1. Interpretation of the problem

We have $5$ distinct persons $A,B,C,D,E$ seated in a fixed circular arrangement.

Each person is to be given a hat of one of $3$ colours: red, blue, green.

We must count the number of colourings of these $5$ seats such that adjacent persons have different colours.

Since the seating arrangement is already fixed, we only need to count valid colour assignments on a cycle of length $5$ using $3$ colours.

2. Model as colouring a cycle $C_5$

Let the seats around the circle be in order: $A \to B \to C \to D \to E \to A$

We want the number of proper colourings of this cycle with $3$ colours.

A standard result for colouring a cycle of length $n$ with $k$ colours is: $N=(k-1)^n+(-1)^n(k-1)$

Here, $n=5$ and $k=3$. So, $N=(3-1)^5+(-1)^5(3-1)$ $N=2^5-2$ $N=32-2=30$

So the required number is: $\boxed{30}$

3. Direct counting verification

We can also verify this without using the formula.

• Choose colour of $A$: $3$ ways.
• Choose colour of $B$ different from $A$: $2$ ways.
• Choose colour of $C$ different from $B$: $2$ ways.
• Choose colour of $D$ different from $C$: $2$ ways.

Now choose colour of $E$ such that it differs from both $D$ and $A$.

So we count according to whether $D=A$ or $D\ne A$.

Case 1: $D=A$

Fix $A$.

• $B$: $2$ choices.
• $C$: $2$ choices.
• For $D$ to equal $A$ and differ from $C$, this happens exactly when $C$ is the third colour not equal to $A$ or $B$.
• Then $E$ must differ from both $D(=A)$ and $A$, so it can be any of the other $2$ colours except $A$, but must also differ from $D$ which is same as $A$; thus $E$ has $2$ choices.

For fixed $A,B$, this gives $1\cdot 2=2$ valid completions.

Case 2: $D\ne A$

Then since $E$ must differ from both $D$ and $A$, and there are only $3$ colours, $E$ is forced to be the third colour. So $E$ has $1$ choice.

For fixed $A,B$, among the $2$ choices of $C$, one leads to $D=A$ and one leads to $D\ne A$. Hence this case contributes $1$ valid completion.

So for fixed $A,B$, total valid completions: $2+1=3$

Thus total number: $3 \times 2 \times 3 = 18$

This seems inconsistent, so let us carefully correct the direct counting.

4. Correct direct counting

Fix colour of $A$: $3$ ways. Fix colour of $B$: $2$ ways. Fix colour of $C$: $2$ ways. Fix colour of $D$: $2$ ways.

So along the path $A-B-C-D-E$, before checking $E$, there are: $3\cdot 2\cdot 2\cdot 2=24$ assignments for $A,B,C,D$.

Now count how many of these allow $E$.

For each colouring of $A,B,C,D$:

• If $A=D$, then $E$ can be any colour different from $A$, so $2$ choices.
• If $A\ne D$, then $E$ must be different from both, so $1$ choice.

Thus we need counts of assignments of $A,B,C,D$ with $A=D$ and with $A\ne D$.

Fix $A$.

• $B$: $2$ ways.
• $C$: $2$ ways.
• $D$: must differ from $C$.

For fixed $A,B$:

• If $C=A$, then $D$ has $2$ choices, neither equal to $A$ necessarily.
• If $C\ne A$, then among the $2$ choices for $D$ (excluding $C$), exactly one is $A$ and one is the third colour.

Let us list for fixed $A$.

Choose $B$ in $2$ ways.

For each fixed $B$:

• $C$ can be $A$ or the third colour.

If $C=A$:

• $D\ne A$, so $D$ has $2$ choices, both with $D\ne A$.

If $C$ is the third colour:

• $D$ has $2$ choices: one is $A$, one is $B$.
• So one gives $D=A$, one gives $D\ne A$.

Hence for fixed $A,B$:

• Number of assignments with $D=A$: $1$
• Number with $D\ne A$: $3$

Therefore for fixed $A$:

• Since $B$ has $2$ choices, $\#(D=A)=2, \qquad \#(D\ne A)=6$

Now including $E$:

• From $D=A$: each gives $2$ choices for $E$.
• From $D\ne A$: each gives $1$ choice for $E$.

So for fixed $A$: $2\cdot 2 + 6\cdot 1 = 10$

Finally, $A$ has $3$ choices, so total: $3\cdot 10 = 30$

Thus the required number is: $\boxed{30}$

5. Comparison with stored answer

Stored correct answer = $30$.

Our derived answer = $30$.

So the stored answer is correct.