How many ways are there for a horse race with four horses to finish if...

1. No ties:
The number of permutations is ${\displaystyle P(4,4)=4\ne 24}$
2. Two horses tie:
There are ${\displaystyle C(4,2)=6}$ ways to choose the two horses that tie. There are ${\displaystyle P(3,3)=6}$ ways for the groups to finish A group is either a single horse or the two tying horses. By the product rule, there are ${\displaystyle 6\cdot 6=36}$ possibilities for this case
3. Two groups of two horses tie
There are ${\displaystyle C(4,2)=6}$ ways to choose the two winning horses. The other two horses tie for second place.
4.Three horses tie with each other:
There are ${\displaystyle C(4,3)=4}$ ways to choose the two horses that tie. There are ${\displaystyle P(2,2)=2}$ ways for the groups to finish. By the product rule, there are ${\displaystyle 4\cdot 2=8}$ possibilities for this case.
5. All four horses tie:
There is only one combination for this. By the sum rule, the total is ${\displaystyle 24+36+6+8+1=75}$.

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Result can be like this: A, B, C, D are horses and 1-4 are the final positions. 1 2 3 4 - 4! = 24 ways 1 2 3 3 - 4!/2! = 12 ways 1 2 2 3 - 4!/2! = 12 ways 1 2 2 2 - 4!/3! = 4 ways 1 1 2 3 - 12 ways 1 1 2 2 - 4!/2!2! = 6 ways 1 1 1 2 - 4 ways 1 1 1 1 - 1 way So, total = 75 ways