Choose 2 positions for the two prime rolls: $ \binom{4}{2} = 6 $
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- So only 2a: 2 (prime assignments) × 2 (odd non-prime) × 2 (even non-prime) = 8
- Plus position arrangement: the even prime and odd prime are in different positions: $ \binom{4}{2} = 6 $ ways to assign which two slots get primes — but in this count, we are assigning to specific positions, so the $ \binom{4}{2} $ places the prime positions, then assign values.
- For Case 2:
- Assign one prime as 2, one as odd (3 or 5): 2 ways (which is which)
- Odd prime choice: 2
- Non-primes: one must be odd (1), one even (4 or 6): 2 choices
