Circular Coloring Distribution of Hats for Five Persons

To find the number of ways to distribute hats of $3$ distinct colours (red, blue, green) to $5$ persons $A, B, C, D$, and $E$ seated in a circular arrangement such that no two adjacent persons have hats of the same colour, we can model this as finding the chromatic polynomial of a 5-cycle graph $C_5$ with $k = 3$ colours.

Method 1: Using the Chromatic Polynomial of a Cycle Graph

For a cycle graph $C_n$ on $n$ vertices with $k$ available colours, the number of valid proper colourings is given by: $P(C_n, k) = (k-1)^n + (-1)^n(k-1)$

Substituting $n = 5$ and $k = 3$: $P(C_5, 3) = (3-1)^5 + (-1)^5(3-1)$ $P(C_5, 3) = 2^5 - 2 = 32 - 2 = 30$

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Method 2: Combinatorial Case-by-Case Analysis

Since adjacent persons cannot have the same colour, and the cycle has an odd length ($5$), it is impossible to colour the cycle using only $1$ or $2$ colours. Hence, all $3$ colours must be used.

The only possible partition of $5$ hats using $3$ colours such that no colour is used more than twice (a colour used $\ge 3$ times in a 5-cycle would force two hats of that colour to be adjacent) is: $\text{Distribution of colours: } 2, 2, 1$

1. Choose which colour is used once: $\binom{3}{1} = 3 \text{ ways}$ Let this colour be Red. The other two colours (Blue and Green) will each be used twice.

2. Choose the person who gets the single Red hat: $\binom{5}{1} = 5 \text{ ways}$ Without loss of generality, let person $A$ receive the Red hat.

3. Assign the remaining hats to $B, C, D, E$: The remaining 4 persons $B, C, D, E$ must receive 2 Blue hats and 2 Green hats such that adjacent persons have different colours. Since $B, C, D, E$ form a line of 4 positions where adjacent persons cannot have the same colour, they must alternate in colour:

   • Configuration 1: $B = \text{Blue}, C = \text{Green}, D = \text{Blue}, E = \text{Green}$
   • Configuration 2: $B = \text{Green}, C = \text{Blue}, D = \text{Green}, E = \text{Blue}$

   In both configurations, $A$ (Red) is distinct from its neighbors $B$ and $E$, and all other adjacent pairs differ. Thus, there are exactly $2$ valid assignments for the remaining 4 persons.

Multiplying the independent choices together: $\text{Total number of ways} = 3 \times 5 \times 2 = 30$

Final Answer: 30