We have 4 balls of different colours and 4 boxes with colours the same as those or the bals. The number of ways in which the balls can be arranged in the boxes so that no ball goes into a box of its own colour are_______________________.

Answer

To solve the problem of arranging 4 balls of different colors into 4 boxes of the same colors such that no ball goes into the box of its own color, we can use the concept of derangements. Here’s a step-by-step solution:

### Step-by-Step Solution:

1. **Understanding Derangements**:
A derangement is a permutation of a set where none of the elements appear in their original position. In this case, we want to find the number of ways to arrange the balls such that no ball is placed in the box of the same color.

2. **Using the Derangement Formula**:
The formula for the number of derangements (denoted as !n) of n items is given by:
\[ !n = n! \left( \sum_{i=0}^{n} \frac{(-1)^i}{i!} \right) \]
For our case, \( n = 4 \).

3. **Calculating \( 4! \)**:
First, we calculate \( 4! \):
\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \]

4. **Calculating the Sum**:
Now we need to compute the sum:
\[ \sum_{i=0}^{4} \frac{(-1)^i}{i!} \]
This can be calculated as follows:
- For \( i = 0 \): \( \frac{(-1)^0}{0!} = 1 \)
- For \( i = 1 \): \( \frac{(-1)^1}{1!} = -1 \)
- For \( i = 2 \): \( \frac{(-1)^2}{2!} = \frac{1}{2} \)
- For \( i = 3 \): \( \frac{(-1)^3}{3!} = -\frac{1}{6} \)
- For \( i = 4 \): \( \frac{(-1)^4}{4!} = \frac{1}{24} \)

Adding these together:
\[ 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \]
To simplify, we can find a common denominator (which is 24):
\[ = \frac{24}{24} - \frac{24}{24} + \frac{12}{24} - \frac{4}{24} + \frac{1}{24} = \frac{12 - 4 + 1}{24} = \frac{9}{24} = \frac{3}{8} \]

5. **Final Calculation**:
Now substituting back into the derangement formula:
\[ !4 = 4! \left( \frac{3}{8} \right) = 24 \times \frac{3}{8} = \frac{72}{8} = 9 \]

Thus, the number of ways to arrange the balls in the boxes such that no ball goes into the box of its own color is **9**.