To solve this problem, we need to determine how many proper colorings are possible for the graph $K_{3,3}$ using 5 available colors.

Step 1: Understanding the Graph

The graph $K_{3,3}$ is a complete bipartite graph with 3 vertices in each partition. A proper coloring of a graph means no two adjacent vertices share the same color.

• $K_{3,3}$ consists of two disjoint sets of vertices, say $\{v_1, v_2, v_3\}$ and $\{u_1, u_2, u_3\}$, where every vertex in the first set is connected to every vertex in the second set.
• There are no edges between vertices within the same set.

Step 2: Bipartite Graph Coloring

In a bipartite graph, the vertices within one partition can all be the same color, as there are no edges between them. However, the two partitions must have different colors to maintain proper coloring.

• The 3 vertices in set $\{v_1, v_2, v_3\}$ can all be colored the same, and similarly for the 3 vertices in set $\{u_1, u_2, u_3\}$.
• The two sets $\{v_1, v_2, v_3\}$ and $\{u_1, u_2, u_3\}$ must have different colors.

Step 3: Counting the Colorings

1. For the first set $\{v_1, v_2, v_3\}$, you can choose any of the 5 colors.
2. For the second set $\{u_1, u_2, u_3\}$, you can choose any color except the color chosen for the first set. This gives 4 possible choices.

Step 4: Total Number of Proper Colorings

Thus, the total number of proper colorings is:

$5 \times 4 = 20$

Conclusion:

There are 20 proper colorings possible for the graph $K_{3,3}$ with 5 available colors.

Would you like a further breakdown or more details on graph coloring?

Here are 5 related questions:

1. How would the result change if there were 6 colors available?
2. What is the chromatic number of $K_{3,3}$?
3. How do bipartite graphs differ from complete graphs in terms of coloring?
4. What is the number of proper colorings for a general $K_{m,n}$ bipartite graph?
5. How does the number of proper colorings change for a non-bipartite graph?

Tip: In bipartite graphs, the chromatic number is always 2, which simplifies proper colorings.