17) What is the probability that, by randomly selecting three distinct vertices of a regular hexagon, they form the vertices of an equilateral triangle?
(A) $\frac{1}{4}$
(B) $\frac{1}{8}$
(C) $\frac{1}{9}$
(D) $\frac{1}{10}$
(E) $\frac{1}{20}$.
17) The answer is $(\mathrm{D})$

Let A, B, C, D, E, F be the vertices of the hexagon, in counterclockwise order. Choose a vertex at random. By symmetry, we can assume that this vertex is A. To form an equilateral triangle with $\mathrm{A}$, the other two vertices must necessarily be $\mathrm{C}$ and $\mathrm{E}$. The probability that the second vertex is one of them is $\frac{2}{5}$; the probability that the third vertex is the remaining one is $\frac{1}{4}$. Therefore, the required probability is $\frac{2}{5} \times \frac{1}{4}=\frac{1}{10}$.

## SECOND SOLUTION

The possibilities of choosing 3 vertices at random are equal to the number of subsets of 3 elements of a set of 6 elements, which is $\left(\begin{array}{l}6 \\ 3\end{array}\right)=20$. The possibilities that they form an equilateral triangle are only 2, namely that the vertices are $A, C, E$ or $B, D, F$. Therefore, the number of favorable cases divided by the number of possible cases is $\frac{2}{20}=\frac{1}{10}$.