Getting a non natural answer for combinatorics question Question: In how many ways can 3 vertices be selected from a decagon so no two are consecutive?
Once I choose one of the vertices, since the adjacent ones are ruled out, I now have 7 vertices to choose my next one from. Similarly, the next time around, I will have 4 vertices to choose from. Additionally, I accounted for overlapping triangles by dividing my answer with 3.
My final answer looks like this:
(10C1 * 7C1 * 4C1)/3 which isn't even a natural number.
Where am I going wrong?
 A: Let's try your strategy with 6 elements
XXXXXX
In the first case, I select one as you told at random:
XX(X)XXX
Now for the second, suppose I take this choice:
XX(X)XX(X)
Then you see there is no where I can keep the third such that all three are not adjacent.
But suppose, I took this other choice:
XX(X)X(X)X
Clearly I have one more choice left  for third element.

I couldn't think of a faithful notation for circular seating but the idea is the string repeats itself as end, meaning if you ask "what is the next element" pointing at the last element from the left, then I'd point at the first element in the string.
A: Number the vertices clockwise from $1-10$
Make $3$ blocks of  chosen-unchosen, looking clockwise, $\,\boxed{\circ\bullet}$
so there are now $3$ paired units and $4$ single units.
The block formation ensures that two chosen vertices can't now be adjacent, wherever the blocks might be placed.
The blocks can be placed among the $7$ units in $\binom73 = 35$ ways,
but each unit is getting only $7$ starting points instead of $10$,
thus the final answer $= \dfrac{10}{7} \times 35 = 50$
