Number of six-digit numbers such that any digit that appears in the number appears at least twice, where the digits of each number are from the set `{1, 2, 3, 4, 5},` is (Example 225252 is valid but 222133 is not valid)

Options

A. `1500`
B. `1850`
C. `1405`
D. `1205`

Explanation

To find the number of six-digit numbers such that any digit that appears in the number appears at least twice, where the digits are from the set {1, 2, 3, 4, 5}, we can break the problem down into different cases based on how many distinct digits are used in the number.

### Step-by-Step Solution:

**Case 1: All digits are the same.**
- Here, we can have numbers like 111111, 222222, etc.
- Since we have 5 digits (1 through 5), there are 5 ways to form such numbers.

**Total for Case 1:** 5 ways.

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**Case 2: Two digits appear twice and one digit appears twice.**
- We need to select 3 digits from the set {1, 2, 3, 4, 5}. This can be done in \( \binom{5}{3} \) ways.
- After selecting the digits, we can arrange them in the form AABBC, where A, B, and C are the selected digits.
- The arrangement of AABBC can be calculated using the formula for permutations of multiset:
\[ \frac{6!}{2! \times 2! \times 2!} = 90 \]
- Therefore, the total for this case is:
\[ 10 \times 90 = 900 \text{ ways.} \]

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**Case 3: Three digits appear twice.**
- We need to select 2 digits from the set {1, 2, 3, 4, 5}. This can be done in \( \binom{5}{2} \) ways.
- The arrangement of AABBCC can be calculated as:
\[ \frac{6!}{3! \times 3!} = 20 \]
- Therefore, the total for this case is:
\[ 10 \times 20 = 200 \text{ ways.} \]

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**Case 4: One digit appears four times and another digit appears twice.**
- We need to select 2 digits from the set {1, 2, 3, 4, 5}. This can be done in \( \binom{5}{2} \) ways.
- The arrangement of AAAABB can be calculated as:
\[ \frac{6!}{4! \times 2!} = 15 \]
- Therefore, the total for this case is:
\[ 10 \times 15 = 150 \text{ ways.} \]

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**Final Calculation:**
- Now, we sum up all the cases:
\[ 5 \text{ (Case 1)} + 900 \text{ (Case 2)} + 200 \text{ (Case 3)} + 150 \text{ (Case 4)} = 1255 \text{ ways.} \]

Thus, the total number of six-digit numbers such that any digit that appears in the number appears at least twice is **1255**.