There are 3 letters and 3 envelopes. Find the number of ways in which all letters are put in the wrong envelopes

Answer

To find the number of ways to place 3 letters into 3 envelopes such that no letter is placed in its corresponding envelope (i.e., all 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. **Identify the Problem**: We have 3 letters (A, B, C) and 3 envelopes (1, 2, 3). We want to find the arrangements where no letter is in its correct envelope.

2. **Understanding Derangements**: The formula for the number of derangements (denoted as !n) of n items is given by:
\[ !n = n! \sum_{i=0}^{n} \frac{(-1)^i}{i!} \]
For n = 3, we will calculate !3.

3. **Calculate 3!**: First, we calculate the factorial of 3:
\[ 3! = 3 \times 2 \times 1 = 6 \]

4. **Calculate the Sum**: Now, we compute the sum:
\[ \sum_{i=0}^{3} \frac{(-1)^i}{i!} = \frac{(-1)^0}{0!} + \frac{(-1)^1}{1!} + \frac{(-1)^2}{2!} + \frac{(-1)^3}{3!} \]
Breaking it down:
- For i = 0: \(\frac{1}{1} = 1\)
- For i = 1: \(\frac{-1}{1} = -1\)
- For i = 2: \(\frac{1}{2} = 0.5\)
- For i = 3: \(\frac{-1}{6} \approx -0.1667\)

Adding these together:
\[ 1 - 1 + 0.5 - 0.1667 = 0.3333 \]

5. **Calculate Derangement**: Now substitute back into the derangement formula:
\[ !3 = 3! \times 0.3333 = 6 \times 0.3333 = 2 \]

6. **Conclusion**: Therefore, the number of ways to place the letters in the envelopes such that no letter is in its correct envelope is 2.

### Final Answer:
The number of ways in which all letters are put in the wrong envelopes is **2**.