6C3 Mentor’s Guide

The main thing a mentor needs to do for this activity and many mathematical activities

is to convince students to try it. Once students start work encourage them, give them

“high-fives.”

It may be tempting to give students hints. You should avoid giving problem specific hints

wherever possible. Instead, make hints general problem solving stratagems. For example:

Try some examples.

Change the problem.

Work backwards.

What do you wish was true?

Can you make a diagram, or an organized list?

Always ask, “why?” One of the characterizing features of mathematics is that it is possible

to justify the things that are known with deductive reasoning. This process is really more

important than just finding the answer. In this activity the answer to every question is 6C3.

This is read “six choose three.” It represents the number of ways one can choose three items

from six. Because the answer to each question in this activity is the same, it isimportant to ask students to compare the different questions.

Encourage students to make up and explore their own questions. Once someone has answered

a question, they should always start asking new questions. In this activity, it is natural to

change each question to a related one resulting in the answers 10C3, nC3, or nCk. One can

not really claim to understand the problem until they can also answer the related questions.

What this activity is really introducing is the idea of a bijective proof. It is possible to

answer the first question by just listing and counting all possibilities: ABC, ABD, . . . , DEF.

Organization helps here. The answers to the remaining problems could be found by making

similar lists, but it is much better to find bijections.

Tell the participants what a bijection is after they have worked for a bit. A bijec-tion as a correspondence between two sets S and T so that every element of S corresponds

to one element in T so that every element of T is associated to some element of S and no

element of T is associated to more than one element of S. We say S and T have the same

number of elements exactly when there is a bijection between them. The game of musicalchairs is a good way to explain this. (You can actually hum and have students play.)

Let S be the set of students and T be the set of table chairs. If every student is sitting in a

chair and there are one or more chairs left over, then there is no bijection between students

and chairs. Indeed, it is not the case that every element of T (chair) has an element of

S (student) associated to it. On the other hand, if we remove a bunch of chairs and ask

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all the students to sit, we will end up with more than one student per chair. In this case

every chair will have a student, but some elements of T (chairs) will have more than one

associated elements of S (students). When each chair has exactly one student, we have a

bijection between chairs and students, and we can conclude that there are the same number

of chairs as students.

It helps to have objects to manipulate. As an example, to find the bijection relating the

number of ways of dividing seven cookies between four people so that everyone gets one

cookie, make 7 paper cookies, put them in a line and but one “wooden cube” jelly bean

between each pair of cookies in the line. Now think.

cAcBcCcDcEcFc

Thus BDE corresponds to 2 cookies for the first person, 2 for the second, 1 for the third,

and 2 for the fourth.

The previous paragraph is the answer to the last question on the sheet. Youshould not give students the answers, you should encourage them to explore andfind them.

Proofs without words There is a theme in mathematics in which the right models or

figures can demonstrate deep patterns. There are many examples of this type of insight.

The following figure suggests a nice formula for the sum of the odd numbers.

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