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📅 May 2, 2026 👤 scraface

May 02, 2026

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Subcase 2a: One odd non-prime (choice: 1), one even non-prime (4 or 6)...
Subcase 2b: Two odd non-primes → 1 × 1 = 1 choice
Subcase 2c: Both even → 2 × 2 = 4 → contributes 0 odd
Now, only subcases that yield even total odd count:
Case 2a: odd non-primes count = 1 → total odd count = 1 (prime) + 1 =...
Case 2b: odd non-primes = 2 → total odd = 1 + 2 = 3 → odd → invalid
Case 2a valid, others invalid
So only 2a: 2 (prime assignments) × 2 (odd non-prime) × 2 (even non-pr...
Plus position arrangement: the even prime and odd prime are in differe...
For Case 2:

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