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

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 will use the concept of derangements. Here's a step-by-step solution:

### Step 1: Understand the Problem
We need to find arrangements where at least 3 digits are not in their original positions. This means we can have cases where exactly 3, exactly 4, or all 5 digits are not in their original positions.

### Step 2: Define Derangements
A derangement is a permutation of a set where none of the elements appear in their original position. The formula for the number of derangements \( !n \) of \( n \) items is given by:
\[ !n = n! \sum_{i=0}^{n} \frac{(-1)^i}{i!} \]

### Step 3: Calculate Derangements for Each Case

#### Case 1: Exactly 3 Digits Not in Position
1. Choose 3 digits from 5: \( \binom{5}{3} = 10 \)
2. Derangement of 3 digits: \( !3 = 3! \left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!}\right) = 6 \left(1 - 1 + 0.5 - \frac{1}{6}\right) = 6 \left(0.5 - \frac{1}{6}\right) = 6 \left(\frac{3}{6} - \frac{1}{6}\right) = 6 \cdot \frac{2}{6} = 2 \)
3. Total arrangements for this case: \( 10 \times 2 = 20 \)

#### Case 2: Exactly 4 Digits Not in Position
1. Choose 4 digits from 5: \( \binom{5}{4} = 5 \)
2. Derangement of 4 digits: \( !4 = 4! \left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}\right) = 24 \left(1 - 1 + 0.5 - \frac{1}{6} + \frac{1}{24}\right) = 24 \left(0.5 - \frac{1}{6} + \frac{1}{24}\right) \)
- Calculate: \( 0.5 = \frac{12}{24}, \frac{1}{6} = \frac{4}{24}, \frac{1}{24} = \frac{1}{24} \)
- So, \( !4 = 24 \left(\frac{12 - 4 + 1}{24}\right) = 24 \times \frac{9}{24} = 9 \)
3. Total arrangements for this case: \( 5 \times 9 = 45 \)

#### Case 3: All 5 Digits Not in Position
1. Derangement of 5 digits: \( !5 = 5! \left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!}\right) = 120 \left(1 - 1 + 0.5 - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}\right) \)
- Calculate: \( 0.5 - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} = \frac{60 - 10 + 2 - 0.5}{120} = \frac{51.5}{120} = \frac{103}{240} \)
- So, \( !5 = 120 \times \frac{44}{120} = 44 \)

### Step 4: Total Arrangements
Now, we add the total arrangements from all cases:
\[ \text{Total} = 20 + 45 + 44 = 109 \]

### Final Answer
The number of arrangements of the digits of `12345` such that at least `3` digits do not come in their original position is **109**.