How many 4-digit numbers are there with no digit repeated?

Options

A. `=9*9*8*7`
B. `=10*9*8*7`
C. `=9*8*7`
D. None of these

Explanation

To find out how many 4-digit numbers can be formed with no repeated digits, we can follow these steps:

### Step 1: Determine the choices for the first digit
The first digit of a 4-digit number cannot be zero (as it would then be a 3-digit number). Therefore, the possible choices for the first digit are 1 through 9. This gives us a total of **9 options** for the first digit.

### Step 2: Determine the choices for the second digit
Once the first digit is chosen, we have already used one digit. Since we can now use zero as a valid digit for the second position, we have 9 remaining digits to choose from (0 and the 8 digits that are left). Thus, we have **9 options** for the second digit.

### Step 3: Determine the choices for the third digit
For the third digit, we have already used two digits. Therefore, we have 8 remaining digits to choose from. This gives us **8 options** for the third digit.

### Step 4: Determine the choices for the fourth digit
For the fourth digit, we have already used three digits. Hence, there are 7 remaining digits available to choose from. This gives us **7 options** for the fourth digit.

### Step 5: Calculate the total number of combinations
To find the total number of 4-digit numbers with no repeated digits, we multiply the number of choices for each digit together:

\[ \text{Total combinations} = \text{(Choices for 1st digit)} \times \text{(Choices for 2nd digit)} \times \text{(Choices for 3rd digit)} \times \text{(Choices for 4th digit)} \]

Substituting the values we found:

\[ \text{Total combinations} = 9 \times 9 \times 8 \times 7 \]

Calculating this step-by-step:

1. \(9 \times 9 = 81\)
2. \(81 \times 8 = 648\)
3. \(648 \times 7 = 4536\)

Thus, the total number of 4-digit numbers with no repeated digits is **4536**.

### Final Answer
The total number of 4-digit numbers with no digit repeated is **4536**.

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