To find the number of all 6-digit natural numbers such that the sum of their digits is 10 and each of the digits 0, 1, 2, and 3 occurs at least once, we can break down the problem into manageable steps.
### Step 1: Understand the requirements
We need to find 6-digit numbers where:
- The sum of the digits equals 10.
- Each of the digits 0, 1, 2, and 3 appears at least once.
### Step 2: Calculate the minimum sum of required digits
The minimum sum of the digits 0, 1, 2, and 3 is:
\[ 0 + 1 + 2 + 3 = 6 \]
### Step 3: Determine the remaining sum needed
Since the total sum of the digits must be 10, the remaining sum needed after using 0, 1, 2, and 3 is:
\[ 10 - 6 = 4 \]
### Step 4: Identify possible combinations for the remaining digits
We need to find combinations of digits that add up to 4. The possible digits we can use are 0, 1, 2, and 3. The combinations that satisfy this requirement are:
1. **Case 1:** 0, 1, 2, 3, 0, 4
2. **Case 2:** 0, 1, 2, 3, 3, 1
3. **Case 3:** 0, 1, 2, 3, 2, 2
### Step 5: Calculate arrangements for each case
#### Case 1: Digits are 0, 1, 2, 3, 0, 4
- Total arrangements = \( \frac{5!}{2!} \) (since there are two 0's)
- The first digit cannot be 0. We can choose any of 1, 2, 3, or 4 as the first digit (4 choices).
- Thus, the total arrangements for this case:
\[ 4 \times \frac{5!}{2!} = 4 \times 60 = 240 \]
**Hint:** Remember that the first digit cannot be 0, so you have to multiply the arrangements by the number of valid choices for the first digit.
#### Case 2: Digits are 0, 1, 2, 3, 3, 1
- Total arrangements = \( \frac{6!}{2! \times 2!} \) (since 1 and 3 both occur twice)
- The first digit cannot be 0. We need to subtract the arrangements where 0 is the first digit.
- Arrangements with 0 first = \( \frac{5!}{2! \times 1!} = 60 \)
- Valid arrangements = \( \frac{6!}{2! \times 2!} - \frac{5!}{2! \times 1!} = 180 - 60 = 120 \)
**Hint:** Always check for cases where the first digit is invalid and subtract those arrangements.
#### Case 3: Digits are 0, 1, 2, 3, 2, 2
- Total arrangements = \( \frac{6!}{3!} \) (since 2 occurs three times)
- The first digit cannot be 0. We need to subtract the arrangements where 0 is the first digit.
- Arrangements with 0 first = \( \frac{5!}{3!} = 20 \)
- Valid arrangements = \( \frac{6!}{3!} - \frac{5!}{3!} = 120 - 20 = 100 \)
**Hint:** Similar to the previous case, ensure to account for the invalid arrangements where 0 is the leading digit.
### Step 6: Sum the valid arrangements from all cases
Total valid arrangements:
\[ 240 + 120 + 100 = 460 \]
### Final Answer
The total number of 6-digit natural numbers such that the sum of their digits is 10 and each of the digits 0, 1, 2, and 3 occurs at least once is **460**.
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