfrac{{18!}}{{16!2!}}} \right) - \left( {\dfrac{{5!}}{{3!2!}}} \right) + 1\]
We know that \[n! = \left( n \right)\left( {n - 1} \right).........2.1\] .Therefore by applying this in the above step we get
\[ \Rightarrow N = \left( {\dfrac{{18 \times 17 \times 16!}}{{16!\left( {2 \times 1} \right)}}} \right) - \left( {\dfrac{{5 \times 4 \times 3!}}{{3!\left( {2 \times 1} \right)}}} \right) + 1\]
Now the factorial terms in both the numerators and the denominators will cancel out and we have
\[ \Rightarrow N = \left( {\dfrac{{18 \times 17}}{2}} \right) - \left( {\dfrac{{5 \times 4}}{2}} \right) + 1\]
On doing multiplication in the numerator we get
\[ \Rightarrow N = \left( {\dfrac{{306}}{2}} \right) - \left( {\dfrac{{20}}{2}} \right) + 1\]
Now we observe that the numerators are divisible by the denominators. Therefore,
\[ \Rightarrow N = 153 - 10 + 1\]
On doing addition or subtraction in the above step we get
\[ \Rightarrow N = 144\]
Therefore the total number of ways to form different lines are \[144\] .
Also we have to find the number of triangles that can be drawn from the \[18\] points in which five are collinear. We know that three points are required to draw a triangle and the collinear points will lie on the same line and no triangle can be drawn by joining any three points of these collinear points.
Let X be the number of triangles formed. Therefore,
X \[ = \] (total number of lines formed by all \[18\] points) – (number of triangles formed by collinear points)
Therefore, the above step will be
\[ \Rightarrow X = {}^{18}{C_3} - {}^5{C_3}\] ----------- (ii)
We know that \[{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}\] .By applying this formula in the equation (ii) we get
\[ \Rightarrow X = \dfrac{{18!}}{{\left( {18 - 3} \right)!3!}} - \dfrac{{5!}}{{\left( {5 - 3} \right)!3!}}\]
On further solving we get
\[ \Rightarrow X = \dfrac{{18!}}{{15!3!}} - \dfrac{{5!}}{{2!3!}}\]
We know that \[n! = \left( n \right)\left( {n - 1} \right).........2.1\] .Therefore,
\[ \Rightarrow X = \dfrac{{18 \times 17 \times 16 \times 15!}}{{15!\left( {3 \times 2 \times 1} \right)}} - \dfrac{{5 \times 4 \times 3 \times 2!}}{{2!\left( {3 \times 2 \times 1} \right)}}\]
The factorial terms will cancel out present in both the numerator and the denominator
\[ \Rightarrow X = \dfrac{{18 \times 17 \times 16}}{{3 \times 2 \times 1}} - \dfrac{{5 \times 4 \times 3}}{{3 \times 2 \times 1}}\]
On doing multiplication in both the numerator and the denominator we get
\[ \Rightarrow X = \dfrac{{4896}}{6} - \dfrac{{60}}{6}\]
On dividing the numerators by their denominators we get
\[ \Rightarrow X = 816 - 10\]
By doing subtraction we get
\[ \Rightarrow X = 806\]
Hence the total number of triangles formed are \[806\] .

Note: Keep in mind that combinations is a way of selecting items from a collection such that the order of selection does not matter. Remember that a combination is the choice of r things from a set of n things without replacement and where order does not matter. That’s why in this question we use combination.