Answer :
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.
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.
