The number of ways in which four different letters can be put in their four addressed envelopes that

Options

A. at least two of them are in the wrong envelopes, are 23
B. at least two of them are in the wrong envelopes, are 25
C. exactly two of them are in the wrong envelopes, are 6
D. exactly two of them are in the wrong envelopes, are 7

Answer

To solve the problem of finding the number of ways in which four different letters can be put in their four addressed envelopes such that at least two of them are in the wrong envelope, we can follow these steps:

### Step 1: Calculate the total arrangements of letters
First, we need to find the total number of ways to arrange 4 different letters in 4 envelopes. This can be calculated using the factorial of the number of letters.

\[ \text{Total arrangements} = 4! = 24 \]

### Step 2: Calculate arrangements with all letters in the correct envelope
Next, we need to consider the scenario where all letters are in the correct envelopes. There is only 1 way for this to happen, which is when each letter is placed in its corresponding envelope.

\[ \text{All correct arrangements} = 1 \]

### Step 3: Calculate arrangements with exactly 1 letter in the correct envelope
Now, we need to find the arrangements where exactly 1 letter is in the correct envelope. If one letter is in the correct envelope, the remaining 3 letters must be in the wrong envelopes.

To find this, we can choose 1 letter to be in the correct envelope (which can be done in 4 ways), and then we need to find the derangements (arrangements where no letter is in its correct envelope) for the remaining 3 letters. The number of derangements \(D_n\) can be calculated using the formula:

\[ D_n = n! \sum_{i=0}^{n} \frac{(-1)^i}{i!} \]

For \(n = 3\):

\[ D_3 = 3! \left( \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} \right) = 6 \left( 1 - 1 + 0.5 - \frac{1}{6} \right) = 6 \left( 0.5 - \frac{1}{6} \right) = 6 \left( \frac{3}{6} - \frac{1}{6} \right) = 6 \cdot \frac{2}{6} = 2 \]

Thus, the number of ways to have exactly 1 letter in the correct envelope is:

\[ \text{Ways with exactly 1 correct} = 4 \times D_3 = 4 \times 2 = 8 \]

### Step 4: Calculate arrangements with at least 2 letters in the wrong envelope
Now we can find the arrangements with at least 2 letters in the wrong envelope. This can be calculated by subtracting the cases where all letters are correct and where exactly 1 letter is correct from the total arrangements.

\[ \text{At least 2 wrong} = \text{Total arrangements} - \text{All correct} - \text{Exactly 1 correct} \]

Substituting the values we calculated:

\[ \text{At least 2 wrong} = 24 - 1 - 8 = 15 \]

### Final Answer
Thus, the number of ways in which at least two letters can be in the wrong envelope is:

\[ \boxed{15} \]