Question

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.)

Comments

you may take a look at this https://oeis.org/A000670

Answer 1

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

Comments

All possible combination of possibilities are required.Then applied principle of addition.
Thanks.

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if the no. of horses is three then the answer should be 10 . but in rosen the answer is given 13

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The given answer in Rosen is correct . It should be 13.

Case 1: No ties. We have (1st , 2nd, 3rd ranks) and there are 3 horses. So permutation of all 3 horses =3!=6

Case: 2 horse make Ties

Tie at 1 st rank, then there is one single horse comes at 2nd rank

or

one horse comes 1st and rest two horse ties for 2nd rank.

So here we have to make one block of two horses and another block of one horses and permute those blocks as for each combination either that block comes at 1st or 2nd.

So total combination -3C2

for each combination , there are 2 ways=2!

So total arrangements=3C2*2!=6

(1st-2nd or 1st-2nd)

(AB-C or C-AB)

(AC-B or B-AC)

(BC-A or A-BC)

 

Case 3:

All makes tie which is possible only in 1 way.

 

So case1+case2+case3=6+6+1=13