# Counting dissimilar championship outcomes

Let's say we have a football league with 16 teams. At the end of the tournament, we have the first three teams (the gold, silver and bronze medals), the fourth place (let's say this place is also important), the 13th and 14th places that will have to play a few more games to stay in the league, and finally the 15th and 16th places that leave the championship (in this case, the exact position doesn't matter).

We consider tournament outcomes to be similar if the outcomes described above are the same.

What is the number of dissimilar outcomes?

I should note, however, that outcomes are considered similar if the 15th and 16th places are reversed because the order doesn't matter when both teams leave the championship.

I know the answer is: $$\frac{16!}{8!2!}$$

My approach to the problem is that we should count the number of permutations of 16 items 8 of which are considered indistinguishable, as well as 2 that can be reversed.

I am not sure that this is correct, but anyway I want to calculate the number of dissimilar outcomes directly.

Probably, it makes sense to calculate the number of similar items first. We first count the number in which we can choose 6 teams from 16 (respecting the order) then multiply it by the number of ways to choose 2 more items from the remaining 10. This is the total number of ways to fix the "important" positions, which we should also multiply by $$8!$$ (the number of permutations for the non-important positions).

As a result I get a huge number, which is obviously wrong.

Seems like I need an explanation.

Thanks

In your second method, what you should get is $$\binom{16}{6}6!\binom{10}{2}8!$$ according to what you suggested. But why would you permute the other $$8$$ teams with $$8!$$ while the order of them does not matter? If you don't do that, you can see $$\frac{16!}{8!2!} = \binom{16}{6}6!\binom{10}{2}$$
• Which outcomes are you counting? For 6 positions we really calculate $\binom{16}{6}$ since the order matters and then multiply it by the number of ways to choose 2 from the remaining 10 teams. This is clear. But isn't this just the number of ways to fix those 8 teams? Jun 21, 2019 at 12:15
• Yes, and since order of $6$ teams matter, we multiply by $6!$. But after fixing those $8$ teams, ranks of other teams are unimportant right? So even if we permute them, it will give us a "similar" outcome. So multiplying by $8!$ counts each dissimilar outcome $8!$ times. Jun 21, 2019 at 12:18
• What distinguishes an outcome from another is the first $4$ ranks, last $2$ ranks and $13^{th}$ and $14^{th}$ places. So as long as we change them (or permute whenever order matters), it will give us a dissimilar outcome. That's why we are only counting the outcomes for these $8$ ranks. Jun 21, 2019 at 12:26
• Thank you, that's clear now. I had thought that for each of the ways to distinguish one outcome from another we also should take account of all the possibilities for the remaining 8 teams. Now I see this is wrong. Another question, if you don't mind. For practice purposes, I also want to calculate the number of "similar" outcomes, but again my reasoning is off. My understanding is as follows: we have fixed 8 positions and now we can just permute 8 remaining teams and multiply the result by 2 since we have $2!$ more possibilities for the 15th and 16th places Jun 21, 2019 at 12:32