Let $$T_n$$ be the number of all possible triangles formed by joining vertices of an $$n$$-sided regular polygon. If $$T_{n+1} - T_n = 10$$, then the value of $$n$$ is :

Let us first recall a very common combinatorial fact. To form a triangle we need exactly three non-collinear points. In a regular polygon no three vertices are collinear, so every choice of three distinct vertices yields one and only one triangle. Hence, for an $$n$$-sided polygon the total number of possible triangles is given by the combination formula

$$T_n=\binom{n}{3}=\frac{n(n-1)(n-2)}{6}.$$

According to the question we have the relation

$$T_{n+1}-T_n=10.$$

First we write both terms with the above formula. For $$n+1$$ sides we get

$$T_{n+1}=\binom{n+1}{3}=\frac{(n+1)n(n-1)}{6}.$$

Substituting $$T_{n+1}$$ and $$T_n$$ into the given difference, we obtain

$$\frac{(n+1)n(n-1)}{6}-\frac{n(n-1)(n-2)}{6}=10.$$

Since both fractions have the same denominator, we combine the numerators directly:

$$\frac{n(n-1)\big[(n+1)-(n-2)\big]}{6}=10.$$

Inside the square brackets we simplify step by step:

$$(n+1)-(n-2)=n+1-n+2=3.$$

So the expression becomes

$$\frac{n(n-1)\cdot3}{6}=10.$$

We can reduce the fraction $$\frac{3}{6}$$ to $$\frac{1}{2}$$, giving

$$\frac{n(n-1)}{2}=10.$$

Now multiply both sides by $$2$$ to clear the denominator:

$$n(n-1)=20.$$

Expanding the left side leads to a standard quadratic equation:

$$n^2-n-20=0.$$

We solve this quadratic by the quadratic formula $$n=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$ with $$a=1,\,b=-1,\,c=-20$$. Therefore

$$n=\frac{1\pm\sqrt{1+80}}{2}=\frac{1\pm9}{2}.$$

This yields two values, $$n=\frac{10}{2}=5$$ and $$n=\frac{-8}{2}=-4$$. Because the number of sides of a polygon must be positive, we discard the negative solution.

Thus, $$n=5$$.

Hence, the correct answer is Option C.