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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?

Nyx
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    You could have more than 4 vertices to choose from in the third round. Suppose you chose 1,3 then you could choose all of 5-9, and these are 5 vertices. – NL1992 Aug 18 '21 at 06:26
  • Another approach is to reverse the question: Count all ways to choose 3 vertices from a decagon, then count how many of those have consecutive vertices. – Semiclassical Aug 18 '21 at 06:35
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    https://math.stackexchange.com/questions/2971121/selecting-three-non-consecutive-vertices-from-a-regular-decagon – tryst with freedom Aug 18 '21 at 06:46
  • If we have $n$ objects placed around a circle, and want to choose $k$ non adjacent objects, the number of choices is $\frac{n}{n-k}\binom{n-k}{k}$. – Muralidharan Mar 29 '22 at 02:36

2 Answers2

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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.

tryst with freedom
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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 $= \dfrac{10}{7} \times 35 = 50$

true blue anil
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