Sum of units digits in conditional permutations

The 5-digit number positions are 1, 2, 3, 4, 5 (from left to right). Odd positions are 1, 3, 5. Odd digits available: {1, 3, 5} (3 digits); Even digits available: {2, 4} (2 digits). Condition: Exactly two of the odd positions (1, 3, 5) contain odd digits. Thus, one odd position contains an even digit. Case 1: Position 5 (rightmost position) is occupied by an EVEN digit (2 or 4). Positions 1 and 3 get 2 odd digits from {1, 3, 5}: $\binom{3}{2} \times 2! = 6$ ways. The remaining 1 odd digit and 1 even digit fill positions 2 and 4: $2! = 2$ ways. Total numbers with 2 or 4 at unit position $= 6 \times 2 = 12$ each. Sum of unit digits for even unit digits $= 12 \times (2 + 4) = 72$.

Case 2: Position 5 is occupied by an ODD digit (1, 3, or 5). One of positions 1 and 3 gets an odd digit: $\binom{2}{1} = 2$ choices for position, and 2 choices of remaining odd digits $= 4$ ways. Position 5 has 3 choices of odd digits. The remaining 1 odd digit and 2 even digits occupy the remaining 3 spots in $3!$ ways... Evaluating all such valid arrangements gives 16 occurrences for each odd digit at position 5. Sum $= 16 \times (1 + 3 + 5) = 144$. Total sum of unit digits $= 72 + 144 = 216$.