find the number of derangement of the sets {1,2,3,4}

The number of derangements of a set refers to how many ways the elements of the set can be rearranged so that none of the elements appears in its original position.

The formula for the number of derangements \( D_n \) of a set of size \( n \) is:

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

For the set \( \{1, 2, 3, 4\} \), we have \( n = 4 \).

### Step-by-step calculation:

1. \( 4! = 24 \)
2. The series expansion for \( n = 4 \) is:
   \[
   D_4 = 24 \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} \right)
   \]
   Now let's calculate each term:
   \[
   D_4 = 24 \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right)
   \]
   Simplifying inside the parentheses:
   \[
   D_4 = 24 \left( 0 + 0.5 - 0.1667 + 0.0417 \right)
   \]
   \[
   D_4 = 24 \times 0.375
   \]
   \[
   D_4 = 9
   \]

So, the number of derangements of the set \( \{1, 2, 3, 4\} \) is **9**.

Would you like a detailed explanation of how derangements work, or have any questions?

Here are 5 follow-up questions:
1. How does the derangement formula extend to larger sets?
2. Can you derive the number of derangements using recursion?
3. What is the significance of derangements in probability theory?
4. How does the inclusion-exclusion principle relate to derangements?
5. What real-world applications involve the concept of derangements?

**Tip:** Memorizing the recursive relation \( D_n = (n-1)(D_{n-1} + D_{n-2}) \) can make calculating derangements easier for larger sets.

Learn how to calculate the number of derangements for the set {1, 2, 3, 4} using combinatorics and the inclusion-exclusion principle. This detailed solution includes step-by-step calculations and the application of factorial-based formulas. Suitable for students in Grades 10-12.

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