In how many different orders can five runners finish a race if no ties are allowed?

/ Permutations and combinations / By

First, let’s examine the definitions of permutations and combinations.

“A permutation of a set of distinct objects is an ordered arrangement of these objects. We also are interested in ordered arrangements of some of the elements of a set. An ordered arrangement of r elements of a set is called an r-permutation. ”

“An r-combination of elements of a set is an unordered selection of r elements from the set. Thus, an r-combination is simply a subset of the set with r elements.”

Both definitions are from the textbook Discrete Mathematics and its Applications by Rosen.

Notice, that in permutations, the elements have an order; and a combination is an unordered selection.

This distinction is the key to answering this (and other) type of exercise.

Runners can finish a race in a specific order, first, second, third, and so on. We also know from the problem description that no ties are allowed. In other words, there cannot be two runners in the first place.

From the above, we can conclude that we need to calculate the number of 5-permutations in a set of 5 runners.

Applying the formula:

P(5,5)=5!/(5-5)!

=5!/0!

=5!

=120

Another approach to solving this type of exercise is as follows.

The first place among the runners can be chosen in 5 ways (as there are 5 runners, each of them can finish in the first place). For each of the 5 ways of choosing the first place, we have 4 ways to choose the second place, and so on.

So, it follows that we must use the multiplication rule.

5x4x3x2x1=120.

Related exercises: