The number of arrangments of all digits of `12345` such that at least `3` digits will not come in its position is

Options

A. `89`
B. `109`
C. `78`
D. `57`

Explanation

To solve the problem of finding the number of arrangements of the digits of `12345` such that at least `3` digits do not occupy their original positions, we can follow these steps:

### Step 1: Calculate the total arrangements
The total number of arrangements of the digits `1, 2, 3, 4, 5` is given by the factorial of the number of digits:
\[ \text{Total arrangements} = 5! = 120 \]

**Hint:** Remember that the factorial of a number \( n \) (denoted as \( n! \)) is the product of all positive integers up to \( n \).

### Step 2: Count the arrangements with all digits in their original positions
When all digits are in their original positions (i.e., `1` is in position 1, `2` is in position 2, etc.), there is only one arrangement:
\[ \text{Arrangements with all digits in position} = 1 \]

**Hint:** This is a straightforward case where no digits are displaced.

### Step 3: Count the arrangements with exactly 2 digits in their original positions
To find the arrangements where exactly 2 digits are in their original positions, we can choose 2 digits to remain in their positions and the remaining 3 digits must not be in their original positions.

1. Choose 2 positions from 5 to remain fixed. The number of ways to choose 2 positions from 5 is given by:
\[ \binom{5}{2} = 10 \]

2. The remaining 3 digits must be arranged such that none of them is in their original position. This is known as a derangement. The number of derangements \( !n \) for \( n = 3 \) can be calculated using the formula:
\[ !n = n! \sum_{i=0}^{n} \frac{(-1)^i}{i!} \]
For \( n = 3 \):
\[ !3 = 3! \left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!}\right) = 6 \left(1 - 1 + 0.5 - \frac{1}{6}\right) = 6 \left(0.5 - \frac{1}{6}\right) = 6 \left(\frac{3}{6} - \frac{1}{6}\right) = 6 \times \frac{2}{6} = 2 \]

3. Therefore, the total arrangements where exactly 2 digits are in their original positions is:
\[ \text{Total for exactly 2 fixed} = \binom{5}{2} \times !3 = 10 \times 2 = 20 \]

**Hint:** Use the concept of derangements to ensure that the remaining digits are not in their original positions.

### Step 4: Count the arrangements with exactly 1 digit in its original position
If exactly 1 digit is in its original position, we can choose 1 digit to remain fixed and the other 4 digits must not be in their original positions.

1. Choose 1 position from 5 to remain fixed. The number of ways to choose 1 position from 5 is:
\[ \binom{5}{1} = 5 \]

2. The remaining 4 digits must be deranged. The number of derangements \( !4 \) can be calculated similarly:
\[ !4 = 4! \left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}\right) = 24 \left(1 - 1 + 0.5 - \frac{1}{6} + \frac{1}{24}\right) = 24 \left(0.5 - \frac{1}{6} + \frac{1}{24}\right) \]
Simplifying:
\[ = 24 \left(\frac{12}{24} - \frac{4}{24} + \frac{1}{24}\right) = 24 \left(\frac{9}{24}\right) = 9 \]

3. Therefore, the total arrangements where exactly 1 digit is in its original position is:
\[ \text{Total for exactly 1 fixed} = \binom{5}{1} \times !4 = 5 \times 9 = 45 \]

**Hint:** Again, use the derangement concept for the remaining digits.

### Step 5: Calculate the total arrangements with at least 3 digits not in their original positions
To find the number of arrangements where at least 3 digits are not in their original positions, we can use the principle of inclusion-exclusion:
\[ \text{At least 3 not in position} = \text{Total arrangements} - (\text{Arrangements with all in position} + \text{Arrangements with exactly 2 in position} + \text{Arrangements with exactly 1 in position}) \]
Substituting the values:
\[ \text{At least 3 not in position} = 120 - (1 + 20 + 45) = 120 - 66 = 54 \]

### Final Answer
The number of arrangements of the digits `12345` such that at least `3` digits do not come in their original positions is:
\[ \boxed{54} \]