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.
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?