Counting Results in a Competition With Ties

Question

In National Games Championship, a high jump competition is being conducted along with other athletic events. Only n out of m athletes were able to meet the eligibility criteria set by the athletics committee of the games. In this event, multiple athletes can even end up jumping the same height and hence, the awards committee has even permitted tie among the athletes as a valid result. Under these assumptions, how many different results are possible at the end of competition.

Solution

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Here, we should consider the positions being allotted to different athletes. This question based on Fubini numbers or Ordered Bell numbers.

The fubini number sequence is 1, 1, 3, 13, 75, 541, 4683, ….. It is often used in this type of problems. And here, we’re counting results in a competition with ties.

Explanation: Suppose there are five athletes in a competition where ties are possible and only one athlete finished first. There are $\binom 51$ ways to choose that athlete and 75 ways to order the remaining horses.

when there are no ties, there is only one way to order the results.
when there is one tie, there is only one way to order the results.
when there are 2 ties, there are three ways to order the results.
when there are 3 ties, there are 13 ways to order the results.

… and so on.

So, if there are 5 people, the total number of possible results will be,

$\begin{align*} \left = \binom 51 * 75 + \binom 52 * 13 + \binom 53 * 3 + \binom 54 * 1 + \binom 55 * 1 = 541 \end{align*}$

Let Fubini numbers for n be represented by fub(n). Then, for n athletes, number of possible results can be given by,

$\begin{align*} \left = \binom n1 * fub(n-1) + \binom n2 * fub(n-2) + \binom n3 * fub(n-3) + ...... + \binom nn * fub(0) \end{align*}$

This is our desired answer (formula) for n athletes.

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