Tag Archives: counting

Class Opener – Day 14 – Many, Many Meals

Starting one of my favorite units of the year: permutations, combinations and the binomial theorem. I stumbled upon this article proclaiming the arrival of 140 Million Burger Combinations, heading to New York City – and when I see combinations in the news, it’s time to investigate.

burgersThe article comes from 2010 and heralds the opening of an exciting new direction in burger construction (which has since closed). The website for 4food is sill active for now, and has a cool applet where you can build your own burger. There are many decisions to be made, and some exotic choices (a scoop of roasted brussels?).  I offered students the opportunity to create their own burger, with stations set up on my laptop and on my ipad. Much pro-con debate over the appropriateness of burger pickles ensued!

The choices 4food offers (or, offered, as they are closed…) were summarized by my students:

  • 4 choices of bun
  • 4 choices of “add-on”
  • 10 condiment choices
  • 5 cheese choices
  • 3 “slice” options
  • 12 “scoops”
  • 6 patty options

But multiplying these numbers does not get us near 140 million…so what gives? My classes will explore this problem deeper in the coming days, but for now some seeds have been planted. Soon, we will consider the possibility that you could select multiple condiments, cheeses, and scoops, and work to derive the final count.

This scenario brings to mind a counting principle challenge I have provided classes in the past:

The Tastee Donut Shop charges eighty-nine cents for its Mix N Match selection, which allows you to select any three doughnuts from among the following varieties: plain, maple, frosted, strawberry, blueberry, vanilla, chocolate, glazed, and jelly.  How many different Mix N Match selections are possible?

Here is a printable version of this problem you can share with classes.

I enjoy this problem because students need to think beyond a one-step counting problem. This challenge is more sophisticated than many worksheet problems in that we need to consider a number of possibilities – could a customer buy 3 of the same donut? 3 different donuts? 2 and 1? In the end, the solution comes down to the sum of 3 distinct possibilities, each more challenging:

  • Buy 3 of the same donut (easy): 9 ways
  • Buy 3 different donuts (medium): compute 9 choose 3
  • Buy 2 of one type, a 1 of another (hard): we need to pick two flavors. But picking 2 glazed and 1 jelly is distinct from 2 jelly, 1 glazed. Order matters. Compute 9 pick (permuation) 2.


Class Opener – Day 12 – How Big is Big?

We’re coming to the end of our first unit of the year on basic probability, and headed towards the fun world of counting principles, including permutations, combinations and the binomial theorem.  To review ideas regaring factorial and size, students were faced with the following question on the board:


Many students ignored the exclamation point right off the bat, giving replies like “it’s a little bigger than 51”, or “pretty big”, until a student realized that I clearly meant factorial here.  This genrated classroom discussion about what factorial meant, and some side discussion about how big a number this could be, including some calculator experimentation. We’re off to a good start!

But just HOW big is this number?  To get students thinking, I asked them to consider what a quantity that big could represent, being as creative (within reason) as they like. Some of the responses were awesome fun.  Did you know Kanye had THAT much swag?

2014-09-17_0011 2014-09-17_0010

To finish this opener, I played one of my favorite clips: from the British panel show QI, Steven Fry uses a simple deck of cards to do something never before done by man! I’ve dicsussed this clip on the blog in the past, so visit there for more info regarding this card shuffling experiment.  Enjoy.

Counting Principles and “The Price is Right”

I have a confession to make…..it’s really quite embarassing…

I’m a Price is Right nerd.

{sigh} wow, feels so good to get that off my chest.

priceSince I was a little kid, I loved watching the Price is Right.  I know all the games, many of the prices, and of course can name the “back in the day” models without batting an eye (Dian, Holly, Janice).  I even made the pilgramage to Television City a few years back to see Bob Barker in his last years of hosting.

Now, as a stat teacher, I have used a number of Price is Right games in the classroom as probability lessons.  I’ve given a number of talks using Plinko as the centerpiece of a lesson.  Almost all of Price’s games have some probability element.  Here are a few games you can discuss on your classes, starting with basic ideas, and moving up to more complex counting principles.


DoublePricesThis is the most simple probability game on the show.  The contestant is shown a prize, and two possible prices for the prize.  If the contestant guesses blindly, they then have a 50% of choosing the correct price, and winning the prize.  In all of these games, the pricing aspect is a “clue” to the player, which hopefully increases their chance of winning.  But in all of these examples, we will look at the games as random chance experiments.


This game is only slightly more  difficult than Double Prices.  In it, the contestant is shown three prizes, each with a price tag, one of which is an incorrect price.  If the contestant identifies the incorrect price, they win all 3 prizes.  Given random guessing, a contestant has a 1/3 chance of winning.


Here’s where we start looking at some more interesting counting methods.  In this game, a contestant can win a large prize and a smaller prize by correctly giving the price to the smaller prize.

SafeThe smaller prize has 3 distinct digits in its price, which the contestant is given.  They must place the digits in the correct order to find the price.  With 3 digits to place in order, we have 3! = 6 possible prices.  BUT, in this game, the price always ends in zero (they don’t tell you this, but it’s always true), which means this is essentially a 50-50 game.  For example, if the 3 given digits are 0, 9 and 5 – then there are only two possible prices, $950 or $590.


A contestant has a chance to win a prize with 4-digits in its price.  A “dummy” price is given, like $5447, and the contestant must determine if each digit in the actual price of the prize is higher or lower than the digit in the dummy price.

BonkersTo make their guesses, the contestant places markers either above or below each digit in the dummy price.  If they are correct with all 4 digits, they win the prize.  If they are wrong, they can go back and make changes.  A total of 30 seconds is given to make as many guesses as they can, running back and forth between the game board and the guessing buzzer.

Each digit has 2 outcomes, higher or lower.  Since there are 4 digits, there are 2 x 2 x 2 x 2 = 16 different outcomes.  With only 30 seconds to make guesses, this game often comes down to how well the contestant uses their time to maximize the number of guesses out of the 16.


In this game, a contestant can win 3 different prizes: one with 2 -digits in its price, one with 3-digits, and a car with a 5-digit price.

The contestant is first shown 3 digits, 2 of which make up the price of the first prize.  The goal is to use as few “chances” as possible to get the correct price, which allows the player to move on the the next prize.  With 3 digits to choose from, there are in theory 3 x 2 = 6 possibilities.  But like Safecrackers above, the price will always end in zero; so there are only 2 real choices.

For example: if the given digits are 0, 3 and 5, then the only real possibilities are $30 or $50.  Often, the frustration in this game is associated with contestants who don’t know that all the prices always end in zero…which causes me to yell at my television.

ChancesMoving onto the next prize, 4 digits are given, 3 of which make up the next price.  In theory, there would be 4 x 3 x 2 = 24 choices here.  But again, given that the price will always end in zero, there are only 3 x 2 = 6 viable choices.

The goal in this game is to economize your Chances, so that you have a good number left to play for the car.  5 digits are given, all of which must be used in the car price.  In theory, this gives 5 x 4 x 3 x 2 = 120 choices.  But there are 2 ideas at play here: the price will end in zero AND the price will always begin wth 1 or 2.  Depending on the assortment of digits given, this reduces the number of possible prices a player needs to assess.


DisksThis is one of the most difficult games on the show to win, and is often played for a luxury car.  The game is played with 8 wooden disks, which are placed in a bag and shaken.  5 of the disks have number on them; digits in the price of the car.  The other 3 disks have red “strikes” on them.  A disk is drawn, and if a number is drawn, the player must tell which position in the car’s price the number represents.  If they are correct, the disk is removed from circulation.  And if the player is able to complete the price of the car before drawing all of the strikes, they win the car.

This game is a bit tricky to analyze, because often numbers are drawn repeatedly, as the contestant tries to narrow down where digits go in the price.  For an in-class analysis, let’s assume that the player knows the price, and is trying to just draw the digits.

If you are assuming perfect play, then we could simply list all of the possible ways to arrange the numbers and strikes.  Let N indicate a number and X indicate a strike.  So you could have:

NNNXXNXN (Loss, since 3 X’s occur before all N’s are drawn)
NNXNNXXN (Loss)…..

The number of ways to play the game is then 8C3, which is 56.  Now, this may seem like a small number, but I am treating the numbers in the bag as similar objects, since we are assuming the contestant places them correctly.

BoardNow, I could go through and count the number of these 56 that produce wins, but I think it might be simpler than that.  The game really comes down to the last item on the list.  If the last item is an X, then you will have won the game.  If the last item is an N, then you have lost.  The chance that the last item is an X 3/8, so the probability of winning the game, assuming perfect play, is 3/8.

So, this game only has  37.5% chance of victory IF the contestant plays perfectly.  Add in that often the contestant often must struggle to position the digits, and you see why the game is so difficult to win.


LineUpThis game is played for a car, with 3 smaller prizes as well.  The 3 smaller prizes have prices with 3 -digits, 2 digits, and 3-digits.  The prices of these smaller prizes are used to fill in the middle 3-digits in the price of the car, as shown on the game board here.  This give 3 x 2 x 3 = 18 possible outcomes.  The nice part about this game is that the contestant is given a second chance, and is told how many digits they have correct after the first attempt.  If the player needs their second attempt, it would be interesting to analyze how many choices of the 18 remain, given that they have 0, 1 or 2 of the digits correct.

There are plenty of other games on the show which also have basic counting prinicple ideas worth exploring.  Some quick hits:

  • Balance Game – how many different ways can 2 bags be chosen from the given 3.
  • Dice Game – how likely is it to roll correct digits?  When should I choose higher or lower?
  • Golden Road – how difficult is it to advance to and win the big price at the end of the Golden Road?
  • Let Em Roll – how many different ways can the 5 special dice be rolled?
  • Make Your Move – how many different possibilities exist for moving tre sliders?
  • Race Game – how many ways can the price tags be placed?
  • Take Two – how many ways are there to choose 2 prizes from the given 4?

And some in-class ideas:

  • Let students choose a game to analyze.  Create a poster and share with the class.
  • Start each day of your probability unit with “a Game a Day”.  Start with the easy games, and move it to the more complex ones.
  • Have a contest where students design there own pricing games.

Thanks to my friends at Golden-Road.net for the fun pictures.

Encourage Generalization and Communication with these Math Challenges

A comment from a recent post of mine on differentiation asked what I do with students who complete tasks early.  In every course I have ever taught (usually Algebra 2, Prob/Stat, Algebra) I have used weekly problem-solving challenges, no matter what the level of student.  Often, my intent in these problems was to develop written communication skills in mathematics, and have students begin to reflect upon their own writing style.  Have students complete the challenge, critique their writing and provide a path for improvement, and have students turn in their best works as part of a portfolio at the end of the semester.

In this post, I focus on tasks involving counting, number theory, and algebra.  The problems here are ones I have assigned, graded, revised, and enjoyed over the past 15 years.  I’ll have some more tasks in a later post.  Click the title to download the PDF document.

PATHS:  How many ways are there from start to finish?  I love this problem, because there are multiple ways to approach it.  Combinations give the result, but there is also a Pascal’s Triangle approach, or as a permutation with identical items.  And Polya’s strategy of starting small and working your way up is key to this one.

SOME ZEROES:  I have always enjoyed giving this problem, as you you can have rich conversations about simple number facts and the commutative property.  And the student explanations will range from the ridiculous to the intriguing.  When I started giving this problem 15 years ago, some students would use Excel to try to simply compute the answer, which often “broke” Excel and gave a wrong answer.  I have given up on trying to follow the technology, and have given a similar problem as a follow-up on a quiz.

AVERAGE SPEED:  One of my favorites because of its simple premise, and a result that is counter-intuitive.  Also, can be easily differentiated.  For some students, choosing distances and testing serves as a good starting point, while students with advanced algebraic skills can dive right into the abstract.

LAST DIGIT:  A premise simple enough for grade 6, yet complex enough to challenge older students if you ask for a general formula.  It’s also easy to adapt this problem and use it as an opener for class.

INVERSES:  In this challenge, students must find a matrix which is its own inverse, of which there are many, many possibilities.  How will your students ensure that their matrix is unique?

Feel free to contact me for solutions, tips, or more ideas.

Shuffle Up and Deal, and Deal, and Deal….

Take a look at the video below, where Stephen Fry, host of the British panel show QI, alleges to do something never done before:

EDIT:  Seems as though the YouTube folk removed the video clip.  Try this link instead, and let’s hope it lasts:

QI Card Shuffling Clip

What a great “hook” for a probability or counting principles unit. Some thoughts about how to use this in your class.

1.  The result given in the video can be expressed as

If we were to shuffle the cards once every second, with each arrangement occurring once, how long would it take for use to go through every possible arrangement?  A neat example of something “big”, which is accessible and easy to discuss.

2.  The online poker site PokerStars is celebrating it’s 10th anniversary, and is offering a prize to the players who participate in their 100 billionth hand (assumed to occur around the 10th anniversary).  At this rate, how long should it take PokerStars to go through all possible arrangements?

3.  As an extension, challenge your class to find the number of possible arrangements of a deck of Pinochle cards. Cards The main differences with a Pincohle deck are that there are only 48 cards, and each card (like the 9 of diamonds) appears twice in the deck.  This problem introduce the idea of permutations with duplicate items.  In this case, we start with 48!, but then must divide out the double-count which occur with the repeat items.  We divide by two for each instance of a repeat item, and the number of permutations is given by:

4.  Let’s evaluate Mr. Fry’s conjecture:

Were you to imagine if every star in our galaxy had a trillion planets, each with a trillion people living on them, and each of these people had a trillion packs of cards, and somehow they managed to shuffle them all a thousand times a second and they had been doing that since the Big Bang, they would just now begin to repeat shuffles

To summarize, we are looking at this many shuffles per second:

Dividing by the number of possible shuffles yields:

The number of seconds in each year is given by:

Which implies we would have to shuffle for this many years:

Great exercises in laws of exponents for your students.  Share your thoughts and ideas about this fascinating video!