Suppose 4 letters are taken out of 4 different envelopes. In how many ways, can they be reinserted in the envelopes so that no letter goes in to its original envelope?

Answer

To solve the problem of how many ways 4 letters can be reinserted into 4 envelopes such that no letter goes into its original 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 positions.

### Step-by-Step Solution:

1. **Understanding the Problem**: We have 4 letters (L1, L2, L3, L4) and 4 envelopes (E1, E2, E3, E4). We need to find the number of ways to place each letter into an envelope such that no letter is placed in its corresponding envelope.

2. **Derangement 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 our case, n = 4.

3. **Calculate Factorial**: First, we need to calculate \(4!\):
\[ 4! = 24 \]

4. **Calculate the Summation**: Now we calculate the summation part:
\[ \sum_{i=0}^{4} \frac{(-1)^i}{i!} = \frac{(-1)^0}{0!} + \frac{(-1)^1}{1!} + \frac{(-1)^2}{2!} + \frac{(-1)^3}{3!} + \frac{(-1)^4}{4!} \]
Evaluating each term:
- For \(i=0\): \(\frac{1}{1} = 1\)
- For \(i=1\): \(\frac{-1}{1} = -1\)
- For \(i=2\): \(\frac{1}{2} = 0.5\)
- For \(i=3\): \(\frac{-1}{6} \approx -0.1667\)
- For \(i=4\): \(\frac{1}{24} \approx 0.0417\)

Now summing these values:
\[ 1 - 1 + 0.5 - 0.1667 + 0.0417 = 0.375 \]

5. **Final Calculation**: Now, we multiply the factorial by the summation:
\[ !4 = 4! \times 0.375 = 24 \times 0.375 = 9 \]

### Conclusion:
Thus, the number of ways to reinsert the letters into the envelopes such that no letter goes into its original envelope is **9**.