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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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    The question is a little difficult to understand. In particular, it is not clear what is meant by (b). Also, the presence of Question (c) seems to indicate that ties are allowed. In that case, we have a substantially more complicated combinatorial problem, and the answer to (a) is not $n!$.2012-11-22
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    Ok I try to formulate b) in a different way:How many possibilities (of results) are there,when the runner1 has the first place and is followed by the runner2 on the second place.2012-11-22
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    Everything is easy if ties are not allowed. There are then $(n-2)!$ possibilities for (b).2012-11-22
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    I suppose that ties are not allowed.And how can I find out c)?2012-11-22
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    If ties are not allowed, then Question (c) does not make sense. Perhaps you did not translate it correctly?2012-11-22
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    ...then I translated it incorrectly.2012-11-22
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    What do you mean with "ties are allowed", that there is a repetition of events?2012-11-22
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    You have towards the end "share first place." hat seems to mean they are tied. Then other people might also be tied. That's a more complicated counting problem.2012-11-22
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    Ah yes, I was mistaken, they share the first two places...2012-11-22

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