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Choose position for 2: 2 ways
Odd prime at other: 3 or 5 → 2 choices
→ $ 2 \times 2 = 4 $ valid prime value pairs (e.g., (2,3), (2,5), (3,2), (5,2))
Now, total odd primes: 1 → contributes 1 odd number
Odd non-primes: 0 → but wait: the two non-prime rolls: we need total odd count even → current odd count = 1 (odd prime) → so even number of odd terms → need an odd number of odd non-primes
Odd non-primes contribute one odd each, even non-primes none.
Subcase 2a: One odd non-prime (choice: 1), one even non-prime (4 or 6) → 1 × 2 = 2 choices
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:
