Derangement Problem: 3 Letters, 3 Envelopes

This question requires finding the number of ways to arrange 3 letters into their corresponding envelopes such that no letter is placed in the correct envelope. This is a classic example of a derangement.

Derangement Calculation (n=3)

We need to calculate the number of derangements for $n=3$, denoted as $D_3$ or $!3$.

Method 1: Listing Possible Derangements

Consider 3 letters ($L_1, L_2, L_3$) and 3 envelopes ($E_1, E_2, E_3$). We seek arrangements where $L_1 \not\rightarrow E_1$, $L_2 \not\rightarrow E_2$, and $L_3 \not\rightarrow E_3$. The specific arrangements are:

• Way 1: $L_1$ in $E_2$, $L_2$ in $E_3$, $L_3$ in $E_1$.
• Way 2: $L_1$ in $E_3$, $L_2$ in $E_1$, $L_3$ in $E_2$.

There are exactly 2 derangements.

Method 2: Using the Derangement Formula

The number of derangements $D_n$ is given by the formula:

$D_n = n! \sum_{i=0}^{n} \frac{(-1)^i}{i!}$

For $n=3$:

$D_3 = 3! \left( \frac{(-1)^0}{0!} + \frac{(-1)^1}{1!} + \frac{(-1)^2}{2!} + \frac{(-1)^3}{3!} \right)$

$D_3 = 6 \left( \frac{1}{1} - \frac{1}{1} + \frac{1}{2} - \frac{1}{6} \right)$

$D_3 = 6 \left( 0 + \frac{1}{2} - \frac{1}{6} \right)$

$D_3 = 6 \left( \frac{3}{6} - \frac{1}{6} \right) = 6 \times \frac{2}{6}$

$D_3 = 2$

Conclusion: Derangement Count

Both the listing method and the formula confirm that there are 2 ways to place 3 letters into 3 envelopes such that none are in their correct envelope.