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I have the numbers $\{1, 2, 3, 4\}$.

How would you find the number of arrangements in which only $3$ of the numbers are in their original positions? What about only $2$ in their original positions? Only $1$? None?

For the none case, I know this is just same as finding number of derangements. However, is there actually a method to calculate this? I could not find any online. Do you just have to brute force it?

ViHdzP
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1110101001
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1 Answers1

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For exactly $3$ numbers in place the answer is easy: there are $0$ possibilities. I think you can see why.

In general if $D(n)$ denotes the number of derangements of$~n$ objects, you can choose the $k$ required fixed points in $\binom nk$ ways, and what remains must by a derangement of the remaining objects, for a number $D_{n,k}\binom nkD(n-k)$ of possibilities. These numbers are called rencontres numbers.

The obvious fact that $\sum_kD_{n,k}=n!$, together with the given identity, is actually at the basis of one well known approach (among many) to finding a formula for the numbers $D(n)$. Namely $D(n)=D_{n,0}$, and a formula for the remaining terms in the summation can be assumed proven by induction.

Marc van Leeuwen
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  • I noticed on wikipedia that one formula for derangement is $\text{Round}\left[\frac{n!}{e}\right]$. How is this derived? – 1110101001 Apr 20 '14 at 04:12
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    The fundamental formula for $D(n)$ gives the first $n+1$ terms of the development of $n!,e^{-1}$, and the sum of the missing infinitely many terms can be estimated to be less than$~\frac12$ in absolute value, if $n>0$. – Marc van Leeuwen Apr 20 '14 at 04:32