To determine how many 4-digit numbers can be formed without repetitions and ending with the digit 3, we can follow a step-by-step approach. Here's the detailed solution:
1. Identify the structure of the 4-digit number:
We are dealing with numbers of the form ABC3, where A, B, and C are digits, and the digit 3 is fixed as the last digit.
2. Determine the choices for each digit:
- First digit (A): The number must be a 4-digit number, so the first digit A cannot be 0. It can be any digit from 1 to 9, but not 3 (since repetitions are not allowed). This gives us 9 possible choices for A.
- Second digit (B): The second digit can be any digit from 0 to 9, excluding the digit chosen for A and excluding 3. This gives us 8 possible choices for B.
- Third digit (C): The third digit can be any digit from 0 to 9, excluding the digits chosen for A and B, and excluding 3. This gives us 7 possible choices for C.
3. Calculate the total number of combinations:
Since each digit is chosen in sequence and the choices are independent, we multiply the number of choices for each digit:
[tex]\[
\text{Total numbers} = (\text{choices for A}) \times (\text{choices for B}) \times (\text{choices for C})
\][/tex]
[tex]\[
\text{Total numbers} = 9 \times 8 \times 7
\][/tex]
4. Perform the multiplication:
[tex]\[
9 \times 8 = 72
\][/tex]
[tex]\[
72 \times 7 = 504
\][/tex]
Therefore, the total number of 4-digit numbers that can be formed without repetition and ending with 3 is 504.
Hence, the correct answer is:
504
