combinatory

Question

How many ways are there for a horse race with 4 horses to finish if ties are possible?(any number of horses may tie)

Answer 1

Let's break it down into cases:-

1. No horses tie: In this case total number of outcomes will be 4!

2. Two Horses tie: Number of ways to choose 2 horses from four horses 4C2 = 4*3/2 = 6, Now we can consider these 2 horses as one horse since they tie for position, Number of outcome of race with 3 horses is 3!. So total number of outcomes becomes 6*3!

3. Three horses tie: (Number of ways to select 3 horses) * (number of outcomes for 2 horse race without tie) = (4C3)*2! = 8

4: Four horses tie: 1

5. two ties with two horses: 4C2 = 6

So answer should be 4! + 6*3! + 8 + 1 + 6

Comments

2nd case has only one tie 5th case has 2 ties.

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two ties with two horses: 4C2 = 6

shall it not be 6*2 =12.

what is the correct answer?

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In 5th case whether we shouldn't multiply 4C2 with 2 here as two ties can be interchanged in 2! ways

Answer 2

Four horses are competing in race , and any no. of ties are possible.

Let's consider race positions as:

I            II           III        IV

4                                           { Here all 4 finishing in I^st place} =>  # ways = 4C4 =  1

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3            1                             # ways = 4C3 *  1C1 = 4

                                             4C3 for selecting 3 horses for first place. 

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2            2                             # ways = 4C2 * 2C2 = 6

                                            4C2 for selecting 2 horses for I^st place

                                            & 2C2 for II place

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2            1          1                 Similarly # ways = 4C2 * 2C1 * 1C1 = 12 

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1            3                              # ways = 4C1 * 3C3 = 4

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1            2          1                  # ways = 4C1 * 3C2  * 1C1 = 12

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1            1          2                  # ways = 4C1 * 3C1 * 2C2 = 12

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1            1          1        1         # ways = 4C1 * 3C1 * 2C1 * 1C1 = 24

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 Adding up all these cases , we get

         Total # ways = 1 + 4 + 6 + 12 + 4 + 12 + 12 + 24

                              = 75 ways.. 

Answer 3

let's

===> No of horse  = 1

              then no of ways = 1

===> when No of horses  = 2

               then no of ways = 3  (2C2 (Both in a first position)+2C1*1C1 (one of them is acquire  2nd position and              another acquire ist position))

== > When No of Horses = 3

then no of ways =  (all there finishes with 1st place) + (2 of them got place 1st and remaining one got 2nd place) + (one got ist place and 2 of them got  2nd place ) + (1 of them got 1st place , one of them got 2nd place and one of the them got 3 place in the race)

                      = 3C3 + 3C2*1C1+3C1*2C2+3C1*2C1*1C1 = 13

    therefore , no of  ways = 13

 

similarly

there are four horses. If just one horse is in first place, there are (4C1) ways to choose that horse, and 13 ways to order the remaining three horses. If there are two horses in first place, there are (4C2) ways to choose them, and 3 ways to order the rest. And so on. Thus the answer is

4C1*13 + 4C2*3  + 4C3*1 + 4C4 = 75

similarly if No of horses  = 5

(5C1)⋅75+(5C2)⋅13+(5C3)⋅3+(5C4)⋅1+(5C5)⋅1=541