Final answer:
There are 3240 odd integers between 1,000 and 9,999 that have distinct digits, calculated by multiplying the number of options for each digit: 9 (first digit) x 9 (second digit) x 8 (third digit) x 5 (odd, distinct fourth digit).
Step-by-step explanation:
The question revolves around finding the number of odd integers between 1,000 and 9,999 that have distinct digits. To ensure an integer is odd, its last digit must be one of the five odd numbers: 1, 3, 5, 7, or 9. The first digit can be any digit from 1 to 9, and since we require all digits to be distinct, the remaining two must be chosen from the digits that have not been used so far, taking into account that the last digit is already occupied by an odd number.
Let's break it down:
- First digit: 9 options (1-9)
- Second digit: 9 options (0-9 minus the first digit)
- Third digit: 8 options (remaining digits)
- Fourth digit (odd): 5 options (1, 3, 5, 7, 9)
Applying the fundamental principle of counting, we get the total count of such numbers as 9 x 9 x 8 x 5. This results in:
9 x 9 x 8 x 5 = 81 x 40 = 3240
Therefore, there are 3240 odd integers between 1,000 and 9,999 that have distinct digits, which is closest to choice B) 3024 but it is not one of the options provided. However, if the options are incorrect, the student should be aware that the exact answer determined through combinatorial analysis is 3240.