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This is a practice question for derangements. The text I have only has half a page on derangements and that doesn't help me solve this question.

For positive integers $1,2,3,\ldots,(n-1),n$ there are $11\text{,}660$ derangements where $1,2,\ldots,5$ appear in the first five positions. What is the value of $n$?

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  • $\begingroup$ Start by finding the number of derangements of the $5$ integers $1,2,3,4,5$. What is it? $\endgroup$
    – Adriano
    Jul 30, 2013 at 6:03

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It helps to have a list of the number $D_k$ of derangements for the first few $k$. The Wikipedia article has such a list.

The number of derangements of $1,2,3,\dots,n$ in which the numbers $1$ through $5$ appear among the first five numbers (though of course not in that order!) is the number of derangements of the first $5$ numbers times the number of derangements of the remaining $n-5$ numbers.

There are $44$ derangements of a set of $5$ numbers. So the number of derangements of the remaining $n-5$ numbers is $\frac{11660}{44}$. This is $265$. But $265$ is $D_k$ for $k=6$. So $n-5=6$, and therefore $n=11$.

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  • $\begingroup$ You are welcome. $\endgroup$
    – André Nicolas
    Jul 30, 2013 at 6:15

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