Question

Let \(S\) be the set of five-digit numbers formed by the digits 1, 2, 3, 4 and 5, using each digit exactly once such that exactly two odd positions are occupied by odd digits. What is the sum of the digits in the rightmost position of the numbers in \(S\)?

• A. 216
• B. 228
• C. 294
• D. 192

Hint:

The odd positions are the 1st, 3rd and 5th. Split into cases according to whether the rightmost (5th) position holds an even or an odd digit.

Answer 1

The Correct Option is A

Step 1: Set up:
Odd positions: 1st, 3rd, 5th. Odd digits: 1, 3, 5. Even digits: 2, 4. Exactly two of the three odd positions hold odd digits, so the third odd position holds an even digit.

Step 2: Case 1 - the rightmost digit is even (2 or 4):
Then positions 1 and 3 must both hold odd digits: \(3 \times 2 = 6\) ways. The remaining one odd and one even digit fill positions 2 and 4 in 2 ways. For each even last digit there are \(6 \times 2 = 12\) numbers. Sum of last digits: \((2 + 4) \times 12 = 72\).

Step 3: Case 2 - the rightmost digit is odd (1, 3 or 5):
Exactly one of positions 1 and 3 holds an odd digit and the other holds an even digit. Choose which position holds the even digit (2 ways), which even digit (2 ways), and the odd digit for the other position from the two remaining odd digits (2 ways). The remaining odd and even digits fill positions 2 and 4 in 2 ways. For each odd last digit: \(2 \times 2 \times 2 \times 2 = 16\) numbers. Sum of last digits: \((1 + 3 + 5) \times 16 = 144\).

Step 4: Total:
\(72 + 144 = 216\).

Final Answer:
The correct option is (1): 216