Question

If ${t_n}$ denotes the number of triangles formed with $n$ points in a plane no three of which are collinear, and if ${t_{n + 1}} - {t_n} = 36$, then $n$ is equal to
(a) 7
(b) 8
(c) 9
(d) 10

Answer 1

Here, we need to find the value of $n$. A triangle is a line segment formed by joining any 3 non-collinear points. We will use the formula for combinations to find the number of triangles that can be drawn through $n$ points, and $n + 1$ points. Then, using the given equation, we will find a quadratic equation in terms of $n$. Finally, we will solve this quadratic equation to find the required value of $n$.
Formula Used: The number of ways in which the $n$ objects can be placed in $r$ spaces, where order of objects is not important, can be found using the formula for combinations ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.

Complete step by step solution:
We will use combinations to find the number of triangles that can be drawn through $n$ points.
A triangle is formed by joining any 3 of the $n$ points.
Therefore, the number of triangles that can be formed using the $n$ points, taking 3 points at a time, is given by ${}^n{C_3}$.
Substituting $r = 3$ in the formula for combinations, we get
Number of triangles that can be formed using the $n$ points $= {}^n{C_3} = \dfrac{{n!}}{{3!\left( {n - 3} \right)!}}$
Therefore, we get
$\Rightarrow {t_n} = \dfrac{{n!}}{{3!\left( {n - 3} \right)!}}$
The number of triangles that can be formed using the $n + 1$ points, taking 3 points at a time, is given by ${}^n{C_3}$.
Substituting $n + 1$ for $n$ and $r = 3$ in the formula for combinations, we get
Number of triangles that can be formed using the $n + 1$ points $= {}^{n + 1}{C_3} = \dfrac{{\left( {n + 1} \right)!}}{{3!\left( {n + 1 - 3} \right)!}}$
Therefore, we get
$\begin{array}{l} \Rightarrow {t_{n + 1}} = \dfrac{{\left( {n + 1} \right)!}}{{3!\left( {n + 1 - 3} \right)!}}\\\ \Rightarrow {t_{n + 1}} = \dfrac{{\left( {n + 1} \right)!}}{{3!\left( {n - 2} \right)!}}\end{array}$
Now, substituting ${t_{n + 1}} = \dfrac{{\left( {n + 1} \right)!}}{{3!\left( {n - 2} \right)!}}$ and ${t_n} = \dfrac{{n!}}{{3!\left( {n - 3} \right)!}}$ in the given equation , we get
$\Rightarrow \dfrac{{\left( {n + 1} \right)!}}{{3!\left( {n - 2} \right)!}} - \dfrac{{n!}}{{3!\left( {n - 3} \right)!}} = 36$
We know that $\left( {n + 1} \right)!$ can be written as $\left( {n + 1} \right)n!$.
Rewriting $\left( {n + 1} \right)!$ as $\left( {n + 1} \right)n!$ and $\left( {n - 2} \right)!$ as $\left( {n - 2} \right)\left( {n - 3} \right)!$, we get
$\Rightarrow \dfrac{{\left( {n + 1} \right)n!}}{{3!\left( {n - 2} \right)\left( {n - 3} \right)!}} - \dfrac{{n!}}{{3!\left( {n - 3} \right)!}} = 36$
Factoring out the term $\dfrac{{n!}}{{3!\left( {n - 3} \right)!}}$, we get
$\Rightarrow \dfrac{{n!}}{{3!\left( {n - 3} \right)!}}\left[ {\dfrac{{n + 1}}{{n - 2}} - 1} \right] = 36$
Taking the L.C.M. and simplifying, we get
$\begin{array}{l} \Rightarrow \dfrac{{n!}}{{3!\left( {n - 3} \right)!}}\left[ {\dfrac{{n + 1 - n + 2}}{{n - 2}}} \right] = 36\\\ \Rightarrow \dfrac{{n!}}{{6\left( {n - 3} \right)!}}\left[ {\dfrac{3}{{n - 2}}} \right] = 36\\\ \Rightarrow \dfrac{{n!}}{{2\left( {n - 2} \right)\left( {n - 3} \right)!}} = 36\end{array}$
Rewriting $n!$ as $n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)!$, we get
$\Rightarrow \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)!}}{{2\left( {n - 2} \right)\left( {n - 3} \right)!}} = 36$
Simplifying the expression, we get
$\Rightarrow \dfrac{{n\left( {n - 1} \right)}}{2} = 36$
Multiplying both sides by 2, we get
$\Rightarrow n\left( {n - 1} \right) = 72$
Multiplying the terms using the distributive law of multiplication, we get
$\begin{array}{l} \Rightarrow {n^2} - n = 72\\\ \Rightarrow {n^2} - n - 72 = 0\end{array}$
Here, we get a quadratic equation. We will split the middle term and factorise to solve the quadratic equation.
Factoring the quadratic equation, we get
$\begin{array}{l} \Rightarrow {n^2} - 9n + 8n - 72 = 0\\\ \Rightarrow n\left( {n - 9} \right) + 8\left( {n - 9} \right) = 0\\\ \Rightarrow \left( {n + 8} \right)\left( {n - 9} \right) = 0\end{array}$
Therefore, either $n + 8 = 0$ or $n - 9 = 0$, that is $n = - 8$ or $n = 9$.
We know that the number of points cannot be negative.
Thus, we get
$n = 9$
$\therefore$ The value of $n$ is 9.

The correct option is option (c).

Note:
We used combinations instead of permutations to find the number of triangles that can be formed using the points. A triangle is formed using 3 non-collinear points. Suppose three of the $n$ points are A, B, and C. Now, if A, B, C are joined, the triangle is ABC. If B, A, C are joined in that order, the triangle is still triangle ABC. If we used permutations, the triangles formed by joining (in order) ABC, ACB, BAC, BCA, CAB, CBA will be counted as 6 different triangles. Therefore, we have used combinations since it does not matter which points are connected first, as long as the points connected are not changed.