There are 5 letters and 5 addressed envelopes.The number of ways in which the letters can be placed in the envelopes so that none of them goes into the right envelope is

Options

A. a. 44
B. b. 144
C. c. 48
D. d. 119

Answer

To solve the problem of finding the number of ways to place 5 letters into 5 addressed envelopes such that none of the letters goes into the correct envelope, 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. **Understanding the Problem**: We have 5 letters (L1, L2, L3, L4, L5) and 5 envelopes (E1, E2, E3, E4, E5). We need to find the number of ways to arrange the letters in the envelopes such that no letter goes into its corresponding envelope.

2. **Total Arrangements**: First, we calculate the total number of arrangements of the 5 letters in the 5 envelopes without any restrictions. This is given by the factorial of the number of letters (or envelopes):
\[ \text{Total arrangements} = 5! = 120 \]

3. **Derangements 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 n = 5, we can calculate it as follows:
\[ !5 = 5! \left( \frac{(-1)^0}{0!} + \frac{(-1)^1}{1!} + \frac{(-1)^2}{2!} + \frac{(-1)^3}{3!} + \frac{(-1)^4}{4!} + \frac{(-1)^5}{5!} \right) \]

4. **Calculating Each Term**:
- 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: \(\frac{(-1)^5}{5!} = -\frac{1}{120}\)

5. **Summing the Terms**:
\[ \text{Sum} = 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} \]
To combine these fractions, we find a common denominator (120):
\[ \text{Sum} = \frac{120}{120} - \frac{120}{120} + \frac{60}{120} - \frac{20}{120} + \frac{5}{120} - \frac{1}{120} = \frac{44}{120} = \frac{11}{30} \]

6. **Calculating Derangement**:
\[ !5 = 5! \cdot \frac{11}{30} = 120 \cdot \frac{11}{30} = 44 \]

7. **Conclusion**: The number of ways to arrange the letters such that none goes into the correct envelope is:
\[ \boxed{44} \]