Button-Up Some More
How many ways can you find to do up all four buttons on my coat? How about if I had five buttons? Six ...?
Problem
You might like to have a look at Button-up before trying this problem.
I have a jacket which has four buttons.
Sometimes, I do the buttons up starting with the top button. Sometimes, I start somewhere else.
How many different ways of buttoning it up can you find?
Look back at the number of different ways you found for buttoning up three buttons and four buttons.
Can you predict the number of ways of buttoning up a coat with five buttons?
Six buttons ...?
This problem was inspired by an idea from Bernard Murphy.
Getting Started
How do you know you have found all the ways?
How could you use the number of ways to button up three buttons to help you work out the number of ways for four buttons?
How will you record what you're doing?
Student Solutions
Thank you to everybody who sent in their solutions for this problem! We received a lot of very detailed solutions and unfortunately we don't have space to feature all of them on this page.
Some children thought that they'd found all the solutions but had actually missed some out. Quintus Fung from Sha Tin Junior School in Hong Kong came up with a strategy to make sure that no solutions were missed out:
When I started this challenge I just did random combinations but then I thought won't it be easier if I did it more organized? So I tried and I figured out a pattern: there always will be a heading number and depending on how many numbers are below the heading number there will be more possibilities, like if there are three numbers below the heading then there will be 6 columns. The amount of different numbers will also affect the possibilities.
This is a great example of how we can begin to work more systematically, by fixing the first number (or button). Tiago from Victoria Primary School in the UK also looked at what happens when the starting button is fixed:
For each button has 6 ways so 4 buttons has 6×4=24
Thank you for sharing this idea with us, Tiago. It looks like there are six solutions for each starting button because the number of possible ways of doing up the other three buttons is 6. Very interesting!
Shane from Newton Prep in China had a similar strategy, which they explained in slightly more detail:
First, I wrote down 234, 243, 324, 342, 432 and 423. After that, I wrote the number one in front of all of them so you already get 6 possibilities. Then I just changed the number one with the number two. After that, I switched 2 with 3 and 3 with 4. so then you add all of them together so you get 24 possibilities or at the start you can just do 6 times 4 to also get 24 possibilities.
Good idea, Shane - once you have the six possibilities that begin with the number 1, you can swap the digits around to get the rest of the possibilities.
James from Rainworth State School in Australia found something very similar, and they chose to use four colours to represent the four buttons:
I really like the pattern that your coloured buttons make! It makes it very easy to see the system you've used.
We received a lot of excellent solutions from the children at St Robert's Catholic Primary School in Harrogate, UK. Ruvimbo and Wiktoria came up with three different possible methods and they rated each one. You can click on the picture below to make it larger:
Good ideas! I do like your Method 3, but I agree that it can get messy when there are lots of buttons to include.
Lots of children thought about how to extend their solution to a coat with more buttons, the way Ruvimbo and Wiktoria did in their 'generalisation' section above. We received quite a few solutions about this from the children at All Hallows' School in Australia. Lizzie said:
You just find the answer to the four buttons then times by five for the five buttons.
Good idea, Lizzie! I wonder why that works?
Youyou from Sha Tin Junior School in Hong Kong explained why they thought the multiplication method would always work:
It would be obvious that if I button one button, there would be 3 buttons left. If this was so, there would be 4 possible buttons you could button in the first place, followed by 3 possible buttons left to button, and so on. This would mean: 4 multiplied by 3 multiplied by 2 multiplied by 1, with the result of 24.
This would apply to all of the solutions, meaning that 5 buttons would mean 120 possibilities, 6 buttons meaning 720, and 7, 5040. The sequence would be infinite and there would be no end to that.
These are some good ideas, Youyou. Elysia from Milton Road Primary School in the UK had a slightly different way of thinking about it:
If you want to find out how many possibilities there are for 5 buttons then there are 2 ways of doing that. You could work out the number of possibilities before (e.g. for 4 buttons there are 24 possibilities in total) and then multiply this by 5 because you have added an extra button which can go in five different positions (5 × 24 = 120 possibilities). Or, you could do 1 × 2 × 3 × 4 × 5 all the way up to the number of buttons you're trying to find. I found the second method is easier for big numbers.
That's a good point, Elysia - when we add the fifth button, there are five different possibilities for which position it could go in. I wonder why this means we can just multiply our previous number of possibilities by five?
Jakob and Marcus from Milton Road Primary School in the UK had a similar idea, and they also worked out every single possibility for a coat with five buttons:
To figure out the number of possibilities, you need to multiply the number of buttons by one less and less up to 1. E.g. 5 × 4 × 3 × 2 × 1 = 120. This is because the first number has 5 different spaces to go in. When you have placed it, the second number will only have 4 possibilities. Then the third number will only have 3. Then 2, then 1.
That shows that for 3 buttons (3 × 2 × 1 = 6) there are 6 different possibiliites.
4 buttons (4 × 3 × 2 × 1 = 24) has 24 different possibilities.
5 buttons (5 × 4 × 3 × 2 × 1 = 120) has 120 different possibilities.
6 buttons (6 × 5 × 4 × 3 × 2 × 1 = 720) has 720 different possibilities.
First, we did all of this recording but then we realised we didn't need to do it because of the formula above. It was a good way to double check our idea though!
Their teacher sent us a picture of Jakob and Marcus's working and said:
I really wanted to include an image of their careful, systematic recording, which they worked so hard on!
They started this in school and were close to a solution, then went home and carried on thinking and came back the next day with a clear formula to solve the problem. They spent their break times writing up their thinking as they were so desperate for me to submit a solution on their behalf and are rightly proud of themselves! This is the same case for all of the children who have submitted a solution from Milton Road Primary School - I have been blown away by their perseverance and inspired by their curiosity!
Thank you for sharing this with us - it sounds like lots of children have been working incredibly hard to solve this activity.
The children at Waverley Primary School in the UK also found every possible combination for the coat with five buttons, to check that the number of solutions was definitely 120:
We started work on this problem by looking at 'Button-up' first. Before investigating the 3 buttons in that scenario, we agreed on how 1 and 2 buttons would work, with 1 button obviously giving 1 possible order, and 2 buttons giving 2 possible orders. The children then began to explore the number of possibilities with 3 buttons. Some chose to do this with physical 'buttons' which they reordered, some preferred to record the buttons as written digits (ie. top button = 1, bottom button = 3). One pupil, Caitlin, devised a systematic method of finding all the possibilities beginning with 1, then all the possibilities beginning with 2, and so forth. This yielded 6 possible orders, which the other groups in the class ultimately agreed with. I then recorded the results so far in a simple table of 'Number of buttons'-'Number of possible orders' (see image). To investigate 4 buttons, we tried to apply Caitlin's ordered method again, finding the possible orders beginning 1-2, then 1-3, then 1-4, etc. This yielded 24 possibilities. At this stage, we tried to identify a pattern in the results that would enable us to predict the result for 5 buttons. Predictions were wide-ranging, from 30 to 125, but Liam recognised that there was a mathematical principle behind the results: multiplying the number of buttons by the previous number of possibilities (eg. 3×2, 4×6) produces the number of possibilities for the current number of buttons. Liam used this process to calculate the possibilities for 5 buttons, which he found to be 120. As a class, we then put this to the test by recording actual possible combinations. After finding that the number of possible orders beginning with 1 would be 24, we confirmed that Liam was correct (as 24 × 5 = 120). This then enabled the rest of the class to calculate the number of possible orders for 6 buttons as 720. We enjoyed the challenge!
Some children found that there is a quick way of writing something like 4×3×2×1. Ayden from St John's College in Cardiff, Wales sent us this explanation:
If your jacket has 4 buttons, you can button it up in many different ways. You can find out how many different combinations there are by using the factorial. This involves multiplying the number of buttons by all the numbers below it, ending at one. The symbol for the factorial is the number, followed by an exclamation mark (!). If the number was y, it would look like this: y!.
For 4 buttons, you would use 4!=4×3×2×1=24. There are 24 different ways of buttoning up 4 buttons. For 3 buttons, you would use 3!=3×2×1=6. There are 6 different ways of buttoning up 3 buttons. For 5 buttons, you would use 5!=5×4×3×2×1=120. There are 120 different ways of buttoning up 5 buttons. For 6 buttons, you would use 6!=6×5×4×3×2×1=720. There are 720 different ways of buttoning up 6 buttons. Also, if 5!=120, then to find out 6!, you could multiply the answer for the previous factorial (120) by 6, and you would get the answer: 720.
Shaunak from Ganit Manthan in Vicharvatika, India used algebra to find the general solution:
I found 6 ways to button up a coat with three buttons, and 24 ways to button a coat with 4 buttons. I observed the number of ways to button a coat with 2 buttons, 3 buttons, or 4 buttons. I found a formula: if there are n buttons, there are n! (n factorial) ways to button up the coat. The factorial of a number is that number multiplied by all the natural numbers before it. So, I predict the number of ways to button up a coat with 5 buttons should be 5! = 120 ways. For 6 buttons, the number of ways should be 6! = 720 ways. I can also show the above formula (if there are n buttons, then there are n! ways to button the coat) is true. If there are n buttons, then I can choose one out of n buttons to button first, one out of n - 1 buttons to button second... So, I can button the coat in n × n - 1 × ... × 1, which is n!, which is why I can button a coat with n buttons in n! ways.
Shaunak also sent in a video describing the solutions for coats with different numbers of buttons:
Thank you for sharing this with us, Shaunak!
Thank you as well to the following children, who also sent in correct solutions to this activity: Clara and Izzy from Milton Road Primary School in the UK; Brooklyn Chan, Ruby, Chloe, Hailey, Jaden and Kian from Sha Tin Junior School in Hong Kong; Orlando, Zayn, Sam and Josh from Coppice Primary School in England; Maebh, Maria, Chizzy, William, Aidan, Kyle, David, Adrian, Nina, Ella, Vicky, Wiktoria, Gabbie, Bartosz, Olive, Alaya and Cassy from St Robert's Catholic Primary School in Harrogate, UK; Jack from Cramlington Village Primary School in the UK; Kayhan, Ruel, Felix, Jack and William from Clifton Hill Primary School in Australia; Javen from Sri Kdu in Malaysia; Sai from Eastcourt Independent School in England; Seth, Elliot, Macy, Isabel, Izzy, Joshua, Noah and Alfie from Richmond Methodist Primary School in the UK; Felix, Henry, Rhys and Charlie from Husthwaite CE Primary School in the UK; Tyler from Burnside Primary School in Australia; Camryn from Mountains Christian College; Riley, Aazim and Mustafa from British School Al Khubairat in the UAE; Kabir from Inventure Ac in India; Pak Lam from Hong Kong; Chloe from Eastcourt School in the UK; Grace, Elowen and Erin from Alexandra Junior School in England; Dhruv from The Glasgow Academy in the UK; and Matilda and Eliza from Rainworth State School in Australia.
Teachers' Resources
Why do this problem?
Here you can see a response from a school in Canada together with some useful notes that might be helpful to read before watching the video.
Possible approach
Key questions
Possible extension
Possible support