Chapter 17Closing Observations

Arthur B. Powell

In the previous 16 chapters, we have witnessed ordinary students develop extraordi-nary mathematical ideas, forms of reasoning, and heuristics. Extraordinary are thesestudents’ accomplishments since their mathematical behaviors emerged not fromquickly parroting rules and formulae but rather from deliberately engaging their owndiscursive efforts. As Speiser (Chapter 7, this volume) notes, these students builtfundamental mathematical understanding, over time, through extended task-basedexplorations. They created models, invented notation, and justified, reorganized, andextended previous ideas and understandings to address new challenges. That is, theyperformed mathematics: created mathematical ideas and reasoned mathematically.These behaviors – ideating and reasoning – are fundamental human activities andhow they occur in the realm of mathematics, specifically elementary combinatorics,is what this book contributes.

Internationally, a community of mathematics education researchers has recog-nized this how question as substantially important. In January 1983, David H.Wheeler (1925–2000), the founding editor of the international journal, For theLearning of Mathematics, sent a letter to 60 or so mathematics educators invit-ing them to engage a daunting task: “suggest research problems whose solutionwould make a substantial contribution to mathematics education” (Wheeler, 1984,p. 40). The varied and thought-provoking responses of more than 15 educators werepublished, some in each of the three issues of the fourth volume of the journal. OnWheeler’s mind was the famous example of the 23 problems from various branchesof mathematics that David Hilbert (1862–1943) announced in an address deliveredto the Second International Congress of Mathematicians in 1900 at Paris (p. 40)and predicted that “from the discussion of which an advancement of science may

A.B. Powell (B)Department of Urban Education, Rutgers University, Newark, NJ, USAe-mail: powellab@andromeda.rutgers.edu

201C.A. Maher et al. (eds.), Combinatorics and Reasoning, MathematicsEducation Library 47, DOI 10.1007/978-0-387-98132-1_17,C© Springer Science+Business Media, LLC 2010

202 A.B. Powell

be expected” (Hilbert, 1900, p. 5).1 Of all the published responses to Wheeler’schallenge, Tall (1984) offered the briefest list of what he considered to be “the cen-tral questions”: (1) how do we do mathematics? and (2) how do we develop newmathematical ideas? (p. 25, emphasis added).

Hilbert’s 23 problems contributed to more than a century of vigorous, fruitfulresearch activity in physics and mathematics.2 Similarly, considered responses toTall’s two questions require substantial research efforts in different environmentsover extended periods of time. For researchers in mathematics education to enter-tain these questions, we must find ways to observe what learners do as they domathematics as well as to describe and analyze how they develop their mathematicalideas.

It bears noting that 8 years after Tall issued his central questions, Davis et al.(1992) similarly challenged mathematics education researchers to study the emer-gence among learners of what lies at the core of mathematics: mathematical ideas.Expanding on Tall’s second question, Davis noted that “very little research in math-ematics education has focused on the actual ideas in students’ minds or on howwell teachers are able to identify these ideas, interact with them, and help studentsimprove on them” (p. 732). The chapters of this book have presented rich descrip-tions and analyses of actual ideas that students built in the realm of combinatorialreasoning. This work has implications for teaching both in the design and sequenceof effective tasks and in demonstrating how teachers could productively interactwith student ideas.

The global picture depicted in the chapters of this book underscores the needfor time to think deeply and discursively. A special issue of Educational Studies inMathematics collected several analyses concerning discourse in mathematics class-rooms. Commenting on these studies, Seeger (2002) wonders about the possibilityof a grand, panoramic theory of learning. In arguing for a comprehensive theoryof mathematics education, he suggests that such a theory needs to embrace fourmetaphors of learning that form the axes “social – individual” and “construction –acquisition,” and represents them in a two-by-two grid (p. 289).3 In addition, Seegerfurther suggests that “theoretical work has to be balanced by the systematic develop-ment of focal problems for practice, theory, and research in mathematics education”

1According to Gray (2000), Hermann Minkowski (1864–1909), whose metric concept (order-pgeometry) provided the theoretical foundation for non-Euclidean, taxicab geometry, was a closefriend of Hilbert and urged him to accept the invitation to speak at the Congress: “Most alluringwould be the attempt to look into the future, in other words, a characterisation of the problems towhich the mathematicians should turn in the future. With this, you might conceivably have peopletalking about your speech even decades from now” (as quoted in Gray, 2000, p. 1).2See Grattan-Guinness (2000) for a critical appraisal of “the range of Hilbert’s problems againstthe panoply then evident in mathematics.”3Here Seeger (2002) differs from Sfard’s (1998) theorization in which she argues for twometaphors – acquisition and participation or construction – that conceptualizes perspectives onlearning and in which she claims that though complementary they are mutually not amenable tocritique.

17 Closing Observations 203

(p. 289). He proposes two focal problems for mathematics education, one concern-ing ecological validity and representation and the other referring to the question oftime and change.

Besides epistemological concerns, the question of time and change also concernsmethodological issues. Building ideas and understanding are certainly temporal andunbounded. Consequently, there are complex judgments an investigator has to makewhen inquiring into what learners build, understand, or acquire from a discussion orlesson on a particular issue. When does an investigator examine what learners say,do, and write? Should these actions be examined in the immediate proximity of thediscussion or lesson, in some other, more distinct time, or in some combination ofthese times? Ball and Lampert (1999) raise somewhat similar questions in a studyof teaching practice.

Epistemologically and methodologically, this book contributes to understandingthe relation between time and development. As outcomes of individual and collec-tive constructive actions over the course of the longitudinal study, the participantsbuild ideas, reason, and employ heuristics to resolve various tasks. They revealand make salient the important relationship between time and development. Theprocesses by which the participants build their ideas evidence an epistemologicalreality: knowledge construction is often a slow process. Mathematical ideas do notdevelop instantaneously and robustly but rather emerge slowly and in their nascentstate are rather fragile. Ideas dawn and mature over time. To loose fragility, amongother things, ideas need to be reflected on deeply, presented publicly, submitted tochallenge, available for negotiation, and subject to modification. That is, the essenceof developing and understanding mathematical ideas is often a protracted, iterative,and recursive phenomenon, occurring over more time than is usually appreciatedor acknowledged in practice in classrooms and in reports in the literature (Pirie &Kieren, 1994; Seeger, 2002). If learners are to develop deep understandings that areless fragile and more durable than is often witnessed by teachers in schools, theyneed to be offered extended periods of time to wrestle with a problem as well asto debate and negotiate heuristics, to articulate and justify their results, and to havetheir ideas challenged and then defend or modify their ideas.

If we agree that students must be actively and purposely engaged in their learningso that they can take ownership and be proud of their accomplishments, we need tocreate opportunities for this to occur. For example, strands of investigations can beintegrated into the regular curriculum as enrichment. When we eliminate the pres-sure of testing and grades, students can invest in thinking and reasoning for its ownsake and for the intrinsic rewards that knowing deeply entails. Perhaps every severalweeks, within particular strands, investigations can be revisited, and students canbring their more recent, accumulated knowledge to a more sophisticated examina-tion of earlier solutions, and thereby extend their knowing. A focus on justificationas a strand of school mathematics has great potential for building a solid foundationfor the later study in many fields and certainly of mathematics. A focus on rea-soning and sense making is an important requirement for a productive, responsiblecitizenry. Questioning, challenging, analyzing, revisiting all lead to better ways ofknowing. Can we as educators meet the challenge of educating thoughtful students

204 A.B. Powell

who are motivated by sense making and the critical review of ideas? The challengeawaits us.

The Video Mosaic Collaborative at Rutgers University provides a mechanismfor the ongoing building and sharing of knowledge. We invite readers to visitour website at http://www.video-mosaic.org/, view the videos and accompanyingobjects that provided the data for this book and join our expanding community ofresearchers by providing additional study and analysis.

206 Appendix A: Combinatorics Problems

plates. The cups and bowls are blue or yellow. The plates are blue, yellow,or orange. Is it possible for ten children at the party each to have a differentcombination of cup, bowl, and plate? Show how you figured out the answer tothis question.

Each of the two cup choices can be matched with each of the two bowl choices,and each cup-bowl pair can be matched with any of the three different platechoices. Therefore, there are 2×2×3 = 12 possibilities. Therefore, yes, it ispossible for ten children at the party each to have a different combination ofcup, bowl, and plate.

5. Relay Race (October 1991, Grade 4) – This Saturday there will be a 500-mrelay race at the high school. Each team that participates in the race must have adifferent uniform (a uniform consists of a solid colored shirt and a solid coloredpair of shorts). The colors available for shirts are yellow, orange, blue, or red.The colors for shorts are brown, green, purple, or white. How many differentrelay teams can participate in the race?

There are four choices for shirts and four choices for shorts, so there are 4×4= 16 ways to make uniforms. Sixteen different relay teams can participate.

6. Five-Tall Towers (February 1992, Grade 4; December 1997, Grade 10) – Yourgroup has two colors of Unifix cubes. Work together and make as many differ-ent towers five cubes tall as is possible when selecting from two colors. See ifyou and your partner can plan a good way to find all the towers five cubes tall.

There are 25 = 32 towers five cubes tall.

7. Four-Tall Towers with Three Colors (February 1992, Grade 4) – Your group hasthree colors of Unifix cubes. Work together and make as many different towersfour cubes tall as is possible when selecting from three colors. See if you andyour partner can plan a good way to find all the towers four cubes tall.

Since there are three choices for each of four positions, there are 34 = 81possible towers that are four cubes tall when selecting from three colors.

8. A Five-Topping Pizza Problem (December 1992, Grade 5; December 1997,Grade 10) – Consider the pizza problem, focusing on the number of pizza com-binations that can be made when selecting from among five different toppings.

There are 25 = 32 different pizzas.

9. Guess My Tower (February 1993, Grade 5) – You have been invited to par-ticipate in a TV Quiz Show and the opportunity to win a vacation to DisneyWorld. The game is played by choosing one of four possibilities for winningand then picking a tower out of a covered box. If the tower you pick matchesyour choice, you win. You are told that the box contains all possible towers thatare three tall that can be built when you select from cubes of two colors, red,and yellow. You are given the following possibilities for a winning tower:

Appendix A: Combinatorics Problems 207

All cubes are exactly the same color.

There is only one red cube.

Exactly two cubes are red.

At least two cubes are yellow.

Which choice would you make and why would this choice be better than anyof the others?

In order to decide which is the best choice, we need to find the probability ofeach choice. The total number of 3-tall towers is 8. The probabilities are:

All cubes are exactly the same color: There are two ways (all red or allyellow). The probability is 2÷8 = 0.25.

There is only one red cube: There are three ways; the red cube can be on thetop, in the middle, or on the bottom. The probability is 3÷8 = 0.375.

Exactly two cubes are red: This is the same as saying exactly one cube isyellow. The probability is the same as for exactly one red cube: 3÷8 = 0.375.

At least two cubes are yellow: This is equivalent to saying that either exactlytwo cubes are yellow or exactly three cubes are yellow. As discussed above,the probability that exactly two cubes are yellow (the same as the probabilitythat exactly two cubes are red) is 0.375. Since there is one way for exactlythree cubes to be yellow, that probability is 1÷8 = 0.125. The probability ofeither event is therefore 0.375 + 0.125 = 0.5. (We can add because the twoevents are mutually exclusive.)

“At least two cubes are yellow” is the most likely event.

Assuming you won, you can play again for the Grand Prize which means youcan take a friend to Disney World. But now your box has all possible towersthat are four tall (built by selecting from the two colors yellow and red). Youare to select from the same four possibilities for a winning tower. Which choicewould you make this time and why would this choice be better than any of theothers?

The total number of four-tall towers is 24 = 16. The probabilities are:

All cubes are exactly the same color: There are two ways (all red or allyellow). The probability is 2÷16 = 0.125.

There is only one red cube: There are four ways; the red cube can be on thetop, second from the top, second from the bottom, or on the bottom. Theprobability is 4÷16 = 0.25.

Exactly two cubes are red: The number of ways to accomplish this is C(4,2)= 6. The probability is therefore 6÷16 = 0.375.

208 Appendix A: Combinatorics Problems

At least two cubes are yellow: This means that exactly two cubes are yellow,exactly three cubes are yellow, or exactly four cubes are yellow. As discussedabove, the probability that exactly two cubes are yellow (the same as theprobability that exactly two cubes are red) is 6÷16 = 0.375. The probabilitythat exactly three cubes are yellow is the same as the probability that onecube is red: 4÷16 = 0.25. Since there is one way for exactly four cubes tobe yellow, that probability is 1÷16 = 0.0625. The probability of any one ofthe three events is therefore 0.375 + 0.25 + 0.0625 = 0.6875.

“At least two cubes are yellow” is the most likely event.

10. The Pizza Problem with Halves (March 1993, Grade 5) – A local pizza shophas asked us to help them design a form to keep track of certain pizza sales.Their standard “plain” pizza contains cheese. On this cheese pizza, one or twotoppings could be added to either half of the plain pizza or the whole pie. Howmany choices do customers have if they could choose from two different top-pings (sausage and pepperoni) that could be placed on either the whole pizzaor half of a cheese pizza? List all possibilities. Show your plan for determiningthese choices. Convince us that you have accounted for all possibilities and thatthere could be no more.

With two topping choices, there are four possibilities for the first half pizza,because each topping can be either on or off that half of the pizza. The fourchoices are: plain (sausage off, pepperoni off), sausage (sausage on, pepperonioff), pepperoni (sausage off, pepperoni on), and sausage/pepperoni (sausage on,pepperoni on). Consider each of the four possibilities in turn.

Case 1: Plain. There are four possibilities for the other half of the pizza, thefour listed above (plain, sausage, pepperoni, and sausage/pepperoni).

Case 2: Sausage. There are three possibilities for the other half of the pizza:sausage, pepperoni, and sausage/pepperoni. (We omit plain, because wealready accounted for the plain-sausage pizza in Case 1.)

Case 3: Pepperoni. There are two possibilities remaining for the other half ofthe pizza: pepperoni and sausage/pepperoni. (Plain and sausage are alreadyaccounted for.)

Case 4: Sausage/pepperoni. There is only one possibility left for the otherhalf of the pizza; that is sausage/pepperoni.

There are 4+3+2+1 = 10 possible pizzas with halves.

11. The Four-Topping Pizza Problem (April 1993, Grade 5) – A local pizza shophas asked us to help design a form to keep track of certain pizza choices. Theyoffer a cheese pizza with tomato sauce. A customer can then select from thefollowing toppings: peppers, sausage, mushrooms, and pepperoni. How manydifferent choices for pizza does a customer have? List all the possible choices.

Appendix A: Combinatorics Problems 209

Find a way to convince each other that you have accounted for all possiblechoices.

There are 2×2×2×2 = 16 possible pizzas.

12. Another Pizza Problem (April 1993, Grade 5) – The pizza shop was so pleasedwith your help on the first problem that they have asked us to continue ourwork. Remember that they offer a cheese pizza with tomato sauce. A customercan then select from the following toppings: peppers, sausage, mushrooms, andpepperoni. The pizza shop now wants to offer a choice of crusts: regular (thin)or Sicilian (thick). How many choices for pizza does a customer have? List allthe possible choices. Find a way to convince each other that you have accountedfor all possible choices.

Each of the 16 four-topping pizzas has two choices of crust, so there are 32pizzas.

13. A Final Pizza Problem (April 1993, Grade 5) – At customer request, the pizzashop has agreed to fill orders with different choices for each half of a pizza.Remember that they offer a cheese pizza with tomato sauce. A customer canthen select from the following toppings: peppers, sausage, mushroom, and pep-peroni. There is a choice of crusts: regular (thin) and Sicilian (thick). How manydifferent choices for pizza does a customer have? List all the possible choices.Find a way to convince each other than you have accounted for all possiblechoices.

The first half of the pizza can have 24 = 16 possible topping configurations,as described above. Consider each of those configurations in turn. Followingthe procedure described above for the two-topping half-pizza problem, we findthat there are 16+15+14+. . . +3+2+1 possible pizzas; this sum is given by16×17÷2. Since each pizza can have a thick or thin crust, we multiply by 2.The number of possible pizzas is 16×17÷2×2 = 272.

14. Counting I and Counting II (March 1994, Grade 6) – How many different two-digit numbers can be made from the digits 1, 2, 3, and 4? Each of four cards islabeled with a different numeral: 1, 2, 3, and 4. How many different two-digitnumbers can be made by choosing any two of them?

Counting I: Assuming that you are not permitted to reuse digits, there are fourchoices for the first digit and three for the second digit, giving 12 two-digitnumbers. (They are 12, 13, 14, 21, 23, 24, 31, 32, 34, 41, 42, and 43.)

Counting II: There are four choices for the first digit and four choices for thesecond digit. This makes 16 different two-digit numbers. (They are 11, 12, 13,14, 21, 22, 23, 24, 31, 32, 33, 34, 41, 42, 43, and 44.)

15. Towers-Binomial Relationship (March 1996, Grade 8) – In an interview,Stephanie discusses the relationship between the towers problems and thebinomial coefficients.

210 Appendix A: Combinatorics Problems

Binomial coefficients arise in connection with the binomial expansion formula(a+b)n. The following can be shown by induction:

(a + b)n =n∑

r=0

(nr

)an−rbr

The coefficient of an–rbr is given by:

(nr

)= n!

r!(n − r)!This number is the rth entry in the nth row of Pascal’s Triangle, and it givesthe number of towers with exactly r cubes of one color, when building towersthat are n-tall and there are two colors to choose from. Hence, the binomialexpansion and the towers problem are isomorphic, with the number of instancesof a in the rth term being equal to the number of towers having exactly r cubes.

16. Five-Tall Towers with Exactly Two Red Cubes (January 1998, Grade 10) – Youhave two colors of Unifix cubes (red and yellow) to choose from. How manyfive-tall towers can you build that contain exactly two red cubes?

You are selecting two items (the positions of the two red cubes) from fivechoices (the number of cubes in the tower); there are ten ways to do this:

(52

)= 5!

2!(5 − 2)! = 10

17. Ankur’s Challenge (January 1998, Grade 10) – Find all possible towers that arefour cubes tall, selecting from cubes available in three different colors, so thatthe resulting towers contain at least one of each color. Convince us that youhave found them all.

Suppose the colors are red, blue, and green. We are counting the towers in threecases: (1) those with two red cubes, one blue cube and one green cube, (2) thosewith one red cube, two blue cubes, and one green cube, and (3) those with onered cube, one blue cube, and one green cube. The following equation gives thenumber of ways of selecting m groups of objects of size r1 through rm:

(nr1, r2, ..., rm

)= n!

r1! · r2! · ... · rm! , where∑

ri = n

So the number of four-tall towers containing exactly two red cubes, one bluecube, and two green cubes is:

(42, 1, 1

)= 4!

2! · 1! · 1! = 12

Appendix A: Combinatorics Problems 211

Similarly for the other two cases:

(41, 2, 1

)=

(41, 1, 2

)= 12

Hence the number of towers with the required condition is 12+12+12 = 36.

18. The World Series Problem (January 1999, Grade 11) – In a World Series, twoteams play each other in at least four and at most seven games. The first team towin four games is the winner of the World Series. Assuming that the teams areequally matched, what is the probability that a World Series will be won: (a) infour games? (b) in five games? (c) in six games? (d) in seven games?

The number of ways for a team to win the series (four games) in n games isthe number of ways it can win three times in n–1 games (and then win thelast game). This is given by C(n–1,3). The probability of any given set of out-comes for n games is 1÷2n (since there are two equally likely outcomes foreach game). So the probability that one team wins the series in n games isgiven by C(n–1,3) ÷2n, and the probability of a win for either team is doublethat: C(n–1,3) ÷2n–1. The probabilities are:

(a) C(4–1,3) ÷24–1 = C(3,3) ÷23 = 1÷8 = 0.125.

(b) C(5–1,3) ÷25–1 = C(4,3) ÷24 = 4÷16 = 0.25.

(c) C(6–1,3) ÷26–1 = C(5,3) ÷25 = 10÷32 = 0.3125.

(d) C(7–1,3) ÷27–1 = C(6,3) ÷26 = 20÷64 = 0.3125.

19. The Problem of Points (February 1999, Grade 11) – Pascal and Fermat aresitting in a café in Paris and decide to play a game of flipping a coin. If thecoin comes up heads, Fermat gets a point. If it comes up tails, Pascal gets apoint. The first to get ten points wins. They each ante up 50 francs, making thetotal pot worth 100 francs. They are, of course, playing “winner takes all.” Butthen a strange thing happens. Fermat is winning, eight points to seven, when hereceives an urgent message that his child is sick and he must rush to his homein Toulouse. They carriage man who delivered the message offers to take him,but only if they leave immediately. Of course, Pascal understands, but later, incorrespondence, the problem arises: how should the 100 francs be divided?

We can list all the circumstances where Fermat gets two points before Pascalgets three points. He can do this in two flips, three flips, or four flips. (The gamecannot proceed past four flips. As soon as both players get to nine points, thenext flip will produce a winner. It takes three flips for this to happen.)

(a) Two flips: Fermat wins both. Probability =1÷22 = 1÷4.(b) Three flips: Fermat wins one of the first two and the last one. Probability

= C(2,1)÷23 = 1÷4.

212 Appendix A: Combinatorics Problems

(c) Four flips: Fermat wins one of the first three and the last one: Probability= C(3,1)÷24 = 3÷16

Probability of any of these events = 1÷4 + 1÷4 + 3÷16 = 11÷16. ThereforeFermat should get 100 × 11÷16 Francs ≈ 69 Francs and Pascal should get 31Francs.

20. The Taxicab Problem (May 2002, Grade 12) – A taxi driver is given a specificterritory of a town, shown below. All trips originate at the taxi stand. One veryslow night, the driver is dispatched only three times; each time, she picks uppassengers at one of the intersections indicated on the map. To pass the time,she considers all the possible routes she could have taken to each pick-up pointand wonders if she could have chosen a shorter route. What is the shortest routefrom a taxi stand to each of three different destination points? How do youknow it is the shortest? Is there more than one shortest route to each point? Ifnot, why not? If so, how many? Justify your answer.

Using Powell’s et al. (2003) notation to denote coordinates on the taxicab grid,(n,r) indicates a point n blocks away from the taxi stand and r blocks to theright. So the blue dot is at (5,1), the red dot is at (7,4), and the green dot is at(10,6). Taking the shortest route means going in two directions only (down andto the right). Finding the number of shortest paths from the taxi stand (0,0) toany point (n,r) involves the number of ways to select r segments of one kind ofmovement in a path that includes two kinds of movements; i.e., the number ofshortest paths to (n,r) is C(n,r). For the specific cases given above, the shortestpaths are:

Blue: C(5,1) = 5.

Red: C(7,4) = 35.

Green: C(10,6) =210.

Appendix BCounting and Combinatorics Dissertationsfrom the Longitudinal Study

Franciso, J. M. (2004). Students’ reflections on their learning experiences: Lessonsfrom a longitudinal study on the development of mathematical ideas andreasoning. Unpublished doctoral dissertation, Rutgers University, Newark, NJ

Glass, B. H. (2001). Mathematical problem solving and justification with commu-nity college students. Unpublished doctoral dissertation, Rutgers University, NewJersey.

Kiczek, R. D. (2001). Tracing the development of probabilistic thinking: Profilesfrom a longitudinal study. Unpublished doctoral dissertation, Rutgers University,New Jersey.

Martino, A. M. (1992). Elementary students’ construction of mathematical knowl-edge: Analysis by profile. Unpublished doctoral dissertation, Rutgers University,Newark, NJ.

Muter, E. M. (1999). The development of student ideas in combinatorics and proof:A six year study. Unpublished doctoral dissertation, Rutgers University, Newark,New Jersey.

O’Brien, M. (1994). Changing a school mathematics program: A ten-year study.Unpublished doctoral dissertation, Rutgers University, New Jersey.

Powell, A. B. (2003). “So let’s prove it!”: Emergent and elaborated mathematicalideas and reasoning in the discourse and inscriptions of learners engaged in acombinatorial task. Unpublished doctoral dissertation, Rutgers University, NewJersey.

Muter, E. M. (1999). The development of student ideas in combinatorics and proof:A six year study. Unpublished doctoral dissertation, Rutgers University, Newark,New Jersey.

Sran, M. K. (2010). Tracing Milin’s development of inductive reasoning: A casestudy. Unpublished doctoral dissertation, Rutgers University, Newark, NJ.

Steffero, M. (2010). Tracing beliefs and behaviors of a participant in a longitudinalstudy for the development of mathematical ideas and reasoning: A case study.Unpublished doctoral dissertation, Rutgers University, Newark, NJ.

Tarlow, L. D. (2004). Tracing students’ development of ideas in combinatorics andproof. Unpublished doctoral dissertation, Rutgers University, New Jersey.

Uptegrove, E. B. (2005). To symbols from meaning: Students’ investigations incounting. Unpublished doctoral dissertation, Rutgers University, New Jersey.

213

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Index

AAbstraction, 10, 17, 24, 77, 85Addition rule, 13, 73, 77, 82, 111–114,

124–125, 129–130, 133, 136–138, 141,152, 181

Addition rule of Pascal’s Triangle, 181Adults, 171–183Ali, 99–100Alston, Alice S., 5–6, 35, 45, 67, 97,

121, 182Amy-Lynn, 59, 62–63, 66–67, 121–122,

124–126Angela, 97–98, 103, 121, 127–129Angela’s Law of Towers, 98Ankur, 59, 62–64, 66–71, 89–90, 92–94,

105–107, 109–119, 133–135, 138–140,142, 157, 159–160, 162, 164–165,167, 185

Ankur’s Challenge, 89–95, 110, 183, 185,199–200, 210

Another Pizza Problem, 70, 209Argument, 13, 27, 31, 33–34, 35, 37, 40–43,

45–57, 61, 66–67, 72, 75, 89–95,97, 102–103, 175–176, 185, 191,195, 200

BBeliefs, 157–159, 168–169Binary notation, 106, 108, 116, 118–120, 135,

141, 143Binary numbers, 106, 137Binomial, 12, 14, 73, 76–77, 104–105,

108–112, 114, 116, 121–131, 134–136,143, 152, 209–210

Binomial expansion, 12, 14, 108–112, 114,116, 134–136, 210

Bobby (Robert), 45, 59, 62Branches, 103, 201

Brian, 59, 62–63, 66, 68–71, 89–92, 94,105–108, 110, 119, 133–135, 138–139,142, 145–150, 153, 157, 159, 161, 163,166–167

CCases, 12, 14, 33–34, 36–37, 39–43, 45–46,

49, 51, 56, 62, 66, 68, 74, 77, 85, 90, 94,97–100, 104, 110, 119–120, 127–128,130, 143, 145, 153, 158, 174–175, 177,182, 185–192, 194–195, 200

Choices, 11–12, 59–60, 69–71, 103, 107–108,116, 122, 124, 127–128, 130, 143, 151,177, 196

Choose, 12, 43, 59, 98, 107, 123, 127,135–140, 142

Choose notation, 135–136, 140Claims, 145–153, 164, 168, 202Coefficient, 77, 106, 135Collegiate, 183Combinations, 13, 19–23, 25, 28–30, 33,

36–38, 40, 42–43, 45–46, 60–67,69–71, 77, 79, 82, 85, 89–90, 92, 94,101–103, 106, 110, 122, 124, 127–128,130, 137, 172, 175–179, 182, 189, 192,195–196, 200

Combinatorial reasoning, 73, 202Combinatorics, 3, 6, 9–11, 14, 17–18, 24–25,

73, 110, 112–114, 120, 122, 131, 133,140–141, 144–146, 169, 198, 201

Combinatorics problems, 11, 14, 17–18,112–113, 131, 133, 144

Conditions, 3, 28, 142, 158, 168–169, 171,183, 190–191

Conjecture, 99, 147Connections, 3–4, 9, 18, 72, 74, 82, 105–120,

130–131, 141, 143, 171–172,178–182

Consumer math, 173

221

222 Index

Contemporary Mathematics, 172–173Contradiction, 14, 31, 40, 66–67, 97, 99Controlling for variables, 14, 35, 43,

97, 99Convincing arguments, 4, 104Counting I, 209Counting II, 209Counting methods, 172–173Cousin, 31–33, 36, 42, 46Critical events, 9Cups, Bowls, and Plates, 205–206

DDana, 11, 17, 19, 21–25, 27–32, 43, 60Danielle, 171, 174Discourse, 4, 74, 142, 146, 150, 159,

164–165, 202Discursive, 150–151, 164, 167, 201Donna, 171, 186–187Doubling rule, 39–40, 42, 49, 56,

106, 130Duplicates, 19, 29–30, 33, 36, 56, 61, 72,

78–82, 100, 122, 127, 197Durability of ideas, 141Dyadic choice, 151

EEducational Studies in Mathematics, 202Elevator, 31, 34, 37Empirical, 82, 152Epistemological beliefs, 157–159, 168Errol, 171, 175–176, 192, 199Euclidean geometry, 150Explanation, 21, 40, 51–55, 67, 70, 74, 92–95,

97, 101, 106, 109, 112, 115–117, 119,123, 126, 135–136, 141, 172–175,180–181, 197, 200

FFactorials, 122Family tree, 52–53Fermat, 77Fermat’s Recursion, 77–78Final Pizza Problem, 70, 209Five-tall towers with exactly two red cubes, 33,

109, 119Five-tall towers selecting from two colors, 31Five-topping pizza problem, 13, 106, 112, 135,

174, 181Forms of reasoning, 6, 11, 14, 39, 72–73, 185,

200–201Formula, 77–78, 82, 103–104, 108–109, 122,

125, 128, 130, 141, 143, 178, 199For the Learning of Mathematics, 201

Four-tall towers selecting from 2 colors, 28,196–197

Four-tall towers selecting from 3 colors, 195Four-topping pizza problem, 68–70, 123, 127,

174, 181, 208Francisco, John M., 157–169

GGang of Four, 38, 42–43, 51, 75Generalization, 3, 17, 91, 94, 98, 106, 126Generalize, 3, 10, 18, 41, 90, 95, 104–105,

133, 147, 151, 172General rule, 97–98, 101, 124, 135, 141, 143Geometric, 108–109Glass, Barbara, 171–183, 185–200Grade 1, 7Grade 2, 20–22, 24, 73Grade 3, 3, 7, 22–23, 28, 74Grade 4, 5–6, 31, 33–42, 77Grade 5, 45, 59, 77Grade 6, 7Grade 7, 75Grade 8, 33, 73, 75, 85Group work, 25, 31, 36, 70, 103, 121, 162,

165, 177Guess My Tower, 42, 45, 50

HHarding School, 6Heuristic, 11–13, 25, 27, 43, 59, 67, 147,

201, 203High school, 4–8, 14, 89, 97, 105, 133–134,

142, 145–146, 157–158, 161, 169,172–173, 182–183, 185–186, 200

Hilbert, David, 201

IInductive argument, 42–43, 45–57, 67,

97, 102–103, 176–177, 182, 185,191–195, 199

Inductive reasoning, 14, 38–39, 50–56, 97,100, 195

Interlocution, 151Isomorphic relationship, 108, 118, 120Isomorphism, 4, 14, 77, 109, 126, 130, 133,

147, 151–153, 179–180

JJaime, 17, 23Jeff C., 171, 174, 176–177, 179–180Junior year, 134Justification, 11–12, 14, 24, 34, 39, 42, 46,

59, 67, 90, 94, 97, 100, 104, 109–110,118–119, 126, 147, 168, 173, 175–177,185–193, 196, 199–200, 203

Index 223

Justification by cases, 97, 104, 119, 177, 185,192, 199

Justify, 3, 12, 19, 39, 42–43, 49, 61, 65–66,73, 92, 130–131, 146–153, 168, 172,175–176, 183, 194, 200, 203

KKenilworth, 4–7, 12, 59, 69–70, 161, 182Kiczek, Regina, 121, 133Knowing deeply, 203

LLearning environment, 131Lecture classes, 118–120Letter codes, 106, 122, 142Liberal-arts, 171–173, 185Linda, 171Lisa, 171, 175, 180Logic, 39, 172Logical reasoning, 10, 17Longitudinal study, 3–8, 10–11, 14, 18,

24, 27, 75, 86, 89, 98–99, 105, 130,157–161, 167–169, 172, 177, 183,185–200, 203

MMagda, 97–99, 103, 121, 127–130Maher, Carolyn A., 4, 45–57, 59–72, 89–95,

105, 121, 133Making sense, 131, 134, 141, 145Martino, Amy M., 4, 9, 17–21, 27–28, 33, 35,

39, 41, 47, 49, 74–75, 178, 180, 182Mary, 171, 175, 195, 197–198, 200Mathematical beliefs, 157, 159, 168–169Mathematical concepts, 6, 119, 161, 172–173Mathematical ideas, 4–5, 9–11, 17–19, 24, 57,

74–75, 85, 95, 130–131, 133, 142, 145,158, 162, 172, 201–203

Mathematical structure, 3, 147Melinda, 171, 178, 180Metaphor, 13, 85Michael, 17, 19–21, 23, 35Michelle, 27, 38, 41, 45, 50–54, 56, 59–60, 62,

64, 97, 99–103, 121–122, 127Michelle I., 27, 45, 50–54, 56, 59, 64–65Michelle R., 45, 50, 52–53, 56, 59, 62Mike, 59, 62–64, 66–68, 70–71, 89–90, 92–94,

105–110, 114–120, 133–142, 145–152,157, 159–160, 162, 165–167, 171, 177,181–182

Milin, 25, 27, 35–41, 45–52, 55–56, 59–61,64–65, 71, 77

Muter, Ethel M., 59, 89–95, 105–120

NNational Council of Teachers of Mathematics

(NCTM), 5, 17, 144, 168National Science Foundation, 5Night session, 112, 116, 118, 133,

135–143, 181Notation, 13–14, 18–19, 24–25, 27, 42, 59–61,

65, 67, 72–74, 77, 103, 105–108, 110,112, 116, 133–142, 201

n-Tall tower, 12, 27N-tall towers selecting from r-colors,

42, 77

OOpposite, 29, 31, 35–36, 42, 46, 54, 67,

99–100, 150, 175Order, 6, 12, 24, 70, 75, 89–90, 106–107, 113,

118, 124, 127–128, 130, 133, 140,143–144, 153, 173, 175, 182, 189,191, 202

Outfit, 11, 19–21, 23–24, 43Over time, 73, 75, 160, 162, 168, 171,

201, 203

PPantozzi, Ralph, 121, 161–162, 166–167Papert, 74–75Partial cases, 42Pascal, 211–212Pascal’s Identity, 13–14, 105, 111–115, 118,

120–121, 124–125, 127–130Pascal’s Triangle, 12–14, 72–73, 77, 81–83, 86,

104–105, 110–117, 120–121, 123–126,128–131, 133–138, 140–141, 143–144,147, 151–153, 160, 181–182

Pattern, 6, 10, 18, 24, 27–28, 31, 33–34, 37–43,46, 50–54, 73–74, 77, 81–82, 85, 90,99, 101, 104, 106, 119, 123, 128, 130,147, 174–175, 181–183, 190

Penny, 171, 177, 195, 199Permutations, 172, 178, 182, 192Pirie, Susan, 97, 158, 203Pizza

with halves problem, 12–13, 67–69, 174plain, 12, 59, 106–107, 113, 117–118,

122–123, 125, 127, 141, 208problem, 12–13, 59, 68–70, 72–73,

104–107, 111–113, 116–123, 125, 127,129–130, 135–136, 141, 145, 169,172–174, 177–178, 180–181, 183

Powell, Arthur B., 9, 145–154Power of 2, 122, 124Probability, 165, 172

224 Index

Problem of Points, 211Proof

by cases, 41, 72, 119, 130, 175, 191, 194by contradiction, 99

RReasoning, 3–4, 6, 9–11, 14, 17, 24–25, 27, 31,

38–39, 42, 45, 50, 52, 54–55, 59–72,75–76, 84, 94–95, 97, 100, 112, 145,147–153, 158, 168, 171–183, 185, 188,191, 195, 201–203

by contradiction, 97Recursive argument, 72, 199Relay Race, 206Representation, 3–4, 10–14, 17–25, 27, 46, 51,

59, 62, 64, 73, 76, 94–95, 97, 105–120,133–144, 168, 172, 203

Rica, Fred, 4–5Rob, 171, 175, 177–178, 181–182, 187Robert, 121–122, 124–127

SSamantha, 171, 179Scheme, 21, 24, 94, 105–106, 109, 174Second International Congress of

Mathematicians, 201Senior year, 145, 200Sense making, 6, 11–12, 24, 159–162, 200,

203–204Shelly, 121–127Sherly, 99–100, 121, 127–129Shirts and jeans, 11, 17–24, 27, 29, 59–60,

64, 73Shirts and jeans extended, 205Socially emergent cognition, 150Sociomathematical norms, 153Sophomore year, 119, 134–135, 140, 143Specialize, 147–148, 151Speiser, Robert, 33, 73–86, 105–112, 201Sran, Manjit K., 27–43, 45–57, 59–72

Staircase, 33, 40, 42, 174–175Standard notation, 133–144Statistics, 172, 178Stephanie, 11, 17, 19–25, 28–35, 40–42,

45–62, 71, 73–85, 121–127, 171, 178,180, 182

Stephanie C., 171, 178, 180Steve, 171, 175, 181Strategy, 13, 21, 23, 29, 31, 35–36, 43, 45, 51,

67, 69, 72, 92, 94, 98–99, 100, 119,147, 153, 177, 182, 192, 194, 197, 200

TTarlow, Lynn D., 59, 97–104, 121–131Taxicab geometry, 150, 202Taxicab problem, 14, 144–147, 150–153, 212Theory, 51, 74, 102–103, 175, 202Tim, 171, 177, 190Tower

binomial relationship, 209families, 31–32, 34

Tracy, 171, 178Tree diagrams, 122, 127Trial and error, 29, 31, 35–42, 51

UUnderstanding, 4–5, 9, 12, 18, 21, 24, 33,

41, 43, 45–46, 51–54, 57, 73–76, 95,101, 105–106, 109, 121, 134, 140, 158,160–164, 180, 201, 203

Unifix cubes, 12, 27, 42, 50, 97, 126Uptegrove, Elizabeth B., 9–14, 97–104,

105–120, 133–144

WWesley, 171, 175–176, 180

YYankelewitz, Dina, 17–25, 27–43, 45–57,

59–72