Answer to Question #159234 in Combinatorics | Number Theory for Kendra

Question

A student purchases the following lock. There are 40 numbers. The three number key cannot use the same number twice. How many possible 3 digit keys are possible for this lock?

Answer 1

$\boxed{A_k^n=\dfrac {k!} {(k-n)!}}$

We are choosing n = 3 out of k = 40

A = $\dfrac {37!*38*39*40}{37!} = 59280$

Answer: 59280 possible keys