To find the total number of 6-digit numbers in which only the digits 1, 3, 5, 7, and 9 appear, we need to consider a few steps.

Understanding the problem:

• We need 6-digit numbers.
• The digits allowed are: 1, 3, 5, 7, and 9.
• All five digits (1, 3, 5, 7, and 9) must appear, and they appear exactly once each.

Thus, the number must be made up of the digits 1, 3, 5, 7, 9, plus one of these digits must appear twice, because we have 6 digits total, and 5 distinct digits available.

Step 1: Choose the digit that appears twice

We have 5 choices for the digit that will appear twice (since each of the digits 1, 3, 5, 7, or 9 can be chosen).

Step 2: Arrange the digits

Once the repeated digit is chosen, we are left with arranging 6 digits, where 5 are distinct and one repeats twice. The number of ways to arrange these 6 digits is given by the formula for permutations of multiset: $\frac{6!}{2!} = \frac{720}{2} = 360$ (We divide by $2!$ because one digit is repeated twice.)

Step 3: Multiply the choices

Since we have 5 choices for which digit repeats, and for each of these choices there are 360 ways to arrange the digits, the total number of such 6-digit numbers is: $5 \times 360 = 1800$

Final Answer:

There are 1800 six-digit numbers in which only the digits 1, 3, 5, 7, and 9 appear, with each digit appearing at least once.

Would you like further clarification or details?

Here are five related questions to consider:

1. How would the result change if the repeated digit could be chosen from a different set of digits?
2. What if the number needed to be divisible by 5?
3. How would you calculate this if there were 7-digit numbers instead?
4. Can this problem be solved using combinations instead of permutations?
5. What happens if we allow the digits to repeat more than twice?

Tip: Always remember to account for repeating elements when using permutations.