Find the number of numbers of six digits that can be formd with the digits 1,2,3,4 if all the digits are to asppear in the same number at least once.

Answer

To find the number of six-digit numbers that can be formed using the digits 1, 2, 3, and 4, with the condition that each digit must appear at least once, we can use the principle of inclusion-exclusion.

### Step-by-Step Solution:

1. **Calculate Total Combinations Without Restrictions**:
Each digit can be any of the four digits (1, 2, 3, 4). Therefore, the total number of six-digit combinations without any restrictions is:
\[ 4^6 \]

2. **Calculate Combinations Where At Least One Digit is Missing**:
We will use the principle of inclusion-exclusion to subtract cases where one or more digits are missing.

- **Case 1: One Digit is Missing**:
If one digit is missing, we can choose which digit to exclude in \( \binom{4}{1} \) ways. The remaining choices will be from the three digits left. Therefore, the number of combinations in this case is:
\[ \binom{4}{1} \cdot 3^6 \]

- **Case 2: Two Digits are Missing**:
If two digits are missing, we can choose which two digits to exclude in \( \binom{4}{2} \) ways. The remaining choices will be from the two digits left. Therefore, the number of combinations in this case is:
\[ \binom{4}{2} \cdot 2^6 \]

- **Case 3: Three Digits are Missing**:
If three digits are missing, we can choose which three digits to exclude in \( \binom{4}{3} \) ways. The remaining choice will be the one digit left. Therefore, the number of combinations in this case is:
\[ \binom{4}{3} \cdot 1^6 \]

3. **Combine the Cases Using Inclusion-Exclusion**:
The total number of valid six-digit combinations where at least one digit is missing is given by:
\[ \text{Total Missing} = \binom{4}{1} \cdot 3^6 - \binom{4}{2} \cdot 2^6 + \binom{4}{3} \cdot 1^6 \]

4. **Calculate the Required Result**:
Finally, we subtract the total combinations where at least one digit is missing from the total combinations:
\[ \text{Required Result} = 4^6 - \left( \binom{4}{1} \cdot 3^6 - \binom{4}{2} \cdot 2^6 + \binom{4}{3} \cdot 1^6 \right) \]

5. **Calculate Each Component**:
- \( 4^6 = 4096 \)
- \( \binom{4}{1} = 4 \) and \( 3^6 = 729 \) so \( 4 \cdot 729 = 2916 \)
- \( \binom{4}{2} = 6 \) and \( 2^6 = 64 \) so \( 6 \cdot 64 = 384 \)
- \( \binom{4}{3} = 4 \) and \( 1^6 = 1 \) so \( 4 \cdot 1 = 4 \)

6. **Final Calculation**:
\[ \text{Required Result} = 4096 - (2916 - 384 + 4) = 4096 - 2536 = 1560 \]

### Final Answer:
The number of six-digit numbers that can be formed with the digits 1, 2, 3, and 4, where each digit appears at least once, is **1560**.