A postman has to deliver five letters to five different houses. Mischievously, he posts one letter through each door without looling to see if it is the correct address. In how many different ways could he do this so that exactly two of the five houses receive the correct letters?

Answer

To solve the problem of how many different ways the postman can deliver five letters to five different houses such that exactly two of the houses receive the correct letters, we can follow these steps:

### Step 1: Choose the Correct Letters
First, we need to select which two letters will be delivered correctly. Since there are five letters, the number of ways to choose 2 letters from 5 is given by the combination formula:

\[ \text{Number of ways to choose 2 letters} = \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]

### Step 2: Arrange the Remaining Letters Incorrectly
Next, we need to arrange the remaining three letters such that none of them is delivered to the correct house. This is known as a derangement. The number of derangements of \( n \) items, denoted as \( !n \), can be calculated using the formula:

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

For \( n = 3 \):

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

So, there are 2 ways to derange the three letters.

### Step 3: Calculate the Total Arrangements
Now, we multiply the number of ways to choose the correct letters by the number of derangements of the remaining letters:

\[ \text{Total ways} = \binom{5}{2} \times !3 = 10 \times 2 = 20 \]

Thus, the total number of ways the postman can deliver the letters such that exactly two houses receive the correct letters is **20**.

### Summary of Steps:
1. Choose 2 letters to deliver correctly: \( \binom{5}{2} = 10 \)
2. Derange the remaining 3 letters: \( !3 = 2 \)
3. Multiply the results: \( 10 \times 2 = 20 \)