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Surrounded by liars

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$10$ witnesses sit around a round table. Each witness is honest (a truth-teller) or deceptive (a liar). Every witness says: "Both witnesses beside me are deceptive." In how many different ways can the types be assigned to the labelled seats so that all $10$ statements are consistent?

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Hint

Translate each type into a rule about neighbours, then cut the circle into blocks that each start with a truth-teller.

Solution

A truth-teller has $2$ liars next to him, so no $2$ truth-tellers are adjacent; a liar's statement is false, so he has at least one truth-teller next to him, so there are no $3$ liars in a row.
So every truth-teller is followed by $1$ or $2$ liars, and the circle splits into blocks of $2$ or $3$ seats. With $x$ blocks of $2$ and $y$ blocks of $3$ we need $2 x + 3 y = 10$ , so $(x, y) = (5, 0)$ or $(2, 2)$ .
With $k$ blocks in a given cyclic order and $10$ labelled seats, the number of placements is $10 \times \frac{( x k )}{k}$ .
For $(5, 0)$ this is $10 \times \frac{1}{5} = 2$ , and for $(2, 2)$ it is $10 \times \frac{6}{4} = 15$ , so the total is $2 + 15 = 17$ .

Answer: $17$

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