There are four balls of different colors and four boxes of colors same as those of the balls. Find the number of ways in which the balls, one in each box, could be placed in such a way that a ball does not go to box of its own color.

Options

A. 9
B. 11
C. 13
D. 7

Answer

To solve the problem of placing four balls of different colors into four boxes of the same colors such that no ball goes into the box of its own color, we can use the concept of derangements.

### Step-by-Step Solution:

1. **Understanding Derangements**: A derangement is a permutation of elements such that none of the elements appear in their original position. In this case, we want to find the number of derangements of 4 balls.

2. **Formula for Derangements**: The number of derangements \( !n \) 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 \( 4! \):
\[ 4! = 24 \]

4. **Calculate the Summation**:
We need to compute the summation:
\[ \sum_{i=0}^{4} \frac{(-1)^i}{i!} \]
- 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 simplify this, we can convert all terms to have a common denominator of 24:
\[ 1 = \frac{24}{24}, \quad -1 = -\frac{24}{24}, \quad \frac{1}{2} = \frac{12}{24}, \quad -\frac{1}{6} = -\frac{4}{24}, \quad \frac{1}{24} = \frac{1}{24} \]

Thus, the summation becomes:
\[ \frac{24 - 24 + 12 - 4 + 1}{24} = \frac{9}{24} = \frac{3}{8} \]

5. **Final Calculation of Derangements**:
Now, substituting back into the derangement formula:
\[ !4 = 4! \cdot \frac{3}{8} = 24 \cdot \frac{3}{8} = 9 \]

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