Inside Collection: Advanced Algebra II: Teacher's Guide
Summary: A teacher's guide to combinations.
Once again, this one is lecture, without an inclass worksheet, walking through a series of questions.
Suppose you have a Daisy, an Iris, a Lily, a Rose, and a Violet. You are going to make a floral arrangement with three of them. How many possible arrangements can you make?
Based on yesterday, it’s tempting to answer
So, let’s start the way we did yesterday: list them all. Give the class a minute to do this. Remind them, as always, that it’s best to be systematic to make sure you list every possibility exactly once. Here’s my list.
First Flower  Second Flower  Third Flower  Arrangement  

Daisy 


Iris 


Lily  Rose  Violet  Lily, Rose, Violet 
10 items in all. I generated the list by using the same kind of systematic approach I used for the permutations, but always moving forward in the list: so after “Lily” I’m allowed to list “Rose” and “Violet,” but not “Daisy” or “Iris.”
This turns out to be such a common and important operation that it gets its own name: “choose.” We would say “5 choose 3 is 10,” sometimes written
The four Beatles are John, Paul, Georg, and Ringo. Suppose you are going to put photographs of three of them on your door. List all the possible combinations. How many are there?
Give the class a minute to list—they should make a diagram like the one I made above!—and count. When they are done, you can point out that the answer (four combinations) is very obvious if you look at it backward: each combination leaves exactly one Beatle out. There is a very important insight here, which can also be applied to the flower example: each arrangement listed leaves exactly two flowers out. So 5 choose 3 (“which flowers should I include?”) is the same number as 5 choose 2 (“which flowers should I leave out?”).
Now we’re going to try something with bigger numbers. A drama teacher looks out at a class of 30 students, and wants to choose 2 of them to run a scene. How many possible pairs of students are there? As before, when the numbers get this large, we can’t reasonably list all the combinations: we need an algebraic way of figuring out how many there are, without actually counting them all.
To start off, let’s turn this combinations (“order doesn’t matter”) problem into a permutations (“order does matter”) problem. The teacher wants to choose one student to play the Child and one to play the Dog. In this version of the problem, “John plays the Child and Susan plays the Dog” is different from “Susan plays the Child and John plays the Dog,” and should be counted separately. So it is a straightforward permutations problem. There are 30 possible actors for the Child, and for each of those, 29 for the Dog.
Now let’s return to the original problem, how many pairs of students are there? In this problem, “JohnSusan” and “SusanJohn” are the same pair. The key insight here is that, when we ran the permutations problem, we counted each pair twice. So the answer is
Take this slow and easy. This is a very general approach to combinations problem. First, you solve the (easier) permutations problem. Then you ask, “How many times did I count every group?” and divide by that.
Ask the class to try this approach on the original (flower) problem. They should answer three questions.
Lay out the process, then have them work the problem. Many of them will get stuck on step (2), incorrectly thinking that we counted each permutation three times. In fact, we counted each one six times:
daisy–iris–rose, daisy–rose–iris, iris–daisy–rose, iris–rose–daisy, rose–daisy–iris, rose–iris–daisy
The permutations are
So...why six times? This is the last question, and it’s a hard one. When we were choosing two items, we counted each combination twice (John–Susan, Susan–John). When we were choosing three items, we counted each combination six times. What’s the pattern? Give them a minute to think about this. Then make sure they understand that it is, in fact...another permutations problem! The list of six I gave above is simply the number of ways you can arrange three items, or (3!). You would divide by (4!) for 4 items, and so on.
“Homework: Permutations and Combinations”
A sample test, and real test, and you’re done!
"This is the "teacher's guide" book in Kenny Felder's "Advanced Algebra II" series. This text was created with a focus on 'doing' and 'understanding' algebra concepts rather than simply hearing […]"