If n things are arranged in a row, the number of way in which they can be arranged so that non occupies its original position is `n!(1-(1)/(1!)+(1)/(2!)-(1)/(3!)+* * * + (-1)^(n)(1)/(n!))`
The number of ways of putting 6 letters into 6 addressed envelopes so that exactly 2 letters are in wrong is

Options

A. `""^(6)C_(4)`
B. 30
C. 45
D. 28

Answer

To solve the problem of finding the number of ways to put 6 letters into 6 addressed envelopes such that exactly 2 letters are in the wrong envelopes, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have 6 letters (L1, L2, L3, L4, L5, L6) and 6 envelopes (E1, E2, E3, E4, E5, E6). We need to find the arrangements such that exactly 2 letters are placed in the wrong envelopes.

2. **Choosing the Letters**:
First, we need to choose which 2 letters will be placed incorrectly. The number of ways to choose 2 letters from 6 is given by the combination formula \( \binom{n}{r} \), where \( n \) is the total number of items, and \( r \) is the number of items to choose.
\[ \text{Number of ways to choose 2 letters} = \binom{6}{2} \]

3. **Calculating \( \binom{6}{2} \)**:
\[ \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \]

4. **Arranging the Chosen Letters**:
Once we have chosen 2 letters, we need to place them in the wrong envelopes. For 2 letters, there is only one way to arrange them incorrectly. For example, if we choose letters L1 and L2, they can only be placed in each other's envelopes (L1 in E2 and L2 in E1).

5. **Final Calculation**:
Since there is only one way to arrange the 2 chosen letters incorrectly, the total number of arrangements where exactly 2 letters are in the wrong envelopes is simply the number of ways to choose the letters:
\[ \text{Total arrangements} = \binom{6}{2} \times 1 = 15 \]

### Conclusion:
Thus, the number of ways of putting 6 letters into 6 addressed envelopes so that exactly 2 letters are in the wrong envelopes is **15**.

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