To determine how many different three-digit codes can be created using the digits 0, 1, 4, 8, and 9 without repetition, we need to calculate the number of permutations of 3 digits chosen from a set of 5 digits.
Here's the step-by-step solution:
1. Identify the total number of digits available:
- The given digits are 0, 1, 4, 8, and 9.
- Therefore, we have 5 digits in total.
2. Select the number of digits to form a code:
- We are forming a three-digit code.
- So, we need to pick 3 digits out of the 5 available digits.
3. Calculate the number of permutations of 3 digits out of 5:
- The formula to calculate permutations (when order matters) is given by the permutation formula \( nPr \), where \( n \) is the total number of items, and \( r \) is the number of items to pick.
- \( nPr = \frac{n!}{(n-r)!} \)
- In this scenario, \( n = 5 \) and \( r = 3 \).
- Therefore, we compute \( 5P3 \) using the formula:
[tex]\[
5P3 = \frac{5!}{(5-3)!} = \frac{5!}{2!}
\][/tex]
- Calculate the factorials:
[tex]\[
5! = 5 \times 4 \times 3 \times 2 \times 1 = 120
\][/tex]
[tex]\[
2! = 2 \times 1 = 2
\][/tex]
- Divide the factorials:
[tex]\[
5P3 = \frac{120}{2} = 60
\][/tex]
4. Conclude with the correct number of permutations:
- The number of different three-digit codes that can be created from the digits 0, 1, 4, 8, and 9 without repetition is 60.
Thus, the answer is C. 60.
