Difference between revisions of "Derangement"
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| − | A '''derangement''' is a [[permutation]] with no [[fixed point]]s. | + | A '''derangement''' is a [[permutation]] with no [[fixed point]]s. That is, a derangement leaves nothing in its original place. For example, the derangements of <math>(1,2,3)</math> are <math>(2, 3, 1)</math> and <math>(3, 1, 2)</math>. |
| + | The number of derangements of a [[set]] of <math>n</math> objects is sometimes denoted <math>!n</math> and is given by the formula: | ||
| + | <math>\displaystyle !n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!}</math> | ||
| − | <math>\ | + | Thus, the number derangements of a 3-[[element]] set is <math>3! \cdot \sum_{k = 0}^3 \frac{(-1)^k}{k!} = 6\cdot(\frac{1}{1} - \frac{1}{1} + \frac{1}{2} - \frac{1}{6}) = 2</math>, which we know to be correct. |
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Revision as of 16:35, 13 November 2006
A derangement is a permutation with no fixed points. That is, a derangement leaves nothing in its original place. For example, the derangements of 
are 
and 
.
The number of derangements of a set of 
objects is sometimes denoted 
and is given by the formula:
Thus, the number derangements of a 3-element set is 
, which we know to be correct.
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