EMMA, EAMM, EMAM
AMME, AEMM, AMEM,
MAME, MMAE, MEMA, MMEA, MAEM, MEAM
There are 12 combinations. Half as many as Lucy
I can put this data in a table.
A name with 3 letters different
MAT, MTA
There are 2 combinations with M at the front.
2x3=6. There are 6 combinations altogether.
3 letters with 2 letters the same
ANA, AAN
NAA
There are 3 combinations altogether, half as many as 3 different letters the same. I can now add this data to the table.
I can work out a name with 2 letters different and a name with letters and 2 letters the same without writing them down.
I have found a pattern between names with all different letters and the names with 2 letters the same. If you look at my table then you will b able to see that names with different letters is double names with 2 letters. This breakthrough will help to speed up the process of working out the combinations, I can now work out the answer for all different letters and now that all I have to do is half that answer and will reach the answer for 2 letters the same.
Now I need to try and find a pattern to work out what the combinations are of whatever name with any number of letters, but first I need to do 5 letters the same and 5 letter with 2 the same.
5 letters with 2 letters the same
SCOTT, SCTOT, SCTTO, SOCTT, SOTCT, SOTTC, STOCT, STCOT, STTOC, STTCO, STOTC, STCTO
12X5=60
There are 60 combinations.
As you can see I have taken a shortcut to reach the answer. I have found out that there are 12 combinations with S at the front. So, that must mean that there are 12 combinations with C, O and T at the front. So if I multiply 12 by 5 I will reach the answer quicker. I also will know how many will be in a 5 letter name with all different letters. All I have to do is multiply the answer I already have by 2. 60x2=120.
I can add this data to my table
I have found a pattern in my table. If you look at it the first number in the “All Different” column is 2 if multiply that by 3 you will get 6, which is the next number. If you then multiply 6 by 4 you will get 24, which again is the next number. So I think to work out the next number in the sequence you multiply your last answer by the by the number of letters in the next name. For example; if I wanted to find how many combinations were in a 4 letter name with all the letters different I would look at my last for a name with 3 letters the same, which is 6, multiply that by 4, which is the next number of letters, and you get 24 which is the correct answer. This process works for names with 2 letters the same too.
I predict that for a name with 6 letters different is 120 (my last answer) x6 (the next number of letters) =720. I will use the name ABCDEF.
To prove my prediction I could work it out the, very, long-winded way of writing down every combination. For example; ABCDEF, ABCDFE, ABCFDE and so on. This method would take a lot of time but would get the correct answer of 720. I have worked out another method which quick and easy. To use this method I need no answers. I am trying to work out the number of combinations with 6 different letter name; I know that the answer is 720 but I need the answer before to work it out at the moment. But if I use the information I already know of times the next number of letters I can get to the next step of working out an answer without any previous answers.
So if I go to the beginning, for a name with 2 different letters there are 2 combinations, to get that I can do 1x2, which gives the correct answer, to get the answer for a name with 3 different letters I do 1x2x3. If I carry on with the old method, I have the answer before 6 and need to answer by 4 to get the answer. If I use the new method I do 1x2x3x4, which will also give the same answer of 24. This method will work every time and is very useful because I can work out the number of combinations without needing any previous answers.
I can use this method for names with 2 letters the same too. As I have said before names with 2 letters the same have half the number of combinations than a name with all different letters with the same number of letters. So using the method I have just explained; I will use it normal way so, 1x2x3x4=24 but then divide the answer by 2 which makes 12 to get to the answer.
Now I have the method I need to work out a formulae.
The method I am using in terms of n is; n x (n-1) x (n-2) x (n-3) etc…
This is an equation but would take a long time to write down and to make this easier I have found on my calculator a button called factorial and symbal is !.
This button means n x (n-1) x (n-2) x (n-3) etc…
Which is the same as my above formulae.
So the formulae for a name with all different letters n!.
The formula for a name with 2 letters the same is n! /2.
Now I have the main formulae I now need to pattern between it and names with 3 letters the same and 2 letters the same.
3 letters with 3 letters the same
AAA = 1 combination
4 letters with 3 letters the same
BAAA
ABAA, AABA, AAAB = 4 combinations
5 letters with 3 letters the same
CBAAA, CABAA, CAABA, CAAAB
BCAAA, BACAA, BAACA, BAAAC,
ABCAA, ABACA, ABAAC, ACBAA, ACABA, ACAAB. AABCA, AACBA = 20 combinations.
I have found a pattern between names with all different letters and a name with 3 letters the same. If you look at my table for a name with 4 different letters has 24 combinations and a name with 4 letters and 3 letters the same has 4 combinations. 24/4 =6. This sum shows that the formula for a word with 3 different letters is n! /6.
A name with 4 letters and 4 letters the same
AAAA = 1 combination
A name with 5 letters and 4 letters the same
BAAAA,
ABAAA, AABAA, AAABA, AAAAB = 5combinations
I have found a pattern between names with all different letters and names with 4 letters the same. If you look at a name with 4 different letters there are 24 combinations and if you look at a name with 4 letters and 4 letters the same there is 1 combination. So the formula for a name with 4 letters the same is n! /24. To show this works; if you look at a name with 5 different letters there are 120 different combinations and if you look at a name with 5 letters and 4 letters the same there are 5 combinations. 120 / 5 =24.
After working out all the formulas for words with all different letters, 2 letters, 3 letters and 4 letters the same I have spotted a pattern between their formulas. To show this pattern if you look at the formulae for a name 2 letters the same, which is n! / 2 you have to divide n! by the number of letters the same factorial. So you could write it n! / 2! which is the same as n! / (2x1) This pattern works for every formulae. For a name with 3 letters the same the formulae is n! / 6 which is the same as n! / 3! Which is the same as n! / (3x2x1). This will help with the overall formulae. If we call the bottom factorial x. The formulae becomes n! / x! x= the number of letters the same (so for 2 letters the same it would be 2! and for 3 letters he same it would be 3! etc…).
The overall formula is n! / x!.
