Let be a 6$$\times$$ 6 matrix over , and let be a simple labeled graph on vertex set . Define to be the set of column indices whose columns in are all-zero.
Count the number of pairs such that
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has no zero rows;
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;
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in , no vertex in is isolated;
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has no edges with one endpoint in and the other in its complement.
[THOUGHT] Choose S subset of {1..6} with |S|=2: \binom{6}{2}=15 ways.
For each S, let T be complement, |T|=4.
A: columns S all zero (fixed), 6x4 submatrix on T: no zero rows, no zero columns. Number N = \sum_{k=0}^4 (-1)^k \binom{4}{k} (2^{4-k}-1)^6
For G: edge within S present (1 way), no edges S-T (1 way), arbitrary on T: 2^{\binom{4}{2}}=64 ways.
Total: 15 * N * 64.
Use Python to compute exactly. [/THOUGHT] [PYTHON] from math import comb
def compute_N(): total = 0 for k in range(5): sign = (-1)k binom = comb(4, k) choices_per_row = 2(4 - k) - 1 total += sign * binom * (choices_per_row ** 6) return total
N = compute_N() num_S = comb(6, 2) num_G = 2**6 total = num_S * N * num_G print(total) [/PYTHON]
[THOUGHT] The Python output is 10487423040, which is the exact count. [/THOUGHT] \boxed{10487423040}
Explanation
The reasoning systematically counts the valid pairs (A, G) by breaking down the conditions.
First, select the set S = Z(A), a subset of {1,…,6} with exactly 2 elements, giving \binom{6}{2} = 15 possibilities. Let T be the complement of S, so |T| = 4.
For matrix A, the columns indexed by S must be entirely zero vectors. The remaining 6×4 submatrix on columns T must satisfy: no zero rows (across the entire A) and no zero columns (since |Z(A)|=2 exactly). The number of such submatrices, denoted N, is computed via inclusion-exclusion principle:
- Start with matrices having no zero rows: (2^4 - 1)^6 = 15^6.
- Subtract cases where at least one specific column in T is zero, and so on: N = ∑_{k=0}^4 (-1)^k \binom{4}{k} (2^{4-k} - 1)^6.
This yields the exact count of valid 6×4 submatrices.
For graph G on vertices {1,…,6}:
- No edges between S and T (fixed: 1 way).
- No vertex in S is isolated. With no edges from S to T, the only possible edge for vertices in S is the single edge between the two vertices in S, which must exist (1 way).
- Edges within T are arbitrary: 2^{\binom{4}{2}} = 2^6 = 64 possibilities.
The total number of pairs is thus 15 × N × 64. The Python code implements this computation precisely, producing the output 10487423040 as the final count, verified as ground truth.