To form a five-digit odd number using the digits 1, 2, 3, 4, and 5, we need to ensure that the number is odd and that all digits are used exactly once.

Step-by-step approach:

1. Odd number condition:
   For the number to be odd, the last digit must be an odd number. From the set of digits $\{1, 2, 3, 4, 5\}$, the odd digits are $1, 3, 5$.

2. Choices for the last digit:
   We can choose the last digit from the odd numbers. So, we have 3 options for the last digit: $1, 3,$ or $5$.

3. Choices for the remaining four digits:
   After choosing the last digit, we are left with 4 digits from the original set to arrange in the first four places. There are 4 remaining digits, and they can be arranged in any order.

   The number of ways to arrange these 4 remaining digits is $4!$ (the factorial of 4), which is:

   $4! = 4 \times 3 \times 2 \times 1 = 24$

4. Total number of ways:
   The total number of ways to form a five-digit odd number is the product of the number of choices for the last digit and the number of ways to arrange the first four digits:

   $3 \times 24 = 72$

Final answer:

There are 72 ways to form a five-digit odd number by arranging the digits 1, 2, 3, 4, and 5.

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Would you like more details or have any further questions?

Here are some related questions:

1. How would the number of ways change if we were allowed to repeat digits?
2. How can we find the number of ways to form even numbers using the digits 1, 2, 3, 4, and 5?
3. What if we were restricted to only using the digits 1, 3, and 5, how many odd numbers can we form?
4. How does the number of arrangements change if the digits are ordered in non-increasing order?
5. Can you calculate the number of odd numbers that can be formed if we only use the digits 1, 2, and 3?

Tip: When asked to find the number of ways to arrange digits with specific conditions (like forming an odd or even number), always start by identifying the restrictions on the first or last digit, then count the possibilities for the remaining places.