To determine how many different three-digit codes can be created from the digits 0, 1, 4, 8, and 9, given that no digit may be repeated in any code, we can follow these steps:
1. Identify the Total Number of Digits Available:
The given digits are 0, 1, 4, 8, and 9. Hence, there are 5 different digits in total.
2. Permutations of 3 Digits:
Since we need to create three-digit codes and each digit cannot be repeated, we use permutations.
The formula to calculate the number of permutations of \( n \) items taken \( r \) at a time is:
[tex]\[
P(n, r) = \frac{n!}{(n - r)!}
\][/tex]
Where:
- \( n \) is the total number of items,
- \( r \) is the number of items to choose, and
- \( ! \) denotes factorial.
3. Substitute the Values:
Here, \( n = 5 \) (since we have 5 digits) and \( r = 3 \) (since we need a three-digit code). So we calculate:
[tex]\[
P(5, 3) = \frac{5!}{(5 - 3)!} = \frac{5!}{2!}
\][/tex]
Calculate the factorials:
[tex]\[
5! = 5 \times 4 \times 3 \times 2 \times 1 = 120
\][/tex]
[tex]\[
2! = 2 \times 1 = 2
\][/tex]
Then, the number of permutations is:
[tex]\[
P(5, 3) = \frac{120}{2} = 60
\][/tex]
Therefore, the number of different three-digit codes that can be created from the digits 0, 1, 4, 8, and 9, without repeating any digit, is 60.
Thus, the correct answer is C. 60.
