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. Express the probability that all 12 edges are colored with pairwise distinct colors as a reduced fraction and calculate .
This problem is based on RUC-AIBOX. OlymMATH: A Dataset of Olympiad-Level Math Problems (en-hard split, test row 31). Available at: https://huggingface.co/datasets/RUC-AIBOX/OlymMATH
[THOUGHT] The problem involves finding the probability that a random proper edge-coloring of the cube (with 12 colors) uses all distinct colors on the 12 edges. The favorable cases are 12! (any permutation works since distinct colors automatically satisfy adjacency condition). The total is the number N of proper 12-edge-colorings.
The cube's edges are partitioned into three groups of 4 parallel edges each (x, y, z directions), each group a perfect matching. Label them as 2x2 matrices X_{b c}, Y_{a c}, Z_{a b} for a,b,c in {0,1}.
The proper coloring conditions translate to:
- Each Y_{a c} ∉ set of row a of Z
- Each X_{b c} ∉ set of column b of Z ∪ set of column c of Y
To compute N:
- Loop over all 12^4 assignments to Z.
- For each Z, compute allowed sets A0, A1 for rows 0,1 of Y (size m0=12-s0, m1=12-s1).
- Compute column sets C0, C1 of Z.
- Compute S = ∑{p∈A0, q∈A1} ∏{b=0,1} (12 - |C_b ∪ {p,q}|)
- Add S² to total N (since two identical columns for Y).
Then, probability = 12! / N, reduce to q/p, compute p + q. [/THOUGHT] [PYTHON] import math
total_N = 0 K = 12 for Z00 in range(K): for Z01 in range(K): for Z10 in range(K): for Z11 in range(K): Z = [[Z00, Z01], [Z10, Z11]] row0_set = set([Z[0][0], Z[0][1]]) row1_set = set([Z[1][0], Z[1][1]]) s0 = len(row0_set) s1 = len(row1_set) A0 = [i for i in range(K) if i not in row0_set] A1 = [i for i in range(K) if i not in row1_set] C0 = set([Z[0][0], Z[1][0]]) C1 = set([Z[0][1], Z[1][1]]) S = 0 for p in A0: for q in A1: sq = {p, q} u0 = len(C0 | sq) u1 = len(C1 | sq) prod = (K - u0) * (K - u1) S += prod total_N += S * S
fact12 = math.factorial(12) g = math.gcd(fact12, total_N) q = fact12 // g p = total_N // g result = p + q print(result) [/PYTHON]
[THOUGHT] The Python code executed successfully and output 6149513, which is p + q for the reduced fraction q/p representing the probability. [/THOUGHT] \boxed{6149513}
Explanation
The reasoning follows a systematic enumeration of valid edge colorings for the cube graph. The cube's 12 edges are grouped into three sets of four parallel edges each, represented as 2x2 matrices: Z for one direction, Y for another, and X for the third. Proper coloring requires that no two edges sharing a vertex have the same color.
To compute the total number of proper 12-edge colorings (N):
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Enumerate all possible assignments of 12 distinct colors to the four edges of Z (12^4 possibilities).
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For each Z assignment:
- Compute forbidden colors for Y's rows: row 0 cannot use colors in Z's row 0 set (size s0), so A0 has 12 - s0 colors; similarly for row 1 (A1).
- Compute column sets of Z: C0 = {Z[0][0], Z[1][0]}, C1 = {Z[0][1], Z[1][1]}.
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For each pair (p from A0, q from A1) for Y's columns:
- For X's edges, compute available colors: for column b of X, exclude C_b ∪ {p, q}, yielding (12 - |C_b ∪ {p, q}|) options per edge.
- Product for both X columns: (12 - |C0 ∪ {p,q}|) * (12 - |C1 ∪ {p,q}|).
- Sum these products over all (p,q) pairs to get S (ways to color one pair of Y columns).
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Since Y has two identical columns, multiply S by S to account for both: add S^2 to total_N.
The number of favorable outcomes is 12!, as any permutation of 12 distinct colors is a proper coloring.
The probability is 12! / total_N. Compute gcd(g) of numerator and denominator, reduce to q/p (q = 12! / g, p = total_N / g), then p + q.
The Python code implements this exactly: nested loops over Z assignments, computes sets and sums as described, accumulates total_N, calculates factorial and gcd, and prints p + q = 6149513, confirming the result.