Question

A horse race has 12 entries. Assuming there are no ties, how many different ways can these
horses finish first, second, and third?​

Answer 1

Answer:

These horses can finish first, second and third in 1320 different ways.

Step-by-step explanation:

The order in which the horses finish is important. That is, horse A finishing 1st, horse B 2nd and horse C 3rd is a different outcome than horse B 1st, horse A 2nd and horse C 3rd.

So the permutations formula is used to solve this question.

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

$P_{(n,x)} = \frac{n!}{(n-x)!}$

Assuming there are no ties, how many different ways can these horses finish first, second, and third?​

12 horses.

3 first positions.

So

$P_{(12,3)} = \frac{12!}{(12-3)!} = 1320$

These horses can finish first, second and third in 1320 different ways.