1looks like the recurrence formula for the number of derangements on a set of length n
in order to derive it we need another relation which can be derived
http://www.math.uiowa.edu/~sokratov/2008m150/derangements.pdf
consider a function f on non-negative integers such that f(0)=1 , f(1)=0 and
f(n) + f(n-1) = nf(n-1) + (n-1)f(n-2) for n≥2 .
Show that ,
\frac{f(n)}{n!} = \sum_{k=0}^{n}{\frac{(-1)^{k}}{k!}}
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3 Answers
71May be we can try some induction at hand. I was wondering if we have any method to solve it directly.
1