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I have a set of documents numbered 1 to n. If I arrange the documents randomly, what is the chance of selecting the last k pages in the correct order if I select them at random from the documents?

Because order is important, I use the following formula: $\displaystyle\frac{n!}{(n-k)!}$

From this I calculate how many ways there are to select k pages from n pages in the correct order. I need to have the last k pages. Which I can do in only one way. So is the chance

1/$\displaystyle\frac{n!}{(n-k)!}$?

It seems unlikely. It seems to me spontaneously that the chance should be smaller.

  • 0
    It's a little counterintuitive that finding the first $k$ pages in order has the same probability, because of the perception that those first pages are "available" whereas the last $k$ pages may already have been stacked. – 2017-02-04
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    So if I select pages at random and I have to find the last 4 pages in a set of documents of 125, the chance of doing so is 1/(125!/121!). – 2017-02-05
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    Maybe I don't understand the question. Are we trying to find the last pages in the last picks, or just in correct order anywhere in the selection? – 2017-02-05
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    The documents have been ordered randomly, and we select each page randomly for that set of pages. What is the chance of selecting the last four pages, correctly order i.e [n-3, n-2, n-1, n] – 2017-02-05

2 Answers 2

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Your result is not unlikely:

If you fix the last $k$ pages and their order then there remain $n-k$ pages to play with. There are exactly $(n-k)!$ possibilities. If you do not fix the last $k$ pages then the total number of possibilities is $n!$. So, the probability that after shuffling, the last $k$ pages will be in the right order and will be the right pages is

$$\frac{(n-k)!}{n!}.$$

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Your value is correct. I think a clearer way to approach it is to ask how many orders there are that leave the last $k$ pages in order. There are $n-k$ other pages which can be put in any of $(n-k)!$ orders. There are $n!$ total orders, so the probability is $\frac {(n-k)!}{n!}$