Let N be the number of 6-digit numbers such that the digits of each number are all form the set {1, 2, 3, 4, 5} and any digit that appears in the number appears atleast twice. Then the number of all 6 digits is

Answer

To solve the problem of finding the number of 6-digit numbers formed from the set {1, 2, 3, 4, 5} where any digit that appears must appear at least twice, we can break the problem down into cases based on how many different digits are used and how many times each digit appears.

### Step-by-Step Solution:

**Step 1: Define the cases based on the repetition of digits.**

1. **Case 1: All digits are the same (6 alike).**
- Here, we can choose any one digit from the set {1, 2, 3, 4, 5}.
- The number of ways to choose 1 digit from 5 is given by \( \binom{5}{1} = 5 \).
- The arrangement of this digit is only 1 way since all digits are the same.
- **Total for Case 1:** \( 5 \)

**Step 2: Case 2: Four digits are the same, and two digits are the same (4 alike, 2 alike).**
- Choose 2 different digits from the set. The number of ways to choose 2 digits from 5 is \( \binom{5}{2} = 10 \).
- The arrangement of these digits (4 of one kind and 2 of another) can be calculated using the formula for permutations of multiset:
\[ \frac{6!}{4! \cdot 2!} = \frac{720}{24 \cdot 2} = 15 \]
- **Total for Case 2:** \( 10 \times 15 = 150 \)

**Step 3: Case 3: Three digits are the same, and three digits are the same (3 alike, 3 alike).**
- Choose 2 different digits from the set. The number of ways to choose 2 digits from 5 is \( \binom{5}{2} = 10 \).
- The arrangement of these digits (3 of one kind and 3 of another) can be calculated as:
\[ \frac{6!}{3! \cdot 3!} = \frac{720}{6 \cdot 6} = 20 \]
- **Total for Case 3:** \( 10 \times 20 = 200 \)

**Step 4: Case 4: Two digits are the same, two digits are the same, and two digits are the same (2 alike, 2 alike, 2 alike).**
- Choose 3 different digits from the set. The number of ways to choose 3 digits from 5 is \( \binom{5}{3} = 10 \).
- The arrangement of these digits (2 of one kind, 2 of another, and 2 of another) can be calculated as:
\[ \frac{6!}{2! \cdot 2! \cdot 2!} = \frac{720}{2 \cdot 2 \cdot 2} = 90 \]
- **Total for Case 4:** \( 10 \times 90 = 900 \)

**Step 5: Calculate the total number of valid 6-digit numbers.**
- Now, we sum the totals from all cases:
\[ \text{Total} = 5 + 150 + 200 + 900 = 1255 \]

### Final Answer:
The total number of 6-digit numbers that can be formed is **1255**.