Question

There are 1111 candidates for three positions at a restaraunt. One position is for a cook. The second position is for a food server. The third position is for a cashier. If all 1111 candidates are equally qualified for the three​ positions, in how many different ways can the three positions be​ filled?

Answer 1

Answer:

1,367,629,890 or 990

Step-by-step explanation:

This is a problem of permutation, because the order does matter. For example:

1 - 2 - 3 isn't the same as 3 - 2 - 1. And also you cant repeat the same candidate in 2 different positions.

The formula for this selection is :

$Permutations = \frac{n!}{(n-r)!}$

Where:

n is the number of things to choose from

r is the number of things you are going to choose

In this case n = 1111 or 11 (depending if you mistype it)

r = 3

$Permutations = \frac{1111!}{(1111-3)!} =1111*1110*1109=1,367,629,890 \\Permutations = \frac{11!}{(11-3)!}=11*10*9=990$