Teaching Counting and Combinatorics

I recently started a unit on counting and combinatorics with my 5th – 8th grade group.

I’m not a fan of having kids memorize formulas, and I’m even less of a fan of handing formulas down from “on high” for kids to randomly plug things into. If they’re going to get formulas, they’re going to see where they come from. (In theory, then, if they forget the formula, they can re-derive it, or just work without it.)

Generally I teach permutations and combinations without using formulas at all. We just use “logic” or “the multiplication principle” or “the counting principle” (particularly for permutations). If we have 5 boys and 5 chairs in a row, and want to find out how many different ways the boys can arrange themselves in the chairs, we have 5 choices of who can sit in the first chair, 4 choices of who can sit in the second chair, then 3, 2 and 1, so we have 5 x 4 x 3 x 2 x 1 possible orders. Why do we multiply them? I use many different ways to illustrate this, hoping that one will click for each student. I draw it out in a tree diagram, write out all the possibilities to show how the permutations multiply with each choice, and we talk about how for each choice of who sits in the first chair, we can then consider all of the choices of who sits in the second chair, and for each pair of first/second boys, we can look at all the possible choices of who sits in the third chair, and then we talk about how “for each” is a key phrase that indicates multiplication — if there are 9 students in the classroom, and I want to bring 3 cookies for each of them, how do I figure out how many cookies to bring?

This works great for Permutations. The kids get it. There’s nothing to memorize. We’re all happy. Even when we do permutations where not every item is used, it makes sense. If we have 30 kids in a science fair, and want to figure out how many different possible ways we can award 1st, 2nd and 3rd place, we have 30 choices of who gets 1st place, 29 choices of who gets 2nd, and 28 choices of who gets 3rd. 30 x 29 x 28 (and I’ll either have them leave the answer like that or allow them to use calculators to compute it).

Combinations are trickier to do this way. But I still prefer to teach them without formulas. Figure out the number of permutations, then figure out how many times each correct answer (each combination) is counted, and divide by that number. So, if we are trying to choose 4 students out of a class of 10 for a MATHCOUNTS team, how many teams can we make? Well, we have 10 choices of who to chose first, 9 choices of who to chose second, then 8 and 7. So we have 10 x 9 x 8 x 7 possibilities. But wait, it doesn’t matter who we chose first, second, third or fourth. So Andrea, Beth, Carrie, David is just the same as Beth, David, Carrie, Andrea. How many different ways would we have chosen this same team? Well, we have 4 choices of who we picked first, 3 choices of who we picked second, then 2 and 1, so each team is included 4 x 3 x 2 x 1 times. So our answer is (10 x 9 x 8 x 7) / (4 x 3 x 2 x 1) or 210. (Fortunately, that works out the same as doing it the “standard” way of using the formula for 10C4 “10 choose 4”, confirming that I didn’t mess it up.)

Well, every few years (often enough for each kid coming through to see it once in his/her time in the middle school program) I do teach the formulas. This is one of those years.

To teach the formulas, first I have to teach factorials. Factorials are fun. The kids love them. They quickly catch on that if they know 4! they can compute 5! by just multiplying 4! by 5. Dividing factorials is fun too. Things cancel like crazy, and hard problems become easy, once they understand cancellation. My goal in teaching them to work/play with factorials is not only for them to realize that 8! / 5! = 8 x 7 x 6, but to recognize when they see 8 x 7 x 6, they can write it as 8! / 5! (which is more of a stretch than the first piece). And then we’re ready to roll…

And so I offer a number of beginning permutation and combination problems as a resource for other instructors, or those exploring the topic on their own. Additional problems with answer key are available from EdHelper

1. There are three balls on the table; each is a different color. How many ways are there to arrange the balls in a line so that no two arrangements have the same color sequence?
2. Mathmom needs to choose 4 students to be on the MATHCOUNTS team. There are twenty students she can choose among (in her dreams!). How many different teams can she make?
3. A certain lottery is played by choosing your own set of 6 winning numbers from among the numbers 1 through 49. How many possible such combinations are there?
4. There are 12 boys and 14 girls in Ms. Brown’s class. She needs to choose 3 boys and 3 girls for a debate team. How many different teams are possible?
5. (a) In a standard deck of 52 cards, how many different 7 card hands are possible?
   (b) If you separate the cards into suits and keep only the hearts, how many different collections of 5 hearts are possible?
6. Betty is about to order dinner at her favorite restaurant. She will order a drink, an appetizer, a main course, 2 different side items, and a dessert. If there are 10 choices for drinks, 5 appetizers, 6 main courses, 8 side items, and 5 desserts, in how many ways can Betty order her meal?

Here are some more from the MATHCOUNTS Counting/Combinatorics Stretch, in their 2002-2003 School Handbook:

1. Six points are drawn on a circle. How many distinct convex pentagons can be drawn using only these points as vertices?
2. A nursery employee wishes to plant six Golden Delicious apple trees and two Bartlett pear trees in one row. How many distinct arrangements are possible?
3. A teacher has made ten statements for a True-False test. Four statements are true and six are false. How many distinct answer keys could there be for the test?

When you’re ready for a little more challenge, try this or this.

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