How many numbers can be formed from the digits 1, 2, 3, 4 when repetition is not allowed?

Explanation

To solve the problem of how many numbers can be formed from the digits 1, 2, 3, and 4 without repetition, we will consider the formation of 1-digit, 2-digit, 3-digit, and 4-digit numbers.

### Step-by-Step Solution:

1. **Count 1-Digit Numbers:**
- We can use any of the 4 digits (1, 2, 3, or 4).
- Therefore, the number of 1-digit numbers = 4.
\[ \text{1-digit numbers} = 4P1 = 4 \]

2. **Count 2-Digit Numbers:**
- For a 2-digit number, we need to choose 2 digits from the 4 available digits.
- The first digit can be any of the 4 digits, and the second digit can be any of the remaining 3 digits.
- Therefore, the number of 2-digit numbers = 4 × 3 = 12.
\[ \text{2-digit numbers} = 4P2 = 4 \times 3 = 12 \]

3. **Count 3-Digit Numbers:**
- For a 3-digit number, we need to choose 3 digits from the 4 available digits.
- The first digit can be any of the 4 digits, the second digit can be any of the remaining 3 digits, and the third digit can be any of the remaining 2 digits.
- Therefore, the number of 3-digit numbers = 4 × 3 × 2 = 24.
\[ \text{3-digit numbers} = 4P3 = 4 \times 3 \times 2 = 24 \]

4. **Count 4-Digit Numbers:**
- For a 4-digit number, we will use all 4 digits.
- The number of ways to arrange 4 digits is 4 × 3 × 2 × 1 = 24.
\[ \text{4-digit numbers} = 4P4 = 4 \times 3 \times 2 \times 1 = 24 \]

5. **Total Numbers:**
- Now, we add all the possibilities together:
\[ \text{Total Numbers} = 4 + 12 + 24 + 24 = 64 \]

Thus, the total number of numbers that can be formed from the digits 1, 2, 3, and 4 without repetition is **64**.