# School fair necklaces

Rob and Jennie were making necklaces to sell at the school fair.

They decided to make them very mathematical.

Each necklace was to have eight beads, four of one colour and four of another.

And each had to be symmetrical, like this.

How many different necklaces could they make?

Can you find them all?

How do you know you haven't missed any out?

What if they had 9 beads, five of one colour and four of another?

What if they had 10 beads, five of each?

What if...?

How will you record what you've found out? You might find it helpful to print out this sheet and to use colouring pens or pencils.

Is there any way of making sure you have them all?

If you cut out individual necklaces from the sheet, you can move them around and arrange them, which might help you find out if any are missing.

We had a number of solutions sent in. Most treated the bracelets as being in a straight line rather than in a circle, which can make it more difficult to check whether you are duplicating ones already found.

We had the following from Lyneham Primary Junior Maths Challenge Group in Australia:

On Friday we worked on the necklaces problem in our maths group.

We used some different coloured counting blocks (unifix) to make symmetrical patterns of eight blocks with two colours.

We figured out that with those blocks we could make six different combinations that were symmetrical (see picture). And then we figured out that it didn't matter if you cut half of it off because they are symmetrical, so the halves are exactly the same.

We found six different towers of four blocks, two of each colour (see at the bottom of the photo). So therefore there are six possible ways you can do it.

We also heard from Cole and Sachi at Randlay Primary School:

I used O,G,G,O,O,G,G,O* for number one.

For number two we did O,G,O,G,G,O,G,O.

Number three: G,O,G,O,O,G,O,G.

Number four we did O,O,G,G,G,G,O,O.

Number five: G,G,O,O,O,O,G,G.

*O stands for orange and G stands for green.

We found over five ways to do the challenge that were different and symmetrical with the same colour.

I wonder whether you can see the sixth way that you've missed out?

Lara-Jade at Bay Primary School wrote:

I started with nine beads and worked out that because it has to be symmetrical the bead at the bottom must be of one colour because the clue says four beads of one colour and five beads of the other colour. So I thought yellow could be at the bottom and my pattern was:

Y,G,Y,G,(Y),G,Y,G,Y.

Then I worked out 13 beads. I thought because it is another odd number the bead at the bottom must be the colour of bead with the most in it. So I worked out the pattern which was

Y,G,Y,G,Y,G,(Y),G,Y,G,Y,G,Y.

Then I tried to work out what 10 was. Then I tried it and thought that it wouldn't work because it was even but I was wrong. So I tried it again and got the pattern right which was:

Y,G,Y,G,(Y)(Y),G,Y,G,Y.

I then moved onto 12. Because 10 worked 12 MUST work so I worked out the pattern which was: Y,G,Y,G,Y,(G)(G),Y,G,Y,G,Y.

Finally, I did 11 beads. And worked out the pattern which was:

G,Y,G,Y,G,(Y),G,Y,G,Y,G.

We had an email sent in from a teacher in Oregon U.S.A. saying:

My students at Springwater Environmental Sciences School found there were zero combinations that worked if you used 5 beads of each colour. I have attached a few explanations.

The teacher later added :-

We spent two class periods looking for a pattern, but could not find one.

One of my students realized that odds and evens should be treated differently, and as a class, we found that multiples of 4 always work, but could not find a pattern of how they increased as you added more beads.

One student had a great idea to start with smaller numbers of beads and work up from there to see the pattern, so we started with one bead but we always either had equal or only one more bead than the other colour.

This task was a great follow up to the three colour block tower (https://nrich.maths.org/137) because the symmetry aspect really changed up the thinking.

Thank you, this was a later solution that was emailed in which dealt very well with the situation when having 10 beads and 5 of each colour. Well done!

Did you find any other ways of making symmetrical necklaces, I wonder? Is there anything general you could say about even and odd numbers of beads?

It's interesting to see the different ways of recording thinking - some of you using cubes, some of you using letters and some of you using squared paper.

If you have worked on this activity, then do send us your thoughts too.

Why do this problem?

This low threshold high ceiling task offers opportunities for children to confirm their understanding of line symmetry, to appreciate why working systematically is useful, and be introduced, or develop for themselves, a proof by exhaustion.

### Possible approach

*This problem featured in an NRICH Primary webinar in March 2022.*

Most children will find some practical apparatus useful for this activity. Beads and string are the obvious choice but can lead to frustration if threading and re-threading is time consuming. A better bet may be interlocking cubes which can be stuck together (and can always be threaded later to make a display) or other similar objects in two colours. If you don't have an IWB (or even if
you do) you may want to make some versions of the necklace on large strips of paper which children can hold up at the front for all to see.

Introduce the task and together with the children, generate a symmetrical necklace. Challenge them to work in pairs to find as many other symmetrical necklaces as possible using eight beads, four each of two colours. When they think they have found as many as possible, they can record them on strips of paper (you could use this sheet cut
into individual strips). Giving separate slips of paper as opposed to a single worksheet will mean the children will be able to arrange and rearrange them.

Can they now arrange the strips to help them to see whether they have found all the necklaces?

Whilst they are doing this, copy the results they have found onto the large strips and then invite children to come and hold them up at the front. Ask "Do we have them all? How do we know? How can we be *sure*?" Encourage individual children to come to the front and rearrange the strips by moving their classmates around.

The most usual reponse to this is that the necklaces fall into pairs, each the inverse of the other (RRBBBBRR and BBRRRRBB, for example). Point out that this helps us to know that there should be an even number of solutions, but it doesn't prove that we have found them all - there might well be another pair that we haven't found yet.

Give some time for the children to rearrange their own strips and if anyone thinks they have found a way to show that they have found them all, invite them to come to the front and rearrange the large strips. If no-one does, you mght want to leave the children for some time - or even ask them to take the problem home - before sharing the proof below.

Focus first on the strips that begin with a red bead.

Arrange them so that the second red bead is in position 2, then position 3 and finally position 4:

RRBB BBRR

RBRB BRBR

RBBR RBBR

This must be the full set of necklaces beginning with a red bead since the second bead has been placed in each of the three other places.

If there are three beginning with red, there must also be three beginning with blue.

This exemplification of every possible solution is known as 'proof by exhaustion'.

### Key questions

Can you make another?

How are these different?

How are these the same?

How do you know that is symmetrical?

Is your pattern the same as ...'s?

Do you think there are any more? Why/why not?

Can you show me why you think you've got them all?

### Possible support

For some children producing more than one necklace will be a challenge, and transferring a pattern accurately onto paper will also be difficult, but worthwhile. Support them in looking for the same pattern using different colour combinations, for example RGRGGRGR is the same underlying pattern as BYBYYBYB. Making generalisations like this is a very important part of thinking like a mathematician.

### Possible extension

Some children will be very interested in justifying their claim that they have found all the possibilities. Listen to their explanations - this is a wonderful opportunity for them to reason mathematically, and is the beginning of understanding mathematical proof.

Once they have convinced you, offer them nine beads, five of one colour and four of another. What difference does this make? What about ten, eleven, etc? Can they state any rules about the number of beads they need to make a symmetrical necklace with two colours?