Madeleine and Luke at Greystoke Primary worked
carefully on four buttons:
We decided that we needed an organised way of working to make
sure that we found all the different ways.
So we started with all the ways when you start with the first
button, then all the ways you start with the second button
In total we found $6$ ways for each so $6 \times 4 =
So we think that there are $24$ ways of buttoning up the
Here are all the ways:
$1234$, $1243$, $1324$, $1342$, $1423$, $1432$, $2134$,
$2143$, $2431$, $2413$, $2314$, $1341$, $3124$, $3142$, $3412$,
$3421$, $3241$, $3214$, $4321$, $4312$, $4213$, $4231$, $4123$,
$6 \times 4 = 24$ ways.
Jordan from Birds Bush Primary gave some
helpful detail about how we can make sure we don't miss any
At the start of a solution start with the basic solution of
Then switch the last two numbers,$1243$.
As you can't swap them again, swap the second number in the
solution to another but not the first number. In this case
it's one eg: $1324$.
Now swap the last two numbers' places - $1342$.
As before you swap the second number in the solution with the only
number which hasn't been second: $1432$.
And again swap the last two numbers' places - $1423$.
To get all solutions just use the same method but change the first
number until all numbers have been in first place and you should
end up with the $24$ different combinations as follows;
$1234$, $1243$, $1324$, $1342$, $1432$, $1423$, $2341$, $2314$,
$2413$, $2431$, $2134$, $2143$, $3214$, $3241$, $3412$, $3421$,
$3124$, $3142$, $4321$, $4312$, $4231$, $4213$, $4123$, $4132$.
Children from La Mare De Carteret
Primary school agreed with 24 being the total number of ways for
Scarlet, Will and Oliver, also from Greystoke Primary, investigate
We looked at what happens when there are $5$ buttons on the
Again we had an organised method and started by looking at all the
ways when starting with the first button.
We found $24$ ways when starting with the first button, so we
didn't need to do any more working, we just calculated $24
\times 5$ which is $120$ ways of buttoning up the coat with
I like the way you decided you didn't
need to work out the number of ways for each button being first as
it would be the same for all buttons. Noticing short-cuts
like that is a great skill for a mathematician.
Ella who goes to North Molton Primary
looked at the number of ways to do up the buttons in a more general
way. Here is what she wrote:
First of all I got a piece of paper and drew all four buttons,
naming them one, two, three and four.
Then, I experimented with the numbers, seeing how many
different ways there were to order them.
I started of by working out how many different ways there were
to do them up starting with button number one. There were six
different ways. I did the same with buttons two, three and four.
Each of them had six different ways.
So, all together there were twenty four different ways to do
the four buttons up.
I tried the same method with three buttons. All together there
were six different ways to do the three buttons up.
I realised that all the different ways for four buttons to be
done up (twenty four) was the number of buttons (four) multiplied
by the amount of ways a jacket could be buttoned up with one fewer
So, the way to work out how many ways there are to do up the
jacket is the amount of buttons it has multiplied by the number of
ways you can do it up with one less button.
For four buttons:
$4 \times 6 = 24$
(number of buttons) $\times$ (the number of ways to do it up
with one less button) = amount of ways to do it up.
Very well noticed, Ella. Dan and G
from St Saviour's also suggested this was a good way to calculate
the number of ways.
William from North Molton looked at it
slightly differently. He said:
I worked out that you have to times all the numbers you are
using then you will get the answers e.g.
$1-4 = 1 \times 2 \times 3 \times 4 =
$1-5 = 1 \times 2 \times 3 \times 4 \times 5 =
$1-6 = 1 \times 2 \times 3 \times 4 \times 5 \times 6 =
Krystof from Uhelny Trh in Prague
used a special symbol to write this down:
$2! = 2 \times1$
$3! = 3 \times 2 \times1$
$4! = 4 \times 3 \times 2 \times1$ etc.
Krystof explained, therefore, that for
$n$ buttons there would be $n!$ different ways of buttoning them
Alex from Maidstone Grammar expressed the
total number of ways of doing up a button slightly differently
There is a pattern to do with the factor of the numbers.
The first amount of sets is $6$, which equals $3 \times 2$.
The second set is a larger figure, $3 \times 8$.
Finally, $5$ buttons is a huge $3 \times 40$.
Each time, the second factor is multiplied larger.
It is first multiplied by $4$, to give $8$, and then $5$ to make
Therefore, $6$ buttons should equal $3 \times (40 \times 6)$ which
is $3 \times 240$, giving $720$ combinations.
I wonder whether you can see how these
different methods of expressing the total number of ways of doing
up the buttons are connected? If I used Alex's or Krystof's
or William's or Ella's method, would I get the same answer, say for
$10$ buttons? Why?
Well done to you all.