*A secretary types  ve letters and the  ve corresponding envelopes. In a hurry,they place at random one letter in each envelope. what is the probability that: (a) at least one letter is in the correct envelope?*

*(b) no letter is in the correct envelope?*

*(c) exactly three letters are in the correct envelop?*

This problem involves understanding permutations and derangements. Let's break it down into parts:

### Part (a): Probability that at least one letter is in the correct envelope

To find the probability that at least one letter is in the correct envelope, we'll use the principle of complementarity. Instead of directly calculating the probability of at least one correct placement, we'll calculate the probability of *no* correct placements (a derangement) and subtract it from 1.

1. **Total number of arrangements:** 
   The total number of ways to arrange 5 letters into 5 envelopes is given by \(5!\).

   \[
   5! = 120
   \]

2. **Number of derangements (no letter in the correct envelope):**
   A derangement is a permutation where none of the objects appear in their "natural" place. The number of derangements \(D_n\) of \(n\) objects is given by:

   \[
   D_n = n! \left(\frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots + (-1)^n \frac{1}{n!}\right)
   \]

   For \(n = 5\):

   \[
   D_5 = 5! \left(\frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!}\right)
   \]

   \[
   D_5 = 120 \left(1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}\right) = 120 \times \left( \frac{44}{120} \right) = 44
   \]

3. **Probability of at least one correct placement:**

   \[
   P(\text{at least one correct}) = 1 - \frac{D_5}{5!} = 1 - \frac{44}{120} = 1 - \frac{11}{30} = \frac{19}{30}
   \]

### Part (b): Probability that no letter is in the correct envelope

This is simply the probability of a derangement, which we calculated earlier:

\[
P(\text{no correct placement}) = \frac{D_5}{5!} = \frac{44}{120} = \frac{11}{30}
\]

### Part (c): Probability that exactly three letters are in the correct envelope

It's impossible to have exactly three letters in the correct envelope and the remaining two not. If three letters are correctly placed, the last two must be either correct or incorrect together, meaning we cannot have exactly three. Therefore, the probability is 0.

\[
P(\text{exactly three correct}) = 0
\]

### Summary of Answers:

- (a) The probability that at least one letter is in the correct envelope is \(\frac{19}{30}\).
- (b) The probability that no letter is in the correct envelope is \(\frac{11}{30}\).
- (c) The probability that exactly three letters are in the correct envelope is 0.

Would you like more details on any part of this solution?

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**Related Questions:**

1. What is the probability that exactly two letters are in the correct envelope?
2. How does the probability change if there are 6 letters and envelopes instead of 5?
3. How do we generalize the formula for the number of derangements?
4. What is the expected number of letters that are in the correct envelope?
5. How do we compute the number of derangements for larger values of \(n\)?
6. Can the problem be solved using inclusion-exclusion principle?
7. What if the secretary places letters randomly but must ensure no two letters are correctly placed?
8. How do derangements relate to real-world scenarios in cryptography?

**Tip:** For problems involving permutations and derangements, breaking down the problem using combinatorial principles often simplifies the calculation.

Understand the probability of correctly placing 5 letters in 5 envelopes randomly. Learn how to calculate probabilities for at least one correct placement, no correct placement, and exactly three correct placements using permutations and derangements.

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