Linda Seales

2022-01-07

How many ways are there for a horse race with four horses to finish if ties are possible? [Note: Any number of the four horses may tie.)

Annie Levasseur

Beginner2022-01-08Added 30 answers

1. No ties:

The number of permutations is$P(4,4)=4\ne 24$

2. Two horses tie:

There are$C(4,2)=6$ ways to choose the two horses that tie. There are $P(3,3)=6$ ways for the groups to finish A group is either a single horse or the two tying horses. By the product rule, there are $6\cdot 6=36$ possibilities for this case

3. Two groups of two horses tie

There are$C(4,2)=6$ ways to choose the two winning horses. The other two horses tie for second place.

4.Three horses tie with each other:

There are$C(4,3)=4$ ways to choose the two horses that tie. There are $P(2,2)=2$ ways for the groups to finish. By the product rule, there are $4\cdot 2=8$ possibilities for this case.

5. All four horses tie:

There is only one combination for this. By the sum rule, the total is$24+36+6+8+1=75$ .

The number of permutations is

2. Two horses tie:

There are

3. Two groups of two horses tie

There are

4.Three horses tie with each other:

There are

5. All four horses tie:

There is only one combination for this. By the sum rule, the total is

ol3i4c5s4hr

Beginner2022-01-09Added 48 answers

The other tutors

karton

Expert2022-01-11Added 613 answers

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

22+64

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