8 persons sit at a round table with 10 seats so that there is exactly one person between the two empty seats. How many possible arrangements are there?

Here's what I have so far:

${10 \choose 1}$ (for choosing the seat of the person to be isolated)

(8-1)! (to permute the group of 3 + remaining 7 people around the round table)

So my solution is 10*7! number of possible arrangements. Is this correct?

  • $\begingroup$ Why are you considering only the "remaining 7" and not all 8 people? What do you mean by "the group of 3"? $\endgroup$ – 79037662 Sep 30 at 14:03
  • $\begingroup$ The group of 3 is the isolated person between 2 empty seats $\endgroup$ – beyuma Sep 30 at 14:07

It doesn't matter how the "seat between the empty seats" is chosen, because we are considering a round table. We simply need to choose one person to sit here, and arrange the remaining 7. The number of possible arrangements is thus:

$$8 \cdot 7! = 8!$$


If the seats are distinguishable then the answer is: $10\cdot8!$.

First choose the isolated seat (factor $10$). This move also determines which $8$ seats are available.

Place the $8$ persons on the available seats (factor $8!$).

If the seats are not distinguishable then the answer is: $8!$.

Factor $10$ falls out.


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