Introduction
To determine the number of five-digit numbers that can be formed using the digits 3, 4, and 5, with each digit used at least once, we can use the principle of inclusion-exclusion.
Step 1: Total Combinations Without Restrictions
- The total number of five-digit combinations using the digits 3, 4, and 5 (with repetitions allowed) is calculated as:
- 3 choices per digit position (3, 4, or 5)
- Total combinations: 3^5 = 243
Step 2: Exclude Combinations Missing One Digit
- We need to exclude cases where at least one digit is missing.
- If we omit one digit, we can use the remaining two digits to form five-digit numbers. The number of combinations for each scenario (missing either 3, 4, or 5) is:
- 2^5 = 32 combinations (using only two digits)
- Since there are three digits, we multiply by 3:
- Excluded combinations: 3 * 32 = 96
Step 3: Include Combinations Missing Two Digits
- We need to add back the cases where only one digit is used (since we've excluded them twice).
- There are three scenarios (using just 3, just 4, or just 5), and each provides 1 combination:
- Included combinations: 3 * 1 = 3
Final Calculation
- Using the inclusion-exclusion principle, the total valid combinations are:
- Total = Total combinations - Excluded combinations + Included combinations
- Total = 243 - 96 + 3 = 150
Conclusion
Thus, the total number of five-digit numbers that can be formed using the digits 3, 4, and 5, with each digit appearing at least once, is 150.

3*3*3*3*3