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Discrete Mathematics >> in how many ways we can have five people finish in a race if a tie is allowed?

this question is about the course Discrete Mathematics and Its Applications

note that it is impossible to have all of them on the 2nd or 3rd or ….
and the sol is not 5^5
thank you for passing here and I appreciate that

Regards
Gold Owner
my friend noliebro i said 5^5 is not the solution thank you for your try

my friend Takfam i am with you about if it not ties

i am interesting in your way of solving
we have 120 and the add it to the ties as you said

so i cannot be 10
see let po=position
po1 po2 po3 po4 po5
5 0 0 0 0
4 1 0 0 0
3 2 0 0 0
3 1 1 0 0
2 1 1 1 0
2 3 0 0 0
2 2 1 0 0
2 1 2 0 0
2 1 1 1 0
1 4 0 0 0
1 3 1 0 0
1 2 2 0 0
1 2 1 1 0
1 1 3 0 0
1 1 1 2 0
so it 120 + 15
i don’t know if i am right or not
i write it quickly
and i found this solution because of you
Am I right

i don’t know how to use the option here ?
i will look for it ?
thank you all

and waiting for more
i miss

1 1 2 1 0

Hi Gold Owner (got some spare to send here? )

I’m with a bad cold tonight so I hope my mind isn’t too boggled to answer this… You ask if 120+10 is correct. Sorry but I think it isn’t, because those 10 were scenarios, not solutions. Each scenario includes many solutions. For example, when you say “4,1,0,0,0″, it means four tie in the first place, but which four? There are five ways to pick those four, so that scenario includes five solutions. You would need to count the number of solutions in each scenario, which is almost as solving your original question.

Let me discuss another solution to your original problem. I use a recursive approach this time. I will call the solution for k competitor “y_k” (read that as “y sub k”). What we need to find is, of course, y_5

When there are five competitors, there are five possibilities with regards to the first place: either one competitor was first alone, or two tied in the first place, or three tied #1, or four or five. How many solutions would you have if only one was first alone?

y_5 = 5 * y_4 + …

This is so because there are 5 possible winners, and for each winner there are y_4 ways to order the rest of the competitors that came behind. Do you get the use of recursion here?

Lets add the second possibility (two people tying in the first place):

y_5 = 5 * y_4 + combinatorial(5;2) * y_3 + …

This is so because there are “combinatorial(5;2)” ways to pick two competitors from the pool of five if the order is not important, and there are three runners left so we use y_3 this time. Similarly, to include the third and fourth case:

y_5 = 5 * y_4 + combinatorial(5;2) * y_3 + combinatorial(5;3) * y_2 + combinatorial(5;4) * y_1 + …

The fifth case (everyone tying) is just one possibility, so the last term to add is simply a one. We can clear up this formula by writing it as a summatory. Notice that the first case can be sent into the summatory too because 5 = combinatorial(5;1), so it follows the same pattern. The general formula would then be:

y_k = 1 + summatory with i from 1 to (k-1) of (combinatorial(k;i) * y_(k-i))

I put the 1 in front to avoid confusion, this formula is very short if written properly in mathematical notation.

I didn’t bother to see if this recursion formula can be simplified or solved, what I did instead is to construct a table as I calculate from y_2 to y_5 (by the way, the initial value is y_1 = 1, as it means only one guy running). This is a technique that in algorithmics is called “dynamic programming”, which is simply building an array with intermediate results that are needed repeatedly as you go bottom-up towards your desired solution. The ideal thing would be to make a computer program or spreadsheet, but as five is not too high a number, I used a calculator and paper instead. The table I calculated reads as follows:

k; y_k
====
1; 1
2; 3
3; 13
4; 75
5; 541

So, if I made no mistakes, 541 should be the answer. Hope it helps. If anyone knows a way to solve the formula analytically please let us know.

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