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30 December, 07:01

How many ways are there for a horse race with three horses to finish if ties are possible?

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  1. 30 December, 08:56
    0
    There are 13 ways. I don't have a super clean counting theory formula to give you but I'll lay out my reasoning and show all 13 ways.

    Two rules come to mind:

    1. In every permutation someone must be first

    2. If any horse finishes 3rd, there can't be a tie (e. g. you can't have one finish third and the others tie for first. The other would then be for 2nd, not 3rd)

    So there are 6 ways for a horse to finish 1st. For example if the first horse finishes first, the following permutations (1, x, x) are possible:

    (1 1 1); (1 1 2); (1 2 1); (1 2 2); (1 2 3); (1 3 2)

    This list however, counts some of the others finishing first also so to avoid recounting any we can make similar lists for the first horse finishing 2nd and 3rd and that will suffice for the list. So for the first horse finishing 2nd (2 x x), we get

    (2 1 1); (2 1 2); (2 2 1); (2 1 3); (2 3 1)

    For the first horse finishing 3rd, we can't have ties so we get

    (3 2 1); (3 1 2)

    This add up to 13 ways. You can go back and count that each horse has:

    6 ways to finish 1st

    5 ways to finish 2nd

    2 ways to finish 3rd

    While these overlap, there are none the less 13 distinct permutations.
  2. 30 December, 09:25
    0
    I think 13 ways too ...
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