Answer to Question #143534 in Statistics and Probability for John

Question

Douglas is going to visit 6

 cities this summer. He will choose from 12

 different cities and the order in which he visits the cities does not matter. How many different city combinations are possible for the summer travelling?

Answer 1

Selection of k items from a collection with n elements, such that the order of selection does not matter

$\dbinom{n}{k} = \frac {n!} {k!(n-k)!}$

$\dbinom{12}{6} = \frac {12!} {6!(12-6)!}=924$