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# Counting Problems, Combinations and Permutations

Lesson 3 of 7

## Objective: SWBAT use permutations and combinations to compute probabilities of compound events and solve problems.

## Big Idea: Why do combinations and permutations exist? To help us count all the possibilities of what might happen!

*75 minutes*

The desks in my room are usually arranged in groups of four, with students facing each other. I often remind my students to try to be the best resources they can for their classmates. My students know that I want everyone to discuss problems and try to figure things out together before getting me involved.

Today, I rearrange the desks into pairs, all facing the front of the room. **When students arrive, the notable change in routine shows students that today's lesson is going to be different than usual.** I'm going to give everyone a lot of notes, which is easier if everyone is facing me. Then students will work in these smaller groups of two instead of the typical groups of four. I consider it important use classroom set up productively. On some occasions, like today, making a move, even in small way like this, is a teaching move that I can use to make the most out of the lesson.

I should also note that like many of my other lessons, there might be more here in today's plan than I can finish in one 75-minute class period. That's ok -- this is the kind over-planning I do for introductory lessons, and then follow-up days are used for finishing work, reviewing, and brief assessments.

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Today's opener is structured just like those of the previous two lessons. There is a set of problems about drawing playing cards from a deck on each table (and the same picture of a deck of cards on the second slide of the lesson notes). I've prepared twice as many copies of the handout today, because students are seated in groups of two instead of four. Because there are fewer students per handout today, there's more responsibility on each student to try to figure out what's going on with these problems.

The real purpose of those opening problems in the previous two lessons was to get us to where we are today. It was important to review some basic probability ideas at the start of those two lessons. Today, we're going to dig into some new ideas, as students develop an understanding of why combinations and permutations exist and how they work. By using similar openers on the days leading up to this one, I hope to have gotten students thinking about these sorts of problems in the backs of their minds, so that we're not starting from scratch today. The ideas that students see now, about how drawing two cards at once is more complicated than drawing one at a time, will make more sense because foundations have been laid.

Just like in the last two lessons, I give students a few minutes to get started on their own. I greet everyone, circulate to see how everyone is doing, and encourage conversations among pairs of students. Unlike the last two, where the lesson following the opener was different from the rest of the class, this IS the lesson today. I'm going to use these problems to dig into vocabulary:

**simple vs. compound probability****dependent vs. independent events****combinations, and permutations**

I'll follow this structure:

- Run through solutions to these problems, fully but rather informally
- Give some guided notes about what we've just seen
- Introduce students to a few example problems
- Give students a chance to practice with what they've seen.

In each of these "rounds" of learning, students will see the same ideas, developing a deeper understanding of the content with each. On this, the first pass, I'm going to move quickly. In the next part of the lesson, I'll acknowledge that I've moved fast, and slow down.

I run our fast-paced look at these opening problems like I did the last two days, but I add new ideas as we go. I tell students that it's ok just to think about these problems. If they want to write everything down they can, but I want them to prioritize thinking about what they're seeing.

After everyone has had a chance to work for five minutes or so, I post the first problem of the opener: "What is the probability of drawing two diamonds?" I ask students what they've come up with, and I hope that some students will be able to get at the idea that drawing one card *changes* the deck from which the second card is drawn. In other words, the probability of drawing a diamond when we pull out the second card *depends* on what we drew first. I listen to student thinking and gauge where they're at. Then, before giving an answer to that first problem, I switch to slide #4, which asks students to imagine a bucket of six marbles: three black and three white. I ask, "What's the probability of pulling two black marble from this bucket?" Sure enough, everyone can see that we've got a 50/50 chance of pulling a black marble out first, but then what happens? "If I remove one black marble from the bucket, do I still have a 50% chance of pulling another one?" In this small sample space, it's pretty clear that now I have a 2-in-5 chance of pulling another black marble. I flip to the next slide, which now shows the solution to the first problem, and we compare this to the idea of the marbles. I say, "I had the same 13/52 - or 1/4 - chance of pulling one diamond from the deck, but after that, there were just 12 diamonds left in my deck of 51 cards." Just like yesterday, we can now multiply these simple probabilities to get the compound probability of pulling two diamonds.

Problems #2-4 are similar: drawing two aces (#2), the impossibility of drawing two of the same card (#3), and drawing two face cards (#4). The third problem, in which we see that there's a 1/52 chance of drawing a card on the first pick, but then a 0/51 chance of drawing it again, yielding a probability of 0 is a big moment for a lot of kids. I can promise some "Oh wow!"'s, as students really get the idea.

Then, problem #5 adds a new wrinkle. I post slide #9 with the problem and the solution that most kids will expect. Then, just as they're nodding like they get it, I post slide #10, which shouts, "That's not the whole story!" and it's back to the bucket of marbles on slide #11. "Imagine this bucket of marbles again," I say. "What are your chances of pulling one of each colored marble from this bucket?" Kids think about it and share their ideas. It's easy enough to think that we've still got that 1/2 chance of pulling a black marble, then a 3/5 chance of pulling a white. "But once those two marbles are in your hand," I say, "does it matter what order you pulled them out?" I ask everyone to consider the idea that we might pull the white marble "first" and then the black marble. Would that be a different situation?

I leave that question hanging, because we'll come back to it in the next part of the lesson. Then I post slide #12, which is the fifth problem again, with a way to think about the solution. We can calculate the probability of pulling a face card then an even-numbered card, and we can calculate the probability of pulling the same cards in the opposite order. As with the marbles, the result of either order would "look the same" once the cards are in our hands. So we can add the two probabilities together to get the answer to this question. Slide #13 and #14 help me frame the rest of the lesson: we need a way to determine the size of the ** sample space**, or the number of all possible outcomes, for problems like these. It's clear enough that there 12 face cards * 20 even-numbered cards will yield 240 possible face+even pairs...but 240 out of how many total pairs? Note that students have now seen the "answer" to this problem, but I'm still asking questions about it. That's because in addition to showing kids that combinations and permutations exist as mathematical ideas, I want to try to get them to an intuitive understanding of how they work. We'll circle back to each of these problems as we move ahead. Finally, the sixth question serves to reiterate how we're framing today's work. The two of clubs and three of spades constitute just one of all the pairs of cards we might draw, but once again, how many pairs are there?

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"Whew!" I say dramatically. "Do you guys have all that?"

I tell everyone that I'm impressed by their ability to think through these ideas with me, and that I know this is a lot of new information. So now we're going to slow down and take another look at what we've just seen. In order to do that, I've prepared this guided note-taking handout. I distribute it, then use slides #17 through 26 of the lesson notes to work through each section.

You can get the gist of how this goes by looking at my completed handouts: here is the front side and the back. The key is that all of these notes relate back to what I've just talked about in the opener. I'll share some of how I make those connections here.

We start with defining independent and dependent events, noting that pulling one card, replacing it in the deck, then drawing another is an example of independent events, while drawing two cards at once is an example of dependent events.

Next, we look at the bucket of six marbles. This is an opportunity to formally define "sample space" as the set of all possible outcomes in a probability experiment, then to use that phrase as we consider everything that might happen when we pull two marbles from the bucket. I show students how to use subscript notation to distinguish between each of the same-colored marbles, and then we use this notation to write out all pairs. We note that, really, the result (B1 + W1) is exactly the same as (W1 + B1) once the marbles are in our hand, so we don't have to write both. My goal is to develop conceptually the distinction between combinations and permutations before getting there formally. I ask students if they're satisfied that we've covered all possibilities, and the way we organize the sample space should help to make it clear that we have. When we're done, we have 15 possible outcomes. "So let's do the math," I say. "If there are six marbles in a bucket, that's six possibilities for what we might pick first, which leaves five possibilities for what we'll pick next." We note that that should yield 30 possibilities. "But, order doesn't matter. If picking B1 then W1 is really the same as picking W1 then B1, we can divide those 30 possibilities by two, because each outcome is accounted for twice when we say six times five." And of course, we see that 30/2 gives us the 15 outcomes we've written out by hand. At this point, we've defined combinations intuitively, and I hope that students will recognize this reasoning when I show them the formula for computing combinations.

We repeat the same reasoning in the next section of the guided notes, which gets us to the point of understanding that there are 1,326 different pairs of cards that can be drawn from a deck of 52 cards. Here, we can take it back to the fifth problem of the opener, the probability of picking one face card and one even-numbered card from a deck. There, we saw that

**(12/52)*(20/51) + (20/52)*(12/51) = 480/2652**

Which, partially reduced, gives us 240 pairs that meet our conditions out of a total of 1326.

The next note is about pairs of letters. When we're forming "words", the order does matter, and "AB" should be considered as different from "BA". Once again, we sketch out the entire sample space, then we do the math. There are five possible "first letters" followed by four possible "second letters" yield 20 possible "words". We'll use this idea to distinguish between combinations and permutations.

Finally, we get to the most important question of the day: "Why do combinations and permutations exist?" At this point, kids haven't seen those words yet. I say that this is really what today's lesson is about, and I'm going to show everyone how to use these tools, but first I want everyone to understand why we've got them: **combinations and permutations exist to count all the possible outcomes - to define the sample space - of a compound event**.

From there, it's pretty straight notes. I define factorial, because kids will need that to understand the formulas for combinations and permutations. In addition to showing students what factorial means, I also show them that

**8! = 7! * 8, or more generally, than n! = n * (n-1)!**

This provides a scaffold for how we'll think about canceling impossibly long lists of numbers when we actually use the formulas students are about to see.

I define **combination** and **permutation**, then we use the combination formula to see that it yields the same result that we achieved on the front of the handout: there are 1326 possible pairs of cards you can draw from a deck, when we assume that the order doesn't matter. I try to get students to see the structure of in this application of the formula. After we start with

**52! / (2!*(52-2)!)**

we see that

**52!/50! simplifies to 52*51**

which is then divided by 2!, which is really just 2, and those are exactly the same as the calculations we used on the front of the guided notes.

Finally, one more time, we think about pulling our face card + even-numbered pair from the deck, and we see the probability of that event. For any specific pair, we note as an answer to the last problem from today's opener, there's a 1/1326 chance of drawing it.

**Four More Practice Problems: **Guided notes are followed by just a few more "unguided notes" in the form of example problems: one is a classic permutation problem, then three are letter arrangement problems that are popular on standardized tests. I post each problem on the board and tell students to take notes on what they see. We apply the permutation problem to Example Problem #3 and #4, then I show students how to account for repeated letters on #5 and #6.

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#### Permutations Row Game

*30 min*

That was about as long of a "lecture-style" lesson as I'll ever give - and I really believe in doing that once in a while, because all modes of teaching have their place. With that in the bag, it's time for kids to practice.

You'll recall that students are seated in pairs today. Another reason for that is because the practice exercises I distribute now come in pairs. There are two versions of the problem set, but both have the same solutions. In the thriving community of math teachers sharing best practices online, this is called a Row Game, and you can read about on Kate Nowak's excellent blog about teaching, which is where I originally heard about this idea a few years ago. Kate also put out a call for other row games, because they're such a powerful teaching tool, but rather time-consuming to make, and now she curates a collection of them for dozens of math topics here.

As I describe in this video, the biggest deal about row games is that they're self checking and they naturally encourage students to talk about and to support their own work. My added purpose with this set of problems is to help students see relationships between the results of related combinations and permutations. For example, that for any combination of n objects, k and n-k are the same. Or, for any permutation of n objects, n and n-1 are the same. So why does this happen?

In the classroom, there are various ways to implement this: you can tell kids right away that the answers to each set are the same, or you can you wait a little while to make the big reveal. Personally, I prefer the latter, because it's fun to tell kids that they have different versions at first, before delivering the punchline. Either way, it's pretty straightforward: distribute the handout, and let kids get to work.

I note at the start of today's lesson that this is a lot of ground to cover in 75 minutes. With some classes, it's possible - I've done it - but it's also normal for this lesson to extend into the next one. I make sure I'm ready to do all of this today, and if we only get so far, I have the chance to reflect on that and how to make the most of our next meeting.

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