Given that no two persons sitting adjacent have hats of same colour. Also, hats of different colour cannot be used in 1 + 1 + 3 combination because any three hats cannot be of same colour.

So, only possible combination due to circular arrangement is 2 + 2 + 1.

So, there are following three cases of selecting hats are

2R + 2B + 1G or 2B + 2G + 1R or 2G + 2R + 1B.

To distribute these 5 hats first we will select a person which we can done in $${{}^5{C_1}}$$ ways and distribute that hat which is one of it's colour. And, now the remaining four hats can be distributed in two ways. So, total ways will be 3 $$ \times $$ $${{}^5{C_1}}$$ $$ \times $$ 2 = 3 $$ \times $$ 5 $$ \times $$ 2 = 30