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?
On
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.
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:
$$8 \cdot 7! = 8!$$