Question

Roy is picking his 3 classes for the upcoming term. There are two groups the classes are divided into and he needs to pick at least one from each group (Group 1 has 5 choices and Group 2 has 4 choices). How many ways can Roy pick his classes?

Answer 1

70

Explanation:

Roy needs to choose 3 courses - at least 1 from Group 1 (5 choices total) and at least 1 from Group 2 (4 choices total). How many ways can Roy choose his 3 courses?

If Roy picks 2 courses from the set of 5 in Group 1, he needs to pick 1 from the set of 4 in Group 2. And if he picks 2 courses from Group 2, he needs to pick 1 from Group 1. We can find those combinations (we don't care in what order he picks the classes) and then add the two scenarios together.

The combination general formula is:

`C_(n,k)=(n!)/((k!)(n-k)!)` with `n="population", k="picks"`

We'll have:

`C_(5,2)xxC_(4,1)+C_(5,1)xxC_(4,2)`

`(5!)/((2!)(3!))(4!)/((1!)(3!))+(5!)/((1!)(4!))(4!)/((2!)(2!))`

`(5xx4xx3!)/((2)(3!))(4xx3!)/((1)(3!))+(5xx4!)/((1)(4!))(4xx3xx2!)/((2)(2!))`

`10xx4+5xx6=40+30=70`