Find the number of ways of placing 6 letters in 6 envelopes which have different addresses in such a way that at least two letters are correctly placed the right envelops

Options

A. 191
B. 171
C. 362
D. 121

Answer

To solve the problem of placing 6 letters in 6 envelopes such that at least 2 letters are correctly placed, we can break it down into several cases based on how many letters are correctly placed. We will use the concept of derangements to find the number of ways to place the remaining letters incorrectly.

### Step-by-Step Solution:

1. **Identify the Total Cases**: We need to consider the cases where exactly 2, 3, 4, 5, or 6 letters are correctly placed. However, since we need at least 2 letters correctly placed, we will focus on the cases where 2, 3, 4, 5, and 6 letters are correctly placed.

2. **Case 1: Exactly 2 Letters Correctly Placed**:
- Choose 2 letters from 6 to be correctly placed. The number of ways to choose 2 letters from 6 is given by the combination formula \( \binom{n}{r} \):
\[ \binom{6}{2} = 15 \]
- The remaining 4 letters must be incorrectly placed. The number of derangements (denoted as \( !n \)) for 4 letters is \( !4 = 9 \).
- Therefore, the total for this case is:
\[ 15 \times 9 = 135 \]

3. **Case 2: Exactly 3 Letters Correctly Placed**:
- Choose 3 letters from 6 to be correctly placed:
\[ \binom{6}{3} = 20 \]
- The remaining 3 letters must be incorrectly placed. The number of derangements for 3 letters is \( !3 = 2 \).
- Therefore, the total for this case is:
\[ 20 \times 2 = 40 \]

4. **Case 3: Exactly 4 Letters Correctly Placed**:
- Choose 4 letters from 6 to be correctly placed:
\[ \binom{6}{4} = 15 \]
- The remaining 2 letters must be incorrectly placed. The number of derangements for 2 letters is \( !2 = 1 \).
- Therefore, the total for this case is:
\[ 15 \times 1 = 15 \]

5. **Case 4: Exactly 5 Letters Correctly Placed**:
- Choose 5 letters from 6 to be correctly placed:
\[ \binom{6}{5} = 6 \]
- The remaining 1 letter must be incorrectly placed. However, there is no derangement for 1 letter, so this case contributes:
\[ 6 \times 0 = 0 \]

6. **Case 5: All 6 Letters Correctly Placed**:
- There is only 1 way to place all letters correctly:
\[ 1 \]

7. **Total Ways**: Now, we sum the contributions from all cases:
\[ 135 + 40 + 15 + 0 + 1 = 191 \]

### Final Answer:
The total number of ways to place 6 letters in 6 envelopes such that at least 2 letters are correctly placed is **191**.

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