How many 3 digit numbers can be formed from the digits 1, 2 , 3, 4 and 5 assuming that repetition of the digits is not allowed .

Options

A. `3|(N-1)`
B. `n|(N_1)`
C. `(n+1(|(N-1)`
D. `3n(n+1)|(N-1)`

Explanation

To find how many 3-digit numbers can be formed from the digits 1, 2, 3, 4, and 5 without allowing repetition of the digits, we can follow these steps:

### Step 1: Identify the available digits
The available digits are 1, 2, 3, 4, and 5.

### Step 2: Determine the number of choices for each digit place
- **Hundreds Place**: For the first digit (hundreds place), we can choose any of the 5 digits (1, 2, 3, 4, or 5). So, there are **5 choices**.

- **Tens Place**: For the second digit (tens place), we cannot use the digit that was already used in the hundreds place. Therefore, we have 4 remaining choices.

- **Units Place**: For the third digit (units place), we cannot use the digits that were already used in the hundreds and tens places. Thus, we have 3 remaining choices.

### Step 3: Calculate the total number of combinations
Using the fundamental principle of counting, we multiply the number of choices for each place:

\[ \text{Total combinations} = (\text{Choices for hundreds place}) \times (\text{Choices for tens place}) \times (\text{Choices for units place}) \]

This can be calculated as:

\[ \text{Total combinations} = 5 \times 4 \times 3 \]

Calculating this gives:

\[ 5 \times 4 = 20 \]
\[ 20 \times 3 = 60 \]

### Final Answer
Thus, the total number of 3-digit numbers that can be formed is **60**.

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