$T_m$ denotes the number of triangles that can be formed with the vertices of a regular polygon of $m$ sides…
- $3$
- $6$
- $9$
- $12$
Solution

As we know, ${ }^n C_r+{ }^n C_{r+1}={ }^{n+1} C_{r+1}$ $\therefore \quad{ }^m C_2+{ }^m C_3={ }^{m+1} C_{2+1}$ ${ }^m C_2+{ }^m C_3=15+{ }^m C_3 \quad$ [from eq. (i) $]$ $\begin{aligned} & \therefore \quad{ }^m C_2=15 \\ & \frac{m !}{2 !(m-2) !}=15 \\ & m(m-1)=30 \\ & m^2-m-30=0 \\ & (m-6)(m+5)=0 \\ & m=6-5\end{aligned}$ $\therefore m=6 \quad[\because m \neq-5]$
Asked in: AP EAMCET 2015