SOLUTION: 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.) My answer is 75, But I still doubt it. Can a

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Question 435686: 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.) My answer is 75, But I still doubt it. Can anyone give me a reason?
Found 2 solutions by stanbon, Edwin McCravy:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
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.) My answer is 75,
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Note: Think of a four letter word.
with no repetition
with 2 letters the same
with 2 sets of 2 letters the same
with 3 letters the same
with all letters the same
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If there are no ties the number of orders is 4! = 24
If 2 tie the number of orders is 4!/2! = 12
If 2 tie and the other 2 tie the number of orders is 4!/[2!*2!] = 6
If 3 tie the number of orders is 4!/3! = 4
If 4 tie the number of orders is 4!/4! = 1
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Total # of orders is 24+12+4+1 = 47
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Cheers,
Stan H.
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Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
The other tutor's solution is wrong.

I get 75 also.  Here's how I did it:

There are 8 possible general situations:

1. 4 horses tie for 1st place
2. 3 horses tie for 1st place and 1 comes in 2nd place
3. 2 horses tie for 1st place and 2 tie for 2nd place
4. 2 horses tie for 1st place, 1 horse comes in 2nd and 1 comes in 3rd
5. 1 horse comes in 1st and 3 horses tie for 2nd place  
6. 1 horse comes in 1st, 2 tie for 2nd place, and 1 comes in 3rd place
7. 1 horse comes in 1st, 1 comes in 2nd, and 2 tie for 3rd place.
8. 1 horse comes in 1st, 1 comes in 2nd, 1 comes in 3rd, and 1 comes in 4th.

Situation 1 can be done in 1 way
Situation 2 can be done in C(4,3) or 4 ways 
Situation 3 can be done in C(4,2) or 6 ways
Situation 4 can be done in C(4,2)*2 or 12 ways
Situation 5 can be done in C(4,1) or 4 ways
Situation 6 can be done in C(4,1)*C(3,2) or 4*3 or 12 ways
Situation 7 can be done in C(4,1)*C(3,1) or 4*3 or 12 ways
Situation 8 can be done in P(4,4) or 4! or 24 ways

So that's 1+4+6+12+4+12+12+24 = 75

So you get what I get.

Edwin
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