For Case 1:
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- Number of assignments: $ 1 \times 4 = 4 $
- But wait: for each prime position, we assign specific values. Since both primes are 2: only 1 way
- Non-primes: each has 3 choices → $ 3 \times 3 = 9 $, but we must restrict to even non-primes? No: for sum parity, only the oddness matters.
- Odd primes: 0 → contribution: 0 odd values
- Odd non-primes: 0 → 0
- Even non-primes: 0 → 0
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