The problem in the image asks for the number of 4-digit numbers between 3,000 and 10,000 that can be formed using odd digits, without repeating any digit.

Odd Digits Set

The odd digits available are: $\{1, 3, 5, 7, 9\}$.

Step-by-Step Approach

1. Range Requirement: The numbers must be between 3,000 and 10,000, so the first digit must be either 3, 5, 7, or 9 (since the number must be a 4-digit number starting from one of these digits).

2. First Digit: The first digit must be chosen from $\{3, 5, 7, 9\}$, so there are 4 possibilities for the first digit.

3. Second Digit: The second digit must be chosen from the remaining odd digits (since no digit is repeated), so after picking the first digit, 4 digits remain to choose from.

4. Third Digit: Similarly, after the first two digits are chosen, 3 odd digits remain for the third digit.

5. Fourth Digit: Finally, after the first three digits are chosen, 2 odd digits remain for the fourth digit.

Total Number of Combinations

Now, multiply the number of choices for each digit: $4 \times 4 \times 3 \times 2 = 96$

Thus, the total number of 4-digit numbers between 3,000 and 10,000 that can be formed from the set of odd digits without repetition is 96.

Would you like further details on any part of the solution? Here are 5 related questions for further practice:

1. How many 4-digit numbers can be formed using even digits without repetition?
2. What is the probability of randomly selecting a 4-digit odd number from this set?
3. How many 5-digit numbers can be formed using odd digits without repetition?
4. If repetition were allowed, how many numbers could be formed in this case?
5. What if the range was 1,000 to 9,999? How would the answer change?

Tip:

Always start by identifying any constraints (like the number range or digit repetition) before solving combinatorics problems.