Ravish writes letters to his five friends and addresses the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that at least two of them are in the wrong envelopes?

Options

A. A. 120
B. B. 119
C. C. 145
D. none of these

Answer

To solve the problem of how many ways Ravish can place letters in envelopes such that at least two letters are in the wrong envelopes, 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 need to find the total number of ways to arrange 5 letters in 5 envelopes such that at least 2 letters are in the wrong envelopes.

2. **Total Arrangements**: The total number of arrangements of 5 letters is given by \(5!\) (5 factorial).
\[ 5! = 120 \]

3. **Using the Complement Principle**: Instead of directly calculating the arrangements with at least 2 letters in the wrong envelopes, we can calculate the total arrangements and subtract the arrangements where fewer than 2 letters are in the wrong envelopes (i.e., 0 or 1 letter in the correct envelope).

4. **Calculating Derangements**:
- **0 letters in the correct envelope** (i.e., all letters are in wrong envelopes): This is the number of derangements of 5 letters, denoted as \( !5 \).
- The formula for derangements is:
\[ !n = n! \sum_{i=0}^{n} \frac{(-1)^i}{i!} \]
For \( n = 5 \):
\[ !5 = 5! \left( \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} \right) \]
\[ = 120 \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} \right) \]
\[ = 120 \left( 0 + 0.5 - 0.1667 + 0.0417 - 0.0083 \right) \]
\[ = 120 \left( 0.3667 \right) \approx 44 \]

5. **Calculating Arrangements with Exactly 1 Letter in the Correct Envelope**:
- Choose 1 letter to be in the correct envelope (5 ways), and derange the remaining 4 letters:
\[ 5 \times !4 \]
Where \( !4 \) is calculated as:
\[ !4 = 4! \left( \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} \right) \]
\[ = 24 \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right) \]
\[ = 24 \left( 0 + 0.5 - 0.1667 + 0.0417 \right) \approx 9 \]
Thus, the total arrangements with exactly 1 letter in the correct envelope is:
\[ 5 \times 9 = 45 \]

6. **Final Calculation**: Now, we can find the total arrangements with at least 2 letters in the wrong envelopes:
\[ \text{Total arrangements} - \text{Arrangements with 0 or 1 correct} = 120 - (44 + 45) = 120 - 89 = 31 \]

### Conclusion:
The number of ways Ravish can place the letters in the envelopes such that at least two of them are in the wrong envelopes is \( 31 \).