Question

If ${{T}_{m}}$ denotes the number of Triangles that can be formed with the vertices of a regular polygon of $m$sides. If ${{T}_{m+1}}-{{T}_{m}}=15$, then $m=$ $A. 3$
B. 6$C. 9$
D. 12$$$$

Answer 1

We use the fact that we can draw a triangle with minimum 3 vertices. We use the combination formula to select $r$ objects at a time from $n$ distinct objects ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ to find ${{T}_{m}},{{T}_{m+1}}$ and then put the obtained expressions in $m$ in the given equation ${{T}_{m+1}}-{{T}_{m}}=15$ and then solve for $m$.

Complete step-by-step solution:
We know that the factorial of natural numbers $n$ is the product of the first $n$ natural number. We denote the factorial of $n$ as $n!$ and it is given by
$n!=1\times 2\times 3\times ...\times n$
We can recursively use the above formula for 1 step as,

& n!=n\times \left( n-1 \right)\times \left( n-2 \right)...\times 3\times 2\times 1 \\\ & \Rightarrow n!=n\left( n-1 \right)! \\\ \end{aligned}$$ The above equation is the recursive relation for $n!$ with initial values $1!=1$.We can recursively use the above formula for $r$ steps as, $$n!=n\left( n-1 \right)\left( n-2 \right)...\left( n-r+1 \right)\left( n-r \right)!$$ We also know from elementary combinatorics that the number of ways we can select $r$ objects at a time from $n$ distinct objects without replacement and irrespective of order of repetition is called $r-$combination of $n$ and is denoted as ${}^{n}{{C}_{r}}$ or $\left( \begin{aligned} & n \\\ & r \\\ \end{aligned} \right)$ and is given by $$\begin{aligned} & {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} \\\ & \Rightarrow {}^{n}{{C}_{r}}=\dfrac{n\times \left( n-1 \right)\times ...\times \left( n-r+1 \right)\left( n-r \right)!}{r!\left( n-r \right)!} \\\ & \Rightarrow {}^{n}{{C}_{r}}=\dfrac{n\left( n-1 \right)...\left( n-r+1 \right)}{r\left( r-1 \right)...1} \\\ \end{aligned}$$ We are given the question that ${{T}_{m}}$ denotes the number of triangles that can be formed with the vertices of a regular polygon of $m$ sides. We are also given the equation $${{T}_{m+1}}-{{T}_{m}}=15.......\left( 1 \right)$$ We know that a regular polygon with $m$ sides has $m$ vertices. We can draw a triangle by selecting any 3 vertices out of $m$ vertices and then joining them. So by combination formula we have $${{T}_{m}}={}^{m}{{C}_{3}}=\dfrac{m!}{3!\left( m-3 \right)!}=\dfrac{m\left( m-1 \right)\left( m-3+1 \right)}{3\times 2\times 1}=\dfrac{m\left( m-1 \right)\left( m-2 \right)}{6}$$ Similarly we use the combination formula for $m+1$ sides to have, $${{T}_{m+1}}={}^{m+1}{{C}_{3}}=\dfrac{\left( m+1 \right)!}{3!\left( m+1-3 \right)!}=\dfrac{\left( m+1 \right)\left( m+1-1 \right)\left( m+1-3+1 \right)}{3\times 2\times 1}=\dfrac{\left( m+1 \right)m\left( m-1 \right)}{6}$$ We put the value of ${{T}_{m}},{{T}_{m+1}}$ in equation (1) to have $$\begin{aligned} & \dfrac{\left( m+1 \right)m\left( m-1 \right)}{6}-\dfrac{m\left( m-1 \right)\left( m-2 \right)}{6}=15 \\\ & \Rightarrow \dfrac{m\left( m-1 \right)}{6}\left( m+1-m+2 \right)=15 \\\ & \Rightarrow \dfrac{m\left( m-1 \right)3}{6}=15 \\\ & \Rightarrow {{m}^{2}}-m-30=0 \\\ \end{aligned}$$ We solve the above quadratic equation in the above step by splitting the middle term method and proceed to have, $$\begin{aligned} & \Rightarrow {{m}^{2}}-6m+5m-30=0 \\\ & \Rightarrow m\left( m-6 \right)+5\left( m-6 \right)=0 \\\ & \Rightarrow \left( m-6 \right)\left( m+5 \right)=0 \\\ & \Rightarrow m-6=0\text{ or }m+5=0 \\\ \end{aligned}$$ **So we obtain the roots of the quadratic equation $m=6,-5$ out of which we reject $m=-5$ as the number of sides cannot be negative. So the number of sides is $m=6$ and hence the correct option is B.** **Note:** We note to be careful of the confusion between formulas of combination from permutation where objects arranged in order in ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ ways after selection. We also note the word ‘regular’ polygon which ensures that the polygon is convex and sides do not intersect to decrease the number of vertices. The number of diagonals of the regular polygon with $m$ sides is ${}^{m}{{C}_{2}}-m$.