## 1. Making animals at random then focussing on giraffes
## 3. Justifying the 1-6-12-8 solutionsAsk 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 = 27Often the group answers do not add to 27.
It is usually the 1-part group that has the greatest trouble confidently agreeing on the number 12.
## 4. Adding a kangaroo & naming the Crazy AnimalsHand 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.)
## 5. ChallengeStudents use the booklet and work in groups to address the challenge. Discuss as a class. The answers are:
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 formulaOnce the results for four animals have been agreed, apply the problem solving strategy of making a list or table to organise the data.
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:
- head with body
- head with legs
- body with legs
- 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)
- 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}
- 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:
## ExtensionBy substituting:
Contacts ... © Maths300 ... Site Map |