# Another horse race

Arjun Shastry

Ranch Hand

Posts: 1899

1

posted 11 years ago

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 ]

**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: 1899

1

posted 11 years ago

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: 1899

1

posted 11 years ago

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: 1899

1

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 ]

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 ]

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 ]