# Another horse race

Arjun Shastry

Ranch Hand

Posts: 1893

posted 11 years ago

- 0

Assuming two horses, they can finish:

1-2

2-1

(1,2) dead race

1 (2 DNF)

2 (1 DNF)

(1,2 DNF)

DNF: Did not finish. (Horse lost jockey, for instance).

Instead of DNF disqualification could occur.

But the difference isn't important for betters or price-money.

Do you have a more clean abstract idea of 'finishing' a race.

Of course we should argue, whether 'did not finish' is a way to finish.

Sounds absurd.

1-2

2-1

(1,2) dead race

1 (2 DNF)

2 (1 DNF)

(1,2 DNF)

DNF: Did not finish. (Horse lost jockey, for instance).

Instead of DNF disqualification could occur.

But the difference isn't important for betters or price-money.

Do you have a more clean abstract idea of 'finishing' a race.

Of course we should argue, whether 'did not finish' is a way to finish.

Sounds absurd.

Francis Siu

Ranch Hand

Posts: 867

posted 11 years ago

If I understand what you ask

Um..........

Assume that there are n horse and without any accident within the racing

the permutation will be nx(n-1)....x1 ==n!

If one accident occurs

the permutation will be 2(n)x2(n-1)....x1== 2(n)!-1

If there are k accident

the permutation will be k(n)xk(n-1)x....x1== k(n)!-(k-1)

It may be wrong

Does it require to write a program to show it out all the permutation and generate it out?

[ October 05, 2004: Message edited by: siu chung man ]

- 0

**In how many ways they can finish the race?**

If I understand what you ask

Um..........

Assume that there are n horse and without any accident within the racing

the permutation will be nx(n-1)....x1 ==n!

If one accident occurs

the permutation will be 2(n)x2(n-1)....x1== 2(n)!-1

If there are k accident

the permutation will be k(n)xk(n-1)x....x1== k(n)!-(k-1)

It may be wrong

Does it require to write a program to show it out all the permutation and generate it out?

[ October 05, 2004: Message edited by: siu chung man ]

Francis Siu

SCJP, MCDBA

Arjun Shastry

Ranch Hand

Posts: 1893

posted 11 years ago

- 0

No accident.Assume everything goes well.

Suppose there are two horses,1 and 2 then race can be finished as

1 finished the race followed by 2

1 and 2 finish at the same time.

2 finished the race followed by 1

So there are three ways in which 2 horses can finish the race.

Hope this helps.

Suppose there are two horses,1 and 2 then race can be finished as

1 finished the race followed by 2

1 and 2 finish at the same time.

2 finished the race followed by 1

So there are three ways in which 2 horses can finish the race.

Hope this helps.

MH

Arjun Shastry

Ranch Hand

Posts: 1893

posted 11 years ago

- 0

Yes,dead race means reaching at same time ,right? if we exclude that then n! possibilities.But we want to consider dead race!!.That is interesting.Actually its integer partition problem.

Say there are 4 horses then 4 can be partitioned as

4----Means all arrive at same time

1+1+2(one followed by second and rest arrive at same moment).Now 2 horses can be chosen in 6 ways.and another horse in two ways.So there will be 6*3! ways in which this combination can arrive.

in similar fashion,

1+1+1+1 Each arrives at different time,4!,24 ways

1+3---4*2! ways

2+2---6 ways

In all there are 75 ways

How to generalize for N?I am not sure.

Say there are 4 horses then 4 can be partitioned as

4----Means all arrive at same time

1+1+2(one followed by second and rest arrive at same moment).Now 2 horses can be chosen in 6 ways.and another horse in two ways.So there will be 6*3! ways in which this combination can arrive.

in similar fashion,

1+1+1+1 Each arrives at different time,4!,24 ways

1+3---4*2! ways

2+2---6 ways

In all there are 75 ways

How to generalize for N?I am not sure.

MH

Arjun Shastry

Ranch Hand

Posts: 1893

shankar vembu

Ranch Hand

Posts: 309

posted 11 years ago

Let me try.

1. No dead races(one winner) = n!

2. 2 winners = nc2*(n-2)!

We have nc2 ways to choose 2 winners. The rest (n-2) can be arranged in (n-2)!

So on and so forth.

Total = nc1*(n-1)! + nc2*(n-2)! + nc3*(n-3)! + ....+ ncn(n-n)!

Notes: First term = n! is essentially point# 1 stated above.

Last term = 1 = all horses are winners

Is this OK?

SHankar.

[ October 25, 2004: Message edited by: shankar vembu ]

- 0

Originally posted by Arjun Shastry:

Is there any similarity between this and factorization of a number.?

Say number is 24.Factors are 2,2,2,3.Number of ways in which 24 can be factored are

2*(12),2*2*(6),2*3*(4),2*2*2*3,3*(8),4*6

Let me try.

1. No dead races(one winner) = n!

2. 2 winners = nc2*(n-2)!

We have nc2 ways to choose 2 winners. The rest (n-2) can be arranged in (n-2)!

So on and so forth.

Total = nc1*(n-1)! + nc2*(n-2)! + nc3*(n-3)! + ....+ ncn(n-n)!

Notes: First term = n! is essentially point# 1 stated above.

Last term = 1 = all horses are winners

Is this OK?

SHankar.

[ October 25, 2004: Message edited by: shankar vembu ]

shankar vembu

Ranch Hand

Posts: 309

posted 11 years ago

I assumed that dead race applies only to position 1.

A more general solution would be:

Let A(n) = total no. of arrangements for n horses(where dead races are possible in every position).

So A(n) = nc1.A(n-1) + nc2.A(n-2) + nc3.A(n-3).........

where (recurse....)

A(n-1) = (n-1)c1.A(n-2) + (n-1)c2.A(n-3) + (n-1)c3.A(n-4)......

A(n-2) = (n-2)c1.A(n-3) + (n-2)c2.A(n-4) + (n-2)c3.A(n-5)......

.

.

.

.

.

I guess I am near to the solution. Any other formulations?

Basically, this would be a "bottom-up" computation.

SHankar

[ October 25, 2004: Message edited by: shankar vembu ]

- 0

Originally posted by shankar vembu:

Let me try.

1. No dead races(one winner) = n!

2. 2 winners = nc2*(n-2)!

We have nc2 ways to choose 2 winners. The rest (n-2) can be arranged in (n-2)!

So on and so forth.

Total = nc1*(n-1)! + nc2*(n-2)! + nc3*(n-3)! + ....+ ncn(n-n)!

Notes: First term = n! is essentially point# 1 stated above.

Last term = 1 = all horses are winners

Is this OK?

SHankar.

[ October 25, 2004: Message edited by: shankar vembu ]

I assumed that dead race applies only to position 1.

A more general solution would be:

Let A(n) = total no. of arrangements for n horses(where dead races are possible in every position).

So A(n) = nc1.A(n-1) + nc2.A(n-2) + nc3.A(n-3).........

where (recurse....)

A(n-1) = (n-1)c1.A(n-2) + (n-1)c2.A(n-3) + (n-1)c3.A(n-4)......

A(n-2) = (n-2)c1.A(n-3) + (n-2)c2.A(n-4) + (n-2)c3.A(n-5)......

.

.

.

.

.

I guess I am near to the solution. Any other formulations?

Basically, this would be a "bottom-up" computation.

SHankar

[ October 25, 2004: Message edited by: shankar vembu ]

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