Five persons A, B, C, D and E are seated in a circular arrangement....

Given that no two persons sitting adjacent, have the hat of same colour, so hats of all colour must be used, also $3$ hats of same colour cannot be used as at least two hats of same colour will become consecutive, so the only combination will be 2, 2, 1 .<br/>Three cases are possible<br/>\left$\begin{array}{ll}a ) 2R, 2B, 1G \\ b ) 2B, 2G, 1R \\ c ) 2R, 2G, 1B\end{array}$$\right\} - 3 way
Select are person for single colour hat $= {C}_{1} 5$ (let say green)

Now remaining $4hats can be distributed alternatively in$two ways \left(BRBRorRBRB\right)
Total number of ways$= 3 \times {C}_{1} 5 \times 2$
$= 30$