There are 4 balls of different colours and 4 boxes of same colours as those of the balls. The number of ways in which the balls, one in each box, could be placed such that a ball does not go to box of its own colour is

Options

A. a. 3
B. b. 6
C. c. 9
D. d. none of these

Answer

To solve the problem of placing 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. A derangement is a permutation of elements such that none of the elements appear in their original position.

### Step-by-Step Solution:

1. **Identify the Problem**: We have 4 balls (let's say colors: Red, Blue, Green, Black) and 4 boxes (also colored Red, Blue, Green, Black). We need to find the number of ways to place the balls in the boxes such that no ball is placed in the box of the same color.

2. **Use the Derangement Formula**: The number of derangements (denoted as !n) of n items can be calculated using the formula:
\[ !n = n! \sum_{i=0}^{n} \frac{(-1)^i}{i!} \]
For our case, n = 4.

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

4. **Calculate the Summation**:
- 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 \(i = 4\): \(\frac{(-1)^4}{4!} = \frac{1}{24}\)

Now, summing these values:
\[ 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \]
To add these fractions, we find a common denominator (which is 24):
- \(1 = \frac{24}{24}\)
- \(-1 = -\frac{24}{24}\)
- \(\frac{1}{2} = \frac{12}{24}\)
- \(-\frac{1}{6} = -\frac{4}{24}\)
- \(\frac{1}{24} = \frac{1}{24}\)

Now summing:
\[ \frac{24 - 24 + 12 - 4 + 1}{24} = \frac{9}{24} = \frac{3}{8} \]

5. **Calculate the Derangement**:
Now, substituting back into the derangement formula:
\[ !4 = 4! \times \frac{3}{8} = 24 \times \frac{3}{8} = 9 \]

6. **Final Answer**: Therefore, the number of ways to arrange the balls such that no ball is in the box of its own color is **9**.