'Twenty persons , of which two are brothers, are to be seated around a circular … ' from Permutation and combination

Question

Twenty persons , of which two are brothers, are to be seated around a circular table. Find the number of arrangements in which at least three person between the brothers.

My attempt

Let 'Two brothers + Three others' be person x.

'Three others' from x can be arranged in $$18*17*16$$ ways. And the brothers can be arranged in 2 ways as they can be switched. Hence x can be arranged in $$18*17*16*2$$ ways. In the circular arrangement let us we fix person x in any one of the seat (Actually 5 seats). Then the remaining 15 + 1 people (person x included) can be arranged in $$15!$$ ways. But as person x can be arranged in $$18*17*16*2$$ ways, the total arrangements should be $$18! * 2$$ ways.

I noticed that the question states 'at least' , so we can move the second brother to remaining 13 places. Hence, $$18! * 26$$ ways are possible.

What am I doing wrong?

You can just use the first brother to orient the table and do everything from there. Seating him prohibits seven seats to the other brother, the one he is in and three on either side. Seat the second brother in one of $$13$$ places, then the rest of the people in $$18!$$ ways, for a total of $$13 \cdot 18!$$
Your multiplying by $$2$$ double counts each arrangement, once when you seat each brother first.
• No, it will produce an arrangement you have already seen. A circular table has all the seats equivalent. When you seat brother A, you identify seat $1$. Then brother B can sit in any from $4$ to $16$. If you seat brother B first in the same chair and A in seat $4$, it is the same as seating A in $1$ and B in $16$. All you care about is the clockwise distance from A to B, which has $13$ possible values. – Ross Millikan Aug 6 at 13:51