The number of possibilities to place exactly one letter in its correct envelope.

Question

Assume that there are $6$ color letters $L_1, L_2, L_3, L_4, L_5$ and $L_6$ are to be placed in $6$ same color envelopes $E_1, E_2, E_3, E_4, E_5$ and $E_6$ (one letter for each envelope). The number of possibilities to place exactly one letter in its correct envelope ________.

Answer 1

Using the inclusion-exclusion principle, we can compute the number of derangements of $n$ items as

All possible ways $-$ at least $1$ item in its proper place $+$ at least $2$ items in their proper places $-$ ...

$=n!-\binom{n}{1}(n-1)!+\binom{n}{2}(n-2)!-\binom{n}{3}(n-3)!+\cdots+(-1)^n\binom{n}{n}(n-n)!$

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

This can be denoted as $D_n$, and here you need to compute that one item is in its correct place, and the remaining $(n-1)$ are deranged, so $\binom{n}{1}\times D_{n-1}$