# Permutations and Combinations in Circular Arrangement

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?

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