MATHS300: Whole & Parts Appendix

MATHS300
Lesson 57

Crazy Animals

Whole & Parts Plan

Each MATHS300 lesson serves two purposes. On the one hand it is a professional development experience offering opportunity to try something new, hopefully in conjunction with your staff, in the knowledge that the notes record the successful experiences of your colleagues elsewhere. On the other hand it is a well trialed lesson plan which provides clear information about 'what to do in maths tomorrow'.

Years: 6 - 12

Time: 1 - 2 lessons

Making animals at random and then classifying them into 3-part, 2-part, 1-part or 0-part giraffes creates a fascinating extended investigation involving probability, problem solving and finally algebra.

Lesson Stages

Use the book to make animals at random.
Classify the animals as 3, 2, 1 or zero part giraffes.
Students physically move into one of the 4 groups.
Why are these groups different sizes?
Problem solving segment for each group. How many of 'your type' exist?
Justifying the 1-6-12-8 solutions (from the total of 27 animals).
Adding a 4th animal (eg: kangaroo) to the book. Now how many animals are in the book?
Language: Giving each of the animals a name by breaking up the word kangaroo.
Challenge: How many of each type (0, 1, 2, 3 - part giraffes) exist now?
Generalising patterns into 5, 6, ... 10 ... n animals. Finding and justifying the algebraic formulae for each group.

Features

physical separation into four groups
small group problem solving
suits many year and ability levels (up to step 10)
the somewhat unexpected generalising into algebraic formulae for senior students

Lesson Notes

1. Each student has a booklet and using the following dice rules, makes an animal at random:

1 or 2 = giraffe, 3 or 4 = horse, and 5 or 6 = duck

2. The teacher suddenly declares an interest in, say, giraffes!!

Who has made a perfect giraffe, ie: a 3 part giraffe? Who has made a 2-part giraffe, ie: two of the three parts of the crazy animal are 'giraffe' parts? Who has made a 1-part and who has made a zero part?

3. Ask students to physically move to 4 separate corners of the room - all the 3 parts together, all the 2 parts together etc. Noticing that the groups are different sizes, ask:

Why are these groups different sizes?

4. Problem Solving:
Ask the groups to talk among themselves to consider the question:
How many possible animals exist in your group?
ie: How many 3 part giraffes exist? How many 2-part giraffes exist? etc.

I very much like this moment in the lesson. The groups have a reason to talk to each other. There is often much vigorous discussion and reference to their booklets.

Given that there is only one possibility for it, the 3-part giraffe group is usually very small. If there are any students in this group, I usually challenge them to solve one of the other group's problems.

5. Ask each group to report. How confident they are about their answer? Given there are 27 animals in total, the answers from the four groups must add to this number. The correct answers are:
3 part = 1
2 part = 6
1 part = 12
0 part = 8
Total = 27

Often the group answers do not add to 27.

I point out this fact to encourage groups to recheck their reasoning. If they still disagree I point to the 'offending' group, and the whole class can check the reasoning for that group.

It is usually the 1-part group that has the greatest trouble confidently agreeing on the number 12.

For younger classes, I often repeat the making animals randomly and getting into the four groups, so that they can see the same patterns emerging.

6. The Kangaroo:
Hand out blanks and challenge students to draw a simple kangaroo to add to their booklet. The 'crossover points' where the head meets the body and the body joins the legs are marked on the blanks.

Discuss the number of Crazy Animals that are possible now.
(The correct answer is 4 x 4 x 4 = 64 different crazy animals.)

7. Language:
It is an interesting language segment to partition the letters of the word kangaroo so that each new crazy animal has a 'name'. KANG - A - ROO is the partitioning option usually selected by most students.

8. Challenge:
How many 3-part, 2-part, 1-part and 0-part giraffes exist now? Students use the booklet and work in groups to address the challenge. Discuss as a class.

The answers are:

No. of Animals	3-Part	2-Part	1-Part	0-Part	Total
4	1	9	27	27	64

The justification or logic for these numbers is the essence of seeing the general pattern that can then be described as an algebraic rule.

9. Once the results for four animals have been agreed, apply the problem solving strategy of making a list or table to organise the data.

No. of Animals	3-Part	2-Part	1-Part	0-Part	Total
2	1	3	3	1	8
3	1	6	12	8	27
4	1	9	27	27	64
5
6
...	...	...	...	...	...
10

Invite students to look for and extend patterns to fill in the results for 5, 6, ... 10 animals in the book. This can be done just from the number patterns, but the justification for the number entered in each column is what leads, ultimately, to an algebraic rule:

3-part giraffe
No matter how many animals are used there is only one combination for this condition - each part must be a giraffe.

2-part giraffe
If two parts have to be a giraffe, there are three ways this can happen.

head with body
head with legs
body with legs

With each of these the third part can be any of the other animals. Therefore:

if there are 2 animals, one is a giraffe, so that's 3 choices for the 2 parts and only one choice for the remaining part, ie:
3 x 1 = 3
if there are 3 animals, one is a giraffe, so that's 3 choices for the 2 parts and 2 animals to choose from for the remaining part, ie:
3 x 2 = 6
if there are 4 animals, one is a giraffe, so that's 3 choices for the 2 parts and 3 animals to choose from for the remaining part, ie:
3 x 3 = 9
if there are n animals, one is a giraffe, so that's 3 choices for the 2 parts and (n - 1) animals to choose from for the remaining part, ie:
3 x (n - 1) = 3(n - 1)

1-part giraffe
The one part that is a giraffe can be the head, the body or the legs. So there are three ways this can happen. That leaves two other parts of the body to 'fill' with not-giraffe parts, and both remaining parts could come from the same other animal. Therefore:

if there are 2 animals, one is a giraffe, so that's 3 choices for the 1 giraffe part and only one choice for each remaining part, ie:
3 x 1 x 1 = 3
if there are 3 animals, one is a giraffe, so that's 3 choices for the 1 giraffe part and 2 choices for each remaining part, ie:
3 x 2 x 2 = 12
if there are 4 animals, one is a giraffe, so that's 3 choices for the 1 giraffe part and 3 choices for each remaining part, ie:
3 x 3 x 3 = 27
if there are n animals, one is a giraffe, so that's 3 choices for the 1 giraffe part and (n - 1) choices for each remaining part, ie:
3 x (n - 1) x (n - 1) = 3(n - 1)²

0-part giraffe
If no part of the animal is giraffe, then this is like taking out the giraffe page. Any of the animals that are left can occupy any of the three parts of the body. Therefore:

if there are 2 animals, one is a giraffe which is 'removed from the book', so that's 1 animal left which can only be made in one way.
if there are 3 animals, one is a giraffe which is 'removed from the book', so that's 2 animals left to 'give' their heads, bodies and legs, ie:
2 x 2 x 2 = 8
if there are 4 animals, one is a giraffe which is 'removed from the book', so that's 3 animals left to 'give' their heads, bodies and legs, ie:
3 x 3 x 3 = 27
if there are n animals, one is a giraffe which is 'removed from the book', so that's (n - 1) animals left to 'give' their heads, bodies and legs, ie:
(n - 1) x (n - 1) x (n - 1) = (n - 1)³

So, the generalisation can be expressed as:

No. of Animals	3-Part	2-Part	1-Part	0-Part	Total
n	1	3(n - 1)	3(n - 1)²	(n - 1)³	n³

A real highlight, and a natural conclusion to the lesson, was when students realised that the total of all the groups must equal n³, and then set about expanding the expressions, simplifying and proving that they all add to n³.

Extension

By substituting:
(n + 1) for n
in the summation which derives from the table, the formula for expanding
(n + 1)³
becomes apparent.

MATHS300 is a living site. This lesson will be enriched through further teacher development in classrooms across the world. You are invited to contribute to that process by submitting:

variations

extensions

inspirations

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student work

Please email material to the address below. If it is included it will be acknowledged. You can review current contributions in the Classroom Contributions folder for this lesson.

email: Doug.Williams@curriculum.edu.au