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 at least 3 out of 6 letters in wrong envelops is

Options

A. (a)265
B. (b)40
C. (c)704
D. (d)50

Answer

To solve the problem of finding the number of ways to put at least 3 out of 6 letters in wrong envelopes, we can break it down into steps.

### Step-by-Step Solution:

1. **Understanding the Problem**: We need to find the total number of ways to arrange 6 letters such that at least 3 of them are in the wrong envelopes.

2. **Using Inclusion-Exclusion Principle**: We will calculate the number of arrangements for exactly 3, 4, 5, and 6 letters being in the wrong envelopes and then sum these values.

3. **Calculating for Exactly 3 Letters Wrong**:
- Choose 3 letters out of 6 to be in the wrong envelopes: This can be done in \( \binom{6}{3} \) ways.
- The number of derangements (wrong arrangements) of 3 letters can be calculated using the formula for derangements:
\[ D_n = n! \left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \ldots + \frac{(-1)^n}{n!}\right) \]
- For \( n = 3 \):
\[ D_3 = 3! \left(1 - 1 + \frac{1}{2} - \frac{1}{6}\right) = 6 \left(0 + 0.5 - 0.1667\right) = 6 \times 0.3333 = 2 \]
- Therefore, the total arrangements for exactly 3 letters wrong:
\[ \text{Ways} = \binom{6}{3} \times D_3 = 20 \times 2 = 40 \]

4. **Calculating for Exactly 4 Letters Wrong**:
- Choose 4 letters out of 6: \( \binom{6}{4} \).
- Derangements for 4 letters \( D_4 \):
\[ D_4 = 4! \left(1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24}\right) = 24 \left(0 + 0.5 - 0.1667 + 0.0417\right) = 24 \times 0.375 = 9 \]
- Total arrangements for exactly 4 letters wrong:
\[ \text{Ways} = \binom{6}{4} \times D_4 = 15 \times 9 = 135 \]

5. **Calculating for Exactly 5 Letters Wrong**:
- Choose 5 letters out of 6: \( \binom{6}{5} \).
- Derangements for 5 letters \( D_5 \):
\[ D_5 = 5! \left(1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}\right) = 120 \left(0 + 0.5 - 0.1667 + 0.0417 - 0.0083\right) = 120 \times 0.3667 = 44 \]
- Total arrangements for exactly 5 letters wrong:
\[ \text{Ways} = \binom{6}{5} \times D_5 = 6 \times 44 = 264 \]

6. **Calculating for Exactly 6 Letters Wrong**:
- All letters are in the wrong envelopes: \( \binom{6}{6} = 1 \).
- Derangements for 6 letters \( D_6 \):
\[ D_6 = 6! \left(1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} + \frac{1}{720}\right) = 720 \left(0 + 0.5 - 0.1667 + 0.0417 - 0.0083 + 0.0014\right) = 720 \times 0.3681 = 265 \]
- Total arrangements for exactly 6 letters wrong:
\[ \text{Ways} = \binom{6}{6} \times D_6 = 1 \times 265 = 265 \]

7. **Total Arrangements for At Least 3 Letters Wrong**:
- Now, we sum up all the cases:
\[ \text{Total} = 40 + 135 + 264 + 265 = 704 \]

### Final Answer:
The total number of ways of putting at least 3 out of 6 letters in wrong envelopes is **704**.