There are 5 letters and 5 directed envelopes. The number of ways in which all the letters can be put in wrong engvelope, is

Options

A. 119
B. 44
C. 59
D. 40

Answer

To solve the problem of how many ways 5 letters can be placed in 5 directed envelopes such that none of the letters is placed in the correct envelope, we will 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 (denoted as !n) for 5 letters.

2. **Derangement Formula**:
The formula for the number of derangements of n items is given by:
\[ !n = n! \left( \sum_{i=0}^{n} \frac{(-1)^i}{i!} \right) \]
For our case, \( n = 5 \).

3. **Calculating \( 5! \)**:
First, we calculate \( 5! \):
\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]

4. **Calculating the Summation**:
Now we need to calculate the summation:
\[ \sum_{i=0}^{5} \frac{(-1)^i}{i!} \]
Breaking it down:
- 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 \( i = 4 \): \( \frac{(-1)^4}{4!} = \frac{1}{24} \)
- For \( i = 5 \): \( \frac{(-1)^5}{5!} = -\frac{1}{120} \)

Now, summing these values:
\[ 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} \]

Finding a common denominator (which is 120):
\[ = \frac{120}{120} - \frac{120}{120} + \frac{60}{120} - \frac{20}{120} + \frac{5}{120} - \frac{1}{120} \]
\[ = \frac{0 + 60 - 20 + 5 - 1}{120} = \frac{44}{120} \]

5. **Final Calculation**:
Now substituting back into the derangement formula:
\[ !5 = 5! \times \frac{44}{120} \]
\[ = 120 \times \frac{44}{120} = 44 \]

Thus, the number of ways in which all the letters can be put in the wrong envelopes is **44**.