Return to Free Tour Crazy Animals
Whole & Parts Lesson

 

Summary

 

Appendix Lesson 57

  • Years: 6 - 12
  • Time: 1 - 2 lessons
  • Strands: Probability, Algebra

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. Learning features include:

  • physical separation into four groups
  • small group problem solving
  • suits many year and ability levels
  • the somewhat unexpected generalising into algebraic formulae for senior students

 

Resources

1. Making animals at random then focussing on giraffes

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

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?
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?

2. Problem solving in groups

Ask the groups to talk among themselves to consider the question: How many possible animals exist in your group?, that is, 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.

Lesson Stages

  1. Making animals at random then focussing on giraffes
  2. Problem solving in groups
  3. Justifying the 1-6-12-8 solutions
  4. Adding a kangaroo & naming the Crazy Animals
  5. Challenge: How many of each type (0, 1, 2, 3 - part giraffes) exist now?
  6. Generalising patterns & finding a formula

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.

3. Justifying the 1-6-12-8 solutions

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.

4. Adding a kangaroo & naming the Crazy Animals

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.)

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.

5. 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.

6. Generalising patterns & finding a formula

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.

 

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