1. **Problem Statement:** We have 6 distinct letters and 6 corresponding envelopes. We want to find the number of ways to place exactly 3 letters in their correct envelopes and the remaining 3 letters in incorrect envelopes. 2. **Understanding the problem:** - Total letters/envelopes: 6 - Exactly 3 letters are correctly matched. - The other 3 letters must be placed so that none of them is in the correct envelope (a derangement). 3. **Formula and concepts:** - Choose which 3 letters are correctly placed: $\binom{6}{3}$ ways. - For the remaining 3 letters, count the number of derangements (permutations with no fixed points). 4. **Number of derangements for $n$ items ($!n$):** $$ !n = n! \sum_{k=0}^n \frac{(-1)^k}{k!} $$ For $n=3$: $$ !3 = 3! \left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!}\right) = 6 \left(1 - 1 + \frac{1}{2} - \frac{1}{6}\right) = 6 \times \frac{1}{3} = 2 $$ 5. **Calculate total ways:** $$ \binom{6}{3} \times !3 = 20 \times 2 = 40 $$ 6. **Explanation:** - First, select which 3 letters go correctly. - Then, arrange the remaining 3 letters so none is in the correct envelope. - Multiply these counts to get the total number of valid arrangements. **Final answer:** 40