Selecting three non consecutive vertices from a regular decagon

Question

A regular polygon of $10$ sides is constructed. In how many ways can 3 vertices be selected so that no two vertices are consecutive.

Attempt:

Choose 1 vertex -> mark the two vertices next to it as "forbidden" -> choose 2 from among the 7 remaining vertices -> Account for the cases in which consecutive vertices are chosen from the remaining 7.

Mathematically:

$(^{10}C_1 \times {^{7}}C_2) - 6 = 210 - 6 = 204 \text{ ways}$

But answer given is: $50$.

I am unable to figure out where I have over-counted.

Answer 1

$^{10}C_1 \times( {^{7}}C_2 - 6 )= 150 \text{ ways}$. now each selection of three vertices is being chosen thrice at each individual vertex, so division by 3 to remove over counting.