How many different combinations of questions can be chosen?

Question

On an English test, Jeff must write an essay for three of six questions in Part 1, and two of five questions in Part 2. How many different combinations of questions can be chosen?

Answer 1

This is an example of a combinatorics question.  If you are familiar with n-choose-k notation, you can skip ahead to the punchline.  If not, read on...

 

To figure out how many different ways we can pick unique sets of k items from n distinct choices, we think about putting all n items down in a row, then taking the first k of them as our set.  There are n! unique ways to put the n items down in a row (n choices for the first element times n-1 choices for the second element times n-2 choices for the third element and so on down to just one choice left for the final element in the row).  However, a lot of these rows will result in the same set of k elements being in the first k spots.  In fact, it doesn't matter what order the first k elements are in, so we have to divide our n! rows into n!/k! rows to account for the possible orderings of the first k elements.  Similarly, we aren't concerned with the order of the last n-k elements in the row, since switching these elements around doesn't change which k elements are at the front of the row.  Dividing out the number of ways those n-k elements can be ordered, (n-k)!, gives us (n!)/[(k!)(n-k)!] ways to choose unique sets of k elements from n total elements.  We'll call this amount n-choose-k

 

With this framework in place, we can reason about the number of different combinations of questions that Jeff can answer on his English test.  Since Jeff must choose three of the six questions in Part 1, there are 6-choose-3 combinations of questions to choose.  Similarly in Part 2, Jeff can answer any two of five questions for a total of 5-choose-2 different combinations.  The questions that Jeff chooses to answer for the first part are independent of the questions that can be chosen for the second part, so the total number of answer combinations for the whole test is simply the product of the numbers of combinations for each part.

 

Thus, the total number of answer combinations for the test is

choose(6, 3)*choose(5, 2) = (6!/[3!(6-3)!])*(5!/[2!(5-2)!])

                                           = (720/[6*6])*(120/[2*6])

                                           = (20)*(10)

                                           = 200.

 

We can use a tool like www.wolframalpha.com to evaluate the factorials for us: http://www.wolframalpha.com/input/?i=%28+six+choose+3+%29+*+%28+five+choose+2+%29