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In how many ways can the word "WORD" be rearranged so that no letter is in its original position?

The answer is $9$, but what is the formula for it?

Wii Craft
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    You are referring to a special kind of permutation called a "derangement". See here: http://mathworld.wolfram.com/Derangement.html – Kaj Hansen Jan 24 '15 at 05:00
  • For a $4$-letter word like this, a careful listing will do it. We can use as shortcut that W ends up in one of $3$ places, and count the legal rearrangements in which W is second, and multiply by $3$. – André Nicolas Jan 24 '15 at 05:10

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The keyword here is derangements. The formula for the number of derangements of $n$ things is a bit messy:

$$d_n=n!\sum_{k=0}^n\frac{(-1)^k}{k!}\;.$$

You’ll find some other formulas, less easy to prove but more usable, at the link; perhaps the nicest is

$$d_n=\left\lfloor\frac{n!}e+\frac12\right\rfloor\;,$$

for $n\ge 1$.

Brian M. Scott
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  • As an addendum for the OP, there are also recurrence relations if one is so-inclined. See here for example: http://math.stackexchange.com/questions/203899/combinatorial-argument-to-prove-the-recurrence-relation-for-number-of-derangemen – Kaj Hansen Jan 24 '15 at 05:03
  • So can you describe what is going on here? Like what "d" and "e" stands for. – Wii Craft Jan 24 '15 at 05:11
  • @WiiCraft: $d_n$ is the number of possible derangements of $n$ things; in your problem you found that $d_4=9$. $e$ is the base of the natural logarithms and is approximately $2.718281828459056$. $\lfloor x\rfloor$ is the largest integer less than or equal to $x$. – Brian M. Scott Jan 24 '15 at 05:14