<think> Okay, I'm trying to solve this probability problem about coloring the edges of a cube. Let me break it down step by step.

First, the problem states that we have a regular cube with 12 edges, each colored with one of 12 different colors. The key condition is that any two edges sharing a vertex must have different colors. We need to find the probability that all edges are colored with pairwise distinct colors and express it as a reduced fraction q/p, then compute p+q.

Hmm, so essentially, we're looking at proper edge colorings of a cube where each edge gets a unique color from 12 available colors. The total number of ways to color the edges without any restrictions is 12! since there are 12 edges and 12 colors. But not all of these colorings will satisfy the condition that adjacent edges (sharing a vertex) have different colors.

Wait, but actually, in this case, each edge must be colored with exactly one color, and no two edges sharing a vertex can have the same color. So it's about counting the number of proper edge colorings using all 12 distinct colors.

I remember that for edge coloring, Vizing's theorem states that any simple graph can be edge-colored with either Δ or Δ+1 colors, where Δ is the maximum degree. For a cube, each vertex has degree 3, so Δ=3. Therefore, the cube is Class 1 if it can be colored with 3 colors, but here we're using 12 different colors, which is more than enough.

Wait, no, that's not directly applicable because Vizing's theorem applies to edge colorings where each color class forms a matching. Here, we are assigning distinct colors to edges such that adjacent edges have different colors, but the number of colors used is equal to the number of edges, so it's more about permutations with constraints.

So perhaps I should model this as counting the number of proper colorings where each edge gets a unique color from 12 options, divided by the total number of possible colorings (which is 12!).

But how do we count the number of valid colorings? This seems complex because each edge's color choice affects multiple others due to shared vertices.

Maybe I can approach this using permutations with constraints. Each vertex has three edges meeting at it, so for each vertex, the three edges must have distinct colors. Since all 12 colors are used, each color is assigned exactly once.

Wait a second, but if we're assigning 12 different colors to 12 edges, then each color is used exactly once. So the problem reduces to counting the number of proper edge colorings where each edge has a unique color from 12 options, ensuring that no two adjacent edges share the same color.

But since all colors are distinct and we have more than enough colors (12), it's possible that such colorings exist. However, I'm not sure how to count them directly.

Alternatively, perhaps this is similar to counting derangements but for edge colorings. But I don't recall a direct formula for this.

Wait, maybe another approach: think of the cube as a graph and find the number of proper edge colorings with 12 colors where each color is used exactly once. This would be equivalent to finding the number of perfect matchings in some transformed graph, but I'm not sure.

Alternatively, perhaps we can model this using permutations. Each edge must have a unique color, so it's like assigning a permutation of 12 colors to the edges such that adjacent edges don't share the same color.

But how do we count such permutations? It seems complicated because each edge is connected to others through vertices, creating dependencies.

Maybe I can use inclusion-exclusion. The total number of colorings without any restrictions is 12!. Then subtract the colorings where at least one vertex has two edges with the same color. But this quickly becomes intractable due to the overlapping conditions.

Alternatively, perhaps it's easier to compute the probability by considering the constraints step by step as we assign colors to each edge.

Let me try that approach:

1. Start with an empty cube.
2. Choose a color for the first edge: 12 choices.
3. Move to another edge adjacent to the first one; since they share a vertex, it must have a different color: 11 choices.
4. Continue this process, ensuring that each new edge's color doesn't conflict with its already colored neighbors.

But this seems too vague because the cube has cycles and multiple connections, making it hard to track dependencies without considering the entire structure.

Wait, perhaps I can model this as a graph coloring problem where each edge is a node in a line graph. The line graph of the cube would have nodes representing edges of the cube, with two nodes connected if their corresponding edges share a vertex. Then, the problem reduces to counting the number of proper colorings of this line graph with 12 colors.

But I'm not sure about the properties of the line graph of a cube. It might be complex, but maybe it's bipartite or has some symmetry that can help.

Alternatively, perhaps there is a known result for edge colorings of cubes. Let me think: A cube is a bipartite graph with each vertex having degree 3. The edge chromatic number (the minimum number of colors needed) is equal to the maximum degree when the graph is Class 1. Since the cube is bipartite and regular, it's actually Class 1, meaning it can be edge-colored with Δ=3 colors.

But in our problem, we're using 12 different colors, which is more than enough. So each color will be used exactly once across all edges.

Wait, but if we have to use all 12 colors, that's impossible because the cube only has 12 edges. Wait no, wait: Each edge must be colored with a unique color from 12 options, so each color is used exactly once. So it's equivalent to assigning a permutation of 12 colors to the 12 edges such that adjacent edges have different colors.

But since all colors are distinct and we're using all of them, perhaps this is similar to counting Latin squares or something else.

Alternatively, maybe I can think in terms of derangements for each vertex. Each vertex has three edges; their colors must be distinct. Since all 12 colors are used, the problem might have a specific structure that allows us to compute the number of valid colorings.

But this seems too abstract. Maybe it's better to look for known results or similar problems.

Wait, perhaps I can model this as counting the number of proper edge colorings with exactly 12 colors where each color is used once. Since the cube has 12 edges and we're using 12 distinct colors, each color appears exactly once.

In that case, the problem reduces to finding the number of derangements where no two adjacent edges share the same color, but since all colors are unique, it's more about ensuring that in the permutation of colors assigned to edges, no two adjacent edges (in terms of sharing a vertex) have the same color.

Wait, perhaps another way: The total number of possible assignments is 12! because each edge gets a unique color. Now, we need to count how many of these assignments satisfy that for every vertex, all three edges incident to it have distinct colors.

But since all colors are used exactly once, the condition is automatically satisfied if no two adjacent edges share the same color. Wait, but in this case, each edge has a unique color, so any two edges sharing a vertex will necessarily have different colors because they're assigned different colors from the 12. So actually, every possible assignment where all 12 colors are used exactly once would automatically satisfy the condition that adjacent edges have different colors.

Wait, is that correct? Let me think again: If each edge has a unique color, then any two edges sharing a vertex must have different colors because they're assigned different colors. So in this case, every possible permutation of assigning 12 distinct colors to the 12 edges would automatically satisfy the condition that adjacent edges have different colors.

Wait, but that can't be right because if you assign colors without considering adjacency, it's possible that two edges sharing a vertex could end up with the same color. But wait no: since each edge is assigned a unique color from 12 distinct colors, any two edges (whether adjacent or not) will have different colors. So in this case, all permutations of assigning 12 distinct colors to the 12 edges would automatically satisfy the condition that adjacent edges have different colors.

Wait, but that can't be correct because the problem states "any two edges that share a vertex must receive different colors." But if we're using 12 distinct colors for 12 edges, then any two edges sharing a vertex will indeed have different colors because each edge has its own unique color. So in this case, every possible assignment where all 12 colors are used exactly once would satisfy the condition.

Wait, but that seems counterintuitive because usually, when you assign colors without considering adjacency, some adjacent edges might end up with the same color. But here, since each edge is assigned a unique color from 12 options, no two edges (whether adjacent or not) can have the same color. So in this specific case, all possible assignments of 12 distinct colors to the 12 edges would satisfy the condition that adjacent edges have different colors.

If that's true, then the number of valid colorings is simply 12! because every permutation satisfies the adjacency condition. But wait, that can't be right because in reality, when you assign colors without considering adjacency, some assignments might violate the condition even if all colors are distinct. Wait no: since each edge has a unique color, any two edges sharing a vertex must have different colors because they're assigned different colors from the 12.

Wait, but that's only true if we're using exactly as many colors as there are edges and assigning each color exactly once. So in this case, yes, every possible assignment would satisfy the condition because no two edges can share the same color, regardless of their adjacency.

Therefore, the number of valid colorings is 12! And since the total number of possible assignments is also 12!, the probability would be 1 (since all assignments are valid). But that seems incorrect because usually, such problems have a non-trivial probability. Maybe I'm misunderstanding the problem.

Wait, let me read the problem again: "A regular cube has its 12 edges colored with 12 different colors. Each edge is colored with exactly one color, and any two edges that share a vertex must receive different colors." So we're using all 12 colors, each exactly once, on the 12 edges, ensuring that adjacent edges have different colors.

Wait, but if you use all 12 colors, then by default, no two edges can have the same color, so any assignment would satisfy the condition. Therefore, every possible permutation of assigning 12 distinct colors to the edges is a valid coloring. Hence, the number of valid colorings is 12!.

But that seems too straightforward, and the probability would be 1 because all assignments are valid. But that can't be right because usually, such problems have constraints that reduce the number of valid colorings.

Wait, perhaps I'm misinterpreting the problem. Maybe it's not about using exactly 12 different colors but choosing from 12 colors where each edge is colored with one of the 12, possibly repeating colors, but ensuring adjacent edges are different. But no, the problem states "colored with 12 different colors," so each color is used exactly once.

Wait, perhaps I'm overcomplicating it. Let me think again: If we have 12 edges and 12 distinct colors, assigning each edge a unique color from these 12 will automatically ensure that no two adjacent edges share the same color because they can't; each has its own unique color.

Therefore, every possible assignment of the 12 distinct colors to the 12 edges is valid. Hence, the number of valid colorings is 12!, and the total number of possible assignments is also 12!. Therefore, the probability is 12! / 12! = 1.

But that seems too simple, so perhaps I'm missing something. Maybe the problem is about using exactly 12 colors but allowing some to be unused? No, the problem says "colored with 12 different colors," implying each color is used exactly once.

Wait, maybe I'm misunderstanding the condition. The problem states that any two edges sharing a vertex must have different colors. But if we're assigning all 12 distinct colors, then yes, this condition is automatically satisfied because no two edges can share the same color.

Therefore, the number of valid colorings is indeed 12!, and the probability is 1. Hence, q/p = 1/1, so p+q=2.

But that seems too straightforward for a problem from an Olympiad dataset. Maybe I'm making a mistake here.

Wait, perhaps the problem isn't about using all 12 colors but choosing any number of colors with the constraint that adjacent edges have different colors. But no, the problem explicitly says "colored with 12 different colors," so each edge must be assigned one of the 12 distinct colors exactly once.

Alternatively, maybe I'm misinterpreting the problem as using all 12 colors when it's actually about choosing a subset of colors such that adjacent edges have different colors. But no, the wording is clear: "colored with 12 different colors," meaning each edge gets one color from a set of 12 distinct colors.

Wait, perhaps another approach: The cube has 8 vertices, each connected to three edges. For each vertex, the three edges must have distinct colors. Since we're using 12 distinct colors, each color is used exactly once across all edges.

In this case, for any given vertex, its three edges will have three different colors, which are unique across the entire cube. So every assignment of 12 distinct colors to the edges satisfies the condition that adjacent edges have different colors because they can't share a color in the first place.

Therefore, the number of valid assignments is indeed 12!, and the total possible assignments (without any restrictions) would be P(12,12)=12! as well. Hence, the probability is 1.

But that seems too trivial for an Olympiad problem. Maybe I'm missing a constraint or misinterpreting the problem.

Wait, perhaps the problem allows using up to 12 colors but doesn't require all to be used. But no, it says "colored with 12 different colors," which implies each edge is assigned one of these 12 distinct colors exactly once.

Alternatively, maybe I'm supposed to consider colorings where edges can share colors as long as adjacent ones don't, and then find the probability that all 12 colors are used. But no, the problem states "colored with 12 different colors," meaning each edge must have a unique color from the set of 12.

Wait, perhaps I'm overcomplicating it. Let me try to think differently: The total number of ways to assign colors is 12! because each edge gets one of 12 distinct colors exactly once. Now, how many of these assignments satisfy that no two adjacent edges share the same color? But since all colors are unique, any assignment will automatically satisfy this condition.

Therefore, the probability is 1, and q/p = 1/1, so p+q=2.

But I'm not confident because usually, such problems have a non-trivial solution. Maybe I should double-check my reasoning. </think>

[THOUGHT] The problem involves coloring all 12 edges of a cube with 12 distinct colors, ensuring that any two adjacent edges (sharing a vertex) have different colors. Since each edge is assigned a unique color from the set of 12, no two edges can share the same color, thus automatically satisfying the adjacency condition. Therefore, every possible permutation of assigning these 12 colors to the edges is valid.

The number of valid colorings is equal to the total number of permutations of 12 distinct colors, which is 12!. The probability is thus the ratio of valid colorings to all possible colorings, both being 12!, resulting in a probability of 1. Expressing this as a fraction q/p gives 1/1, leading to p + q = 2. [/THOUGHT] [PYTHON]

Since every permutation satisfies the condition, the probability is 1

q = 1 p = 1 print(f"The reduced fraction is {q}/{p}, so p+q = {p + q}")

<think> Okay, so I'm trying to solve this cube edge coloring probability problem. Let me break it down step by step.

First, the cube has 12 edges, and each edge is colored with one of 12 different colors. The condition is that any two edges sharing a vertex must have different colors. Since there are 12 different colors, assigning each edge a unique color automatically satisfies this condition because no two adjacent edges will share the same color.

So, the total number of ways to color the edges without any restrictions would be 12 factorial (12!), which is the number of permutations of 12 distinct colors. But since every permutation here already meets the adjacency condition, all possible colorings are valid.

Therefore, the probability that all 12 edges are colored with pairwise distinct colors is simply the ratio of valid colorings to total colorings, both of which are 12!. This simplifies to a probability of 1. Expressed as a reduced fraction, this is 1/1. Hence, p = 1 and q = 1, so p + q equals 2.

I think that's the correct approach because every possible way of assigning the colors meets the condition, making the probability 1. </think>

The problem involves coloring all 12 edges of a cube with 12 distinct colors. Since each edge is assigned a unique color from the set of 12, no two adjacent edges will share the same color, satisfying the given condition. The number of valid colorings is equal to the total number of permutations of 12 distinct colors, which is 12!. Thus, every permutation is valid, leading to a probability of 1. Expressed as a reduced fraction, this is (\frac{1}{1}), so (p + q = 1 + 1 = 2).

The final answer is:

\boxed{2}