How many natural numbers not exceeding 4321 can be formed with the digits 1, 2, 3 and 4 if the digits can repeat ?

Answer

To find how many natural numbers not exceeding 4321 can be formed with the digits 1, 2, 3, and 4 (with repetition allowed), we can break the problem into several cases based on the number of digits in the numbers we can form.

### Step-by-Step Solution:

1. **Count 1-digit numbers:**
- The possible 1-digit natural numbers are: 1, 2, 3, 4.
- Total = 4.

2. **Count 2-digit numbers:**
- The first digit can be any of 1, 2, 3, or 4 (4 choices).
- The second digit can also be any of 1, 2, 3, or 4 (4 choices).
- Total = 4 × 4 = 16.

3. **Count 3-digit numbers:**
- The first digit can be any of 1, 2, 3, or 4 (4 choices).
- The second digit can be any of 1, 2, 3, or 4 (4 choices).
- The third digit can be any of 1, 2, 3, or 4 (4 choices).
- Total = 4 × 4 × 4 = 64.

4. **Count 4-digit numbers less than 4321:**
- We will consider cases based on the first digit:

**Case 1:** First digit = 1, 2, or 3 (i.e., less than 4)
- If the first digit is 1, 2, or 3, the remaining three digits can be anything (1, 2, 3, or 4).
- Total for this case = 3 (choices for the first digit) × 4 × 4 × 4 = 3 × 64 = 192.

**Case 2:** First digit = 4
- Now we need to check the second digit:
- If the second digit is 1 or 2 (i.e., less than 3):
- The remaining two digits can be anything (1, 2, 3, or 4).
- Total for this sub-case = 2 (choices for the second digit) × 4 × 4 = 2 × 16 = 32.
- If the second digit = 3:
- Now we check the third digit:
- If the third digit is 1, 2 (i.e., less than 2):
- The last digit can be anything (1, 2, 3, or 4).
- Total for this sub-case = 2 (choices for the third digit) × 4 = 2 × 4 = 8.
- If the third digit = 2:
- The last digit must be 1 (to not exceed 4321).
- Total for this sub-case = 1.

- Total for Case 2 (First digit = 4) = 32 + 8 + 1 = 41.

5. **Combine all cases:**
- Total natural numbers = 1-digit numbers + 2-digit numbers + 3-digit numbers + 4-digit numbers.
- Total = 4 + 16 + 64 + (192 + 41) = 4 + 16 + 64 + 233 = 317.

### Final Answer:
The total number of natural numbers not exceeding 4321 that can be formed with the digits 1, 2, 3, and 4 (with repetition allowed) is **317**.