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What is this question asking for?

How many 4-permutations of the positive integers not exceeding 100 contain three consecutive integers in the correct order?

[My note: 4-permutation = $P(n,4) = n\times(n-1)\times(n-2)\times(n-3)$]

The answer by the source is 18,915.

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    My guess is number of ways of selecting 4 numbers (order matters) such that you don't have 3 consecutive integers appearing in order. Example: 3,4,5,77 is not to be counted. If you have a number as an answer, perhaps we can confirm.2011-06-02
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    It is a little ambiguous; I believe you want to count lists of $4$ distinct elements of $\{1,2,\ldots,100\}$ that contain $3$ consecutive integers in the correct order. So for example, $(17,18,19,72)$ would be counted, and $(6,4,5,99)$ would not be. What is not clear to me is whether $(17,18,72,19)$ should be counted.2011-06-02
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    Is it "exactly $3$" or "at least $3$"? So for example does the quadruple $(11, 12, 13, 14)$ qualify? Also, is something like $(7,8, 44, 9)$ to be counted? As soon as the question is made clear, answers are likely to come quickly.2011-06-02
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    @Jonas: I would say the latter is not to be counted.2011-06-02

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