NIMCET 2007 — Mathematics PYQ

Step 1: Understand the formula for the number of triangles

A triangle is uniquely formed by selecting any $3$ non-collinear points. For a polygon with $n$ vertices (where no three vertices are collinear), the number of triangles that can be formed is given by selecting $3$ vertices out of $n$.

Therefore, the formula for $T_n$ is:

$$T_n=\left(\genfrac{}{}{0ex}{}{n}{3}\right)=\frac{n(n-1)(n-2)}{3\times 2\times 1}$$

Similarly, the number of triangles formed by a polygon with $n+1$ vertices is:

$$T_{n+1}=\left(\genfrac{}{}{0ex}{}{n+1}{3}\right)=\frac{(n+1)n(n-1)}{3\times 2\times 1}$$

Step 2: Use the given condition

We are given the condition:

$$T_{n+1}-T_n=21$$

Substitute the combination expressions into the equation:

$$\left(\genfrac{}{}{0ex}{}{n+1}{3}\right)-\left(\genfrac{}{}{0ex}{}{n}{3}\right)=21$$

Using Pascal's Identity rule ($\left(\genfrac{}{}{0ex}{}{n}{r}\right)+\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{r}\right)$), we can rewrite $\left(\genfrac{}{}{0ex}{}{n+1}{3}\right)-\left(\genfrac{}{}{0ex}{}{n}{3}\right)$ directly as $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$.

Let us also expand it algebraically to verify:

$$\frac{(n+1)n(n-1)}{6}-\frac{n(n-1)(n-2)}{6}=21$$

Take the common terms $\frac{n(n-1)}{6}$ out:

$$\frac{n(n-1)}{6}\cdot [(n+1)-(n-2)]=21$$

$$\frac{n(n-1)}{6}\cdot [n+1-n+2]=21$$

$$\frac{n(n-1)}{6}\cdot 3=21$$

Simplify the fraction:

$$\frac{n(n-1)}{2}=21$$

Step 3: Solve for $n$

Cross-multiply by $2$:

$$n(n-1)=42$$

Expand the left side to form a quadratic equation:

$$n^2-n-42=0$$

Factorize the quadratic equation by splitting the middle term (since $-7\times 6=-42$ and $-7+6=-1$):

$$n^2-7n+6n-42=0$$

$$n(n-7)+6(n-7)=0$$

$$(n-7)(n+6)=0$$

This gives two possible values for $n$:

$$n=7\text{or}n=-6$$

Since the number of vertices ($n$) of a polygon must be a positive integer, we reject $n=-6$. Therefore:

$$n=7$$

Correct Answer

The correct option is (b) 7.