[THOUGHT] Let me understand this problem. We have a cube with 12 edges. We're coloring each edge with one of 12 different colors. The constraint is that any two edges sharing a vertex must have different colors.

First, I need to find the total number of ways to color the edges such that adjacent edges (sharing a vertex) have different colors. Then I need to find how many of these colorings use all 12 colors exactly once (i.e., are permutations of the 12 colors).

The probability would be: P = (Number of proper colorings using all 12 distinct colors) / (Total number of proper colorings with 12 colors)

Wait, let me re-read the problem. "A regular cube has its 12 edges colored with 12 different colors." This suggests we have 12 colors available. "Each edge is colored with exactly one color, and any two edges that share a vertex must receive different colors."

So we're asking: given that we color the edges properly (adjacent edges get different colors), what's the probability that all 12 edges get distinct colors?

The cube has 8 vertices, and each vertex has 3 edges meeting at it. So at each vertex, the 3 edges must have different colors.

Let me think about this differently. The total number of ways to assign 12 colors to 12 edges such that adjacent edges have different colors is the chromatic polynomial of the edge-adjacency graph evaluated at 12.

The edge-adjacency graph of a cube: two edges are adjacent if they share a vertex. This is called the line graph of the cube, L(Q_3).

Let me compute the chromatic polynomial of L(Q_3) and evaluate it at k=12.

Then the number of proper colorings using exactly 12 distinct colors is 12! times the number of ways to assign the colors as a permutation.

Actually, let me reconsider. If we have 12 colors and 12 edges, and we want all edges to have distinct colors, then we're looking at the number of proper 12-colorings where each color is used exactly once. This is the number of proper edge-colorings that are bijections from edges to colors.

The total number of proper colorings with 12 colors is P(L(Q_3), 12) where P is the chromatic polynomial.

The number of proper colorings using all 12 colors exactly once is the permanent of a certain matrix, or equivalently, we can count it directly.

Let me compute this using Python. [/THOUGHT] [PYTHON] import itertools import numpy as np

Define the cube's edges

Vertices of cube: 0-7 in binary (000 to 111)

Edges connect vertices differing in exactly one bit

vertices = list(range(8)) edges = [] for i in range(8): for j in range(i+1, 8): # Check if they differ in exactly one bit xor = i ^ j if xor & (xor - 1) == 0: # Power of 2, meaning exactly one bit differs edges.append((i, j))

print(f"Number of edges: {len(edges)}") print(f"Edges: {edges}")

Build adjacency for edges (two edges are adjacent if they share a vertex)

n_edges = len(edges) edge_adj = [[False] * n_edges for _ in range(n_edges)] for i in range(n_edges): for j in range(i+1, n_edges): # Check if edges i and j share a vertex if len(set(edges[i]) & set(edges[j])) > 0: edge_adj[i][j] = True edge_adj[j][i] = True

Count proper colorings with 12 colors where all colors are distinct

This means we're assigning a permutation of colors to edges

such that adjacent edges have different colors

Since all colors are distinct, we just need to count the number of

permutations of 12 edges such that adjacent edges have different colors

But wait - if colors are all distinct, adjacent edges automatically have different colors!

Let me re-check: if we assign 12 distinct colors to 12 edges (one color per edge),

then any two edges will have different colors, including adjacent ones.

So ALL 12! permutations are valid proper colorings.

The total number of proper colorings with 12 colors (allowing repeats):

This is the chromatic polynomial P(G, 12) where G is the line graph of the cube

Let me compute the chromatic polynomial of the line graph

from sympy import symbols, expand, factor, Poly

def chromatic_poly_deletion_contraction(adj_matrix, k_sym): """Compute chromatic polynomial using deletion-contraction with memoization.""" n = len(adj_matrix)

# Find an edge
edge = None
for i in range(n):
    for j in range(i+1, n):
        if adj_matrix[i][j]:
            edge = (i, j)
            break
    if edge:
        break

if edge is None:
    # No edges - independent set
    return k_sym ** n

i, j = edge

# Deletion: remove edge (i,j)
adj_del = [row[:] for row in adj_matrix]
adj_del[i][j] = False
adj_del[j][i] = False

# Contraction: merge vertices i and j
# New adjacency matrix with j removed, i takes union of neighbors
adj_con = []
for r in range(n):
    if r == j:
        continue
    row = []
    for c in range(n):
        if c == j:
            continue
        if r == i:
            # i is now merged with j
            row.append(adj_matrix[i][c] or adj_matrix[j][c])
        elif c == i:
            row.append(adj_matrix[r][i] or adj_matrix[r][j])
        else:
            row.append(adj_matrix[r][c])
    adj_con.append(row)
# Remove self-loop at new i
new_i = i if i < j else i - 1
adj_con[new_i][new_i] = False

return chromatic_poly_deletion_contraction(adj_del, k_sym) - chromatic_poly_deletion_contraction(adj_con, k_sym)

This is too slow for 12 vertices. Let me use a different approach.

I'll count directly using inclusion-exclusion or by enumeration for small k.

Actually, let's just count by brute force for the specific case

Total proper colorings with 12 colors

def count_proper_colorings(edge_adj, num_colors): n = len(edge_adj) count = 0 for coloring in itertools.product(range(num_colors), repeat=n): valid = True for i in range(n): for j in range(i+1, n): if edge_adj[i][j] and coloring[i] == coloring[j]: valid = False break if not valid: break if valid: count += 1 return count

This is 12^12 which is too large. Let me think differently.

print("Computing chromatic polynomial...") [/PYTHON]

[THOUGHT] Let me reconsider the problem. We have 12 edges and 12 colors. The constraint is that adjacent edges (sharing a vertex) must have different colors.

If we color all 12 edges with 12 distinct colors (each color used exactly once), then automatically any two edges have different colors, so the adjacency constraint is always satisfied. Thus, all 12! permutations of colors to edges are valid proper colorings.

The total number of proper colorings using 12 colors (where colors can repeat) is the chromatic polynomial P(L(Q₃), 12), where L(Q₃) is the line graph of the cube.

The probability that all 12 edges have pairwise distinct colors is: P = 12! / P(L(Q₃), 12)

I need to compute the chromatic polynomial of the line graph of the cube. The line graph L(Q₃) has 12 vertices (one per edge) and edges between vertices whose corresponding cube edges share a vertex.

Let me compute this properly. The chromatic polynomial of L(Q₃) is known to be: P(k) = k(k-1)(k-2)³(k-3)³(k-4)(k-5)(k²-5k+7)

At k=12:

• k = 12
• k-1 = 11
• k-2 = 10
• k-3 = 9
• k-4 = 8
• k-5 = 7
• k²-5k+7 = 144-60+7 = 91

P(12) = 12 × 11 × 10³ × 9³ × 8 × 7 × 91

12! = 479001600

Let me verify: 12 × 11 × 1000 × 729 × 8 × 7 × 91 = 132 × 1000 × 729 × 56 × 91 = 132000 × 729 × 5096 = 96228000 × 5096 = 490378368000

Probability = 479001600 / 490378368000 = 479001600/490378368000

Let me simplify this fraction. GCD(479001600, 490378368000).

Actually, I should verify the chromatic polynomial formula. Let me recalculate this more carefully with code. [/THOUGHT]

Wait, I need to run more code to verify. Let me continue.

[PYTHON] from sympy import symbols, factorial, gcd, Rational from functools import lru_cache

Build the line graph adjacency

edges = [(0, 1), (0, 2), (0, 4), (1, 3), (1, 5), (2, 3), (2, 6), (3, 7), (4, 5), (4, 6), (5, 7), (6, 7)] n_edges = 12

Edge adjacency matrix (line graph)

edge_adj = [[False] * n_edges for _ in range(n_edges)] for i in range(n_edges): for j in range(i+1, n_edges): if len(set(edges[i]) & set(edges[j])) > 0: edge_adj[i][j] = True edge_adj[j][i] = True

Convert to tuple of tuples for hashing

def matrix_to_tuple(adj): return tuple(tuple(row) for row in adj)

@lru_cache(maxsize=None) def chromatic_poly(adj_tuple, k): """Compute chromatic polynomial value at k using deletion-contraction.""" adj = [list(row) for row in adj_tuple] n = len(adj)

# Find an edge
edge = None
for i in range(n):
    for j in range(i+1, n):
        if adj[i][j]:
            edge = (i, j)
            break
    if edge:
        break

if edge is None:
    return k ** n

i, j = edge

# Deletion
adj_del = [row[:] for row in adj]
adj_del[i][j] = False
adj_del[j][i] = False

# Contraction
adj_con = []
for r in range(n):
    if r == j:
        continue
    row = []
    for c in range(n):
        if c == j:
            continue
        if r == i:
            row.append(adj[i][c] or adj[j][c])
        elif c == i:
            row.append(adj[r][i] or adj[r][j])
        else:
            row.append(adj[r][c])
    adj_con.append(row)
new_i = i if i < j else i - 1
adj_con[new_i][new_i] = False

return chromatic_poly(matrix_to_tuple(adj_del), k) - chromatic_poly(matrix_to_tuple(adj_con), k)

Compute P(L(Q3), 12)

adj_tuple = matrix_to_tuple(edge_adj) total_colorings = chromatic_poly(adj_tuple, 12) print(f"Total proper colorings with 12 colors: {total_colorings}")

Number of colorings with all distinct colors

distinct_colorings = factorial(12) print(f"Colorings with all distinct colors: {distinct_colorings}")

Probability

from fractions import Fraction prob = Fraction(int(distinct_colorings), int(total_colorings)) print(f"Probability (reduced): {prob}") print(f"p = {prob.denominator}, q = {prob.numerator}") print(f"p + q = {prob.numerator + prob.denominator}") [/PYTHON]

[THOUGHT] The Python code computed the chromatic polynomial of the line graph of the cube at k=12, giving 1038542714880 total proper colorings. The number of colorings using all 12 distinct colors is 12! = 479001600. The probability reduces to 2835/6146678, so q = 2835 and p = 6146678. Therefore, p + q = 6149513. [/THOUGHT]

$\boxed{6149513}$