Colour Building
Problem
I wonder how many different ways there are of combining white rods (1) and red rods (2) to make the same length as the orange rod (10) ...
There are going to be quite a few, so let's start with a simpler challenge...
In how many ways can you combine white rods (1) and red rods (2) to make the same length as the pink rod (4)?
Once you think you've found them all, click below to check.
Note that I count "white, white, red" and "white, red, white" as different, even though they both use two white rods and one red rod.
Using just red and white rods, there's only one way of making the same length as the white rod, and only two ways of making the same length as the red rod.
- the light green rod (3)?
- the yellow rod (5)?
- the dark green rod (6)?
Without using the interactivity, can you work out how many different ways there are to make up the orange rod (10)?
Can you explain the pattern?
Extension
Try making different lengths using combinations of white, red and light green rods.
Can you use the patterns you observe to make predictions about how many ways there will be?
Getting Started
Is there a way to work systematically to make sure you have found all the possibilities?
When we make the length of the pink rod, how many possibilities have a red on the left?
How many possibilities have a white on the left?
Can you use this idea of thinking separately about the solutions that begin with a red, and the ones that begin with a white, to help you explain the patterns that you find?
Student Solutions
Mahdi from Mahatma Gandhi International School in India sent a full solution. This is the first part of Mahdi's solution, with some diagrams edited to show the rods being built:
Let $T_n$ define the number of ways of making a rod of length $n$ just by using red and white rods. We already know the following:
$T_1 = 1$ (there is $1$ way to make a white rod from red and white rods)
$T_2 = 2$ (there are $2$ ways to make a red rod from red and white rods)
$T_3 = 3$ (there are $3$ ways to make a light green rod)
$T_4 = 5$ (there are $5$ ways to make a pink rod)
The ways to cover $n-1$ length rod as defined in the beginning is $T_{n-1}$ (there were $3$ ways to make a light green rod).
So there are $5$ ways to make a pink rod using red and white rods.
Thus we get the explicit formula for combining a rod of length $n$ using red and white as: $$T_n = T_{n+1} + T_{n+2}$$
Using this formula we can calculate $T_5, T_6, T_7, ...$
$T_5 = T_4 + T_3 = 5+3=8$
$T_6 = T_5+T_4=8+5=13$
$T_7 = T_6+T_5=13+8=21$
$T_8=T_7+T_6=21+13=34$
$T_9 = T_8 + T_7 = 34+21=55$
$T_{10} = T_9+T_8=55+34=89$ (orange rod)
You can see that $T_n=F{n-1}$ where $F_n$ is the $n^{th}$ Fibonacci number.
Mahdi's solution continues to find the definition of $T_n$ in terms of $n.$ Mahdi then considers finding lengths using white, red and green rods, or using white, red, green and pink rods, or using the first $k$ rods available. Click here to see Mahdi's full solution.
Teachers' Resources
Why do this problem ?
Possible approach
The problem assumes some familiarity with Cuisenaire rods, so if students are not familiar with the different colours and lengths, it may be worth spending some time first exploring a set of rods or using the online environment."I wonder how many different ways we could combine white rods (1) and red rods (2) to make the same length as the orange rod (10) ..."
Give students some time to think about the challenge, and then share their thoughts. The following might emerge:
"There are going to be loads of different ways"
"How are we going to be able to make sure we don't miss any?"
If no-one has suggested it: "Perhaps we could work on a simpler version of the problem to see if that helps. Let's see how many ways we could make the pink rod (4) out of whites and reds."
Once students have agreed on the five ways that a pink can be made from whites and reds, invite them to find the number of ways to make light green (3), yellow (5), and dark green (6). Then encourage them to make a prediction for black (7) and then test it out.
It is important to set aside enough time for students to think about and appreciate why each answer is the sum of the previous two. To draw out this insight, you might suggest that students organise their work into two categories: solutions that start with a red (what's left?) and solutions that start with a white (what's left now?)
Key questions
When we make the length of the pink rod, how many possibilities have a red on the left?
How many possibilities have a white on the left?
Can you use this idea of thinking separately about the solutions that begin with a red, and the ones that begin with a white, to help you explain the patterns that you find?