How many possibilities are there for the win, place, and show (first, second, and third) positions in a horse race with 12 horses if all orders of finish are possible?

Before starting to solve the exercise, let’s start with the definitions.

“A permutation of a set of distinct objects is an ordered arrangement of these objects.”

It is also important to know beforehand how many of these ordered arrangements are there. So, we don’t forget anyone when asked this type of question.

Theorem: “If n is a positive integer and r is an integer with 1 ≤ r ≤ n, then there are P (n, r ) = n(n − 1)(n − 2) · · · (n − r + 1) r-permutations of a set with n distinct elements.”

Corollary: “If n and r are integers with 0 ≤ r ≤ n, then P(n,r) = n! / (n−r)!”

The previous definition, theorem, and corollary are from the textbook Discrete Mathematics and its Applications by Rosen.

In this case, we need to find out how we can choose the first, second, and third place (3 elements) from a total of 12 horses.

The solution to this exercise is straightforward. We just need to calculate how many 3-permutations of a set with 12 distinct objects are there.

For that, just use the formula from the corollary.

P(12,3) = 12!/(12-3)! = 12*11*10*9!/9!=12*11*10 = 1320.

Like with many other similar types of exercises, we can use the previous knowledge we have of counting to solve this exercise without knowing the formula for an r-permutation.

We have 12 horses.

We can choose the first place in 12 ways.

For each of the 12 ways, we can choose the second place in 11, and the third place in 10 ways.

By the multiplication rule, we can choose the three winners in 12*11*10=1320 ways.

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