Exercise $2\mathrm{.}31.(?)$

This question has three parts.

$($a$)$ The subfactorial of a number, namely $$n$,$ is the number of permutations of n objects such

that no object appears in its natural spot. For example, take the collection of objects

fa; b; cg$.$ There are $6$ possible permutations $($because we choose arrangements for three

items, and $3!=6)$: $\{$a; b; c$\},\{$a; c; b$\},\{$b; c; a$\},\{$c; b; a$\},\{$c; a; b$\},\{$b; a; c$\},$ but only two

of these are derangements: $\{$b; c; a$\}$ and $\{$c; a; b$\},$ because no element is in the same spot

as the original collection. Therefore, we say that $!3=2.$ We can describe subfactorial as

a recursive formula:

$!0=1$

$!1=0$

$!$n $=($n$-1)*(!($n$-1)+!($n$-2))$