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If n e N with n >= 2 . A running competition takes place witn n numbered runners.
With (number of possible results)=PR.
a)How many running results are there?
b)In how many results does the runner with the number 1, has the first place before the runner with the number 2?
(In other words: How many possibilities (of results) are there,when the runner1 has the first place and is followed by the runner2 on the second place.)
c)When is the number of possible results in the two following cases equal?
i)the runner with number 1 wins.
ii) the runners with number 1 and 2 share the first two places.

a) I have n! = PR.
b) I have n!/(n-1)! = PR.
c) n!/(n-1)!X = n!/(n-1)!*2! ...

I am not sure about my results.
Are these results true?If not, please correct me.

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    Ah yes, I was mistaken, they share the first two places...2012-11-22

1 Answers 1

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If ties are not allowed, then your answer to (a) is correct.

For (b), again if ties are not allowed, places $1$ and $2$ have been assigned. There are $(n-2)!$ different orders of the runners other than the first two.

If ties are allowed, the problems become vastly more complicated. The answers then involve the ordered Bell numbers. Unless this problem comes from a relatively advanced combinatorics course, it seems unlikely that ties, including multiple ties, are allowed.

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    Ok and if ties are allowed what will change?2019-04-24