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 $3$ D $6)$: fa; b; cg; fa; c; bg; fb; c; ag; fc; b; ag; fc; a; bg; fb; a; cg$,$ but only two

of these are derangements: fb; c; ag and fc; a; bg$,$ because no element is in the same spot

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

$!0=1$

$!1=0$

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

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