(i) To keep the odd digits together, treat them as a single unit. The odd digits are 1, 3, 5, 7, 9. Arrange these 5 digits in a block: 5! ways. The remaining two digits (4 and 8) can be arranged with this block in 3! ways. Thus, the total number of arrangements is:
\(5! \times 3! = 120 \times 6 = 720\)
(ii) For even numbers between 3000 and 5000, the first digit must be 3 or 4, and the last digit must be 4 or 8. Consider two cases:
- Case 1: Start with 3 and end with 4. The middle two digits can be any of the remaining 5 digits. Thus, there are:
\(5 \times 4 = 20\)
- Case 2: Start with 3 and end with 8. The middle two digits can be any of the remaining 5 digits. Thus, there are:
\(5 \times 4 = 20\)
- Case 3: Start with 4 and end with 8. The middle two digits can be any of the remaining 5 digits. Thus, there are:
\(5 \times 4 = 20\)
Adding these cases gives:
\(20 + 20 + 20 = 60\)
(iii) For multiples of 5 less than 1000, the number must end in 5. Consider numbers with 1, 2, or 3 digits:
- 1-digit: Only 5 is possible.
- 2-digit: The first digit can be any of 1, 3, 4, 7, 8, 9 (6 choices).
- 3-digit: The first two digits can be any combination of the remaining 6 digits (6 choices for the first, 5 for the second).
Thus, the total is:
\(1 + 6 + 6 \times 5 = 1 + 6 + 30 = 57\)
