All nursing students of Kolej Manjakini are assigned identity codes. The code consists of the first two letters of a student's name followed by four digits. How many different identity codes are there that start with letter \( M \) and end with an even number if repetitions are not allowed?

To determine the number of different identity codes that start with the letter \( M \) and end with an even number, we can break down the problem into manageable parts. 1. **Structure of the Code**: The identity code consists of: - The first two letters of the student's name (the first letter is fixed as \( M \)). - Four digits. 2. **Choosing the Second Letter**: Since the first letter is \( M \), we need to choose the second letter. Assuming the second letter can be any letter from the English alphabet (A-Z), there are 25 possible choices (since \( M \) is already used). 3. **Choosing the Digits**: The last digit must be an even number. The even digits available are \( 0, 2, 4, 6, \) and \( 8 \), giving us 5 options for the last digit. 4. **Choosing the Remaining Digits**: After selecting the second letter and the last digit, we need to choose the remaining two digits from the digits \( 0-9 \) (which gives us 10 options initially). However, since we cannot repeat digits, the choices will be reduced based on what has already been selected. - **Total Digits**: The digits available are \( 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 \). - **Last Digit**: After choosing the last digit (one of the 5 even digits), we have 9 digits left to choose from for the first of the two remaining digits. - **First Remaining Digit**: After selecting the first remaining digit, we will have 8 digits left for the second remaining digit. 5. **Calculating the Total Combinations**: - **Choices for the Second Letter**: 25 choices. - **Choices for the Last Digit**: 5 choices. - **Choices for the First Remaining Digit**: 9 choices (after choosing the last digit). - **Choices for the Second Remaining Digit**: 8 choices (after choosing the first remaining digit). Putting it all together, the total number of different identity codes can be calculated as follows: \[ \text{Total Codes} = (\text{Choices for Second Letter}) \times (\text{Choices for Last Digit}) \times (\text{Choices for First Remaining Digit}) \times (\text{Choices for Second Remaining Digit}) \] Substituting the values: \[ \text{Total Codes} = 25 \times 5 \times 9 \times 8 \] Now, we calculate this step-by-step: 1. \( 25 \times 5 = 125 \) 2. \( 125 \times 9 = 1125 \) 3. \( 1125 \times 8 = 9000 \) Thus, the total number of different identity codes that start with the letter \( M \) and end with an even number, with no repetitions allowed, is \[ \boxed{9000} \]