There are $6$ letter$\{L_1,L_2,L_3,L_4,L_5,L_6\}$ and $6$ addressed envelope $\{E_1,E_2,E_3,E_4,E_5,E_6\}$. The correct place for $L_i$ is $E_i$. Find total cases in which $L_1$ goes to $E_3$ and no other letters goes to its correct envelope.
My approach:
This is similar to dearrangement. In $D(6)$ let $x$ be the cases in which $L_1$ goes to $E_3$. An also it is same when.$L_1$ goes to $E_{ \{2,4,5,6\}}$. So,
$5x=D(6)$
This gives, $x= \frac{D(6)}{5}$
But the textbook states that the answer is both $\frac{D(6)}{5}$ and $D(5)+D(4)$.
The value of answer matches but I am unable to find different logic for the second one. Would you help me finding?
$$D(n)=n!\cdot \sum_{i=0}^n \frac{(-1)^i}{i!}$$
For every derangement of the six letters into six envelopes, such that $L_1$ goes into $E_3$, there are two cases.
Suppose that $L_3$ also goes into $E_1$. We know that $L_1$ and $L_3$ went into each other's envelopes, and that letters numbered $2,4,5,6$ are deranged into their envelopes. There are $D(4)$ ways to choose the derangement for $L_2,L_4,L_5,L_6$.
Otherwise, there is some other number, $k$, such that $L_k$ goes into $E_1$, where $k\notin \{1,3\}$. That is, we have $$ L_k\to E_1,\qquad L_1\to E_3, $$ Imagine switching the contents of $E_1$ with $E_3$. The result is, now, $$ L_k\to E_3,\qquad L_1\to E_1 $$ Note that $L_1$ is now in its correct envelope, but every other letter is still in the wrong envelope. This means we now have a derangement of the five letters $\{L_2,L_3,L_4,L_5,L_6\}$ into their envelopes. This is why $D(5)$ appears.
To prove this works, we need to establish that the switching procedure above is actually a bijection. It is a bijection because it is reversible. Given a derangement of $\{L_2,L_3,L_4,L_5,L_6\}$, then you can introduce a new letter and envelope $L_1$ and $E_1$ where $L_1$ is in $E_1$, and then switch the contents of $E_1$ with $E_3$. The result will be a derangement of all six letters such that $L_1$ goes into $E_3$, but $L_3$ does not go into $E_1$.