From these 2 investigation I worked out a method: I used numbers to represent the letters,
Step1: 1234---Do the last two number first then you get 1243.
1243---Do the last three numbers and try the possibility. 1423. 1432. 1342. 1324, because the number 2 has been the first number of last three numbers, so we don’t do it again.
Step2: we have list all arrangements of 1 go front, so we do 2 go front. 2134 and we do same thing to it, it will like this:
2134---2143, 2143---2431,2413,2314,2341
Step3: We have finished 2 go first, then let’s do 3 go ahead.
3124---3142, 3142---3241,3214,3412,3421
Step4: We have finished 3 go ahead, then try 4
4123---4132, 4132---4231,4213,4312,4321
I have listed all the arrangements of 1234, I will use this method so I can arrange the letter which have 4 figures or more.
Once I have discovered these formulae I am going to investigate, other combinations of letters and different amounts of letters. I will then try to discover a link between the formulae to enable me to find a formula for the general case.
What if there were 4 letters with 2 different?
Arrangements for anna:
anna anan aann
nana naan nnaa
There are 6 arrangements for aabb. From 2 different letters to all different letters in a 4-letter word I have found a pattern of 6, 12 and 24. Maybe it would be easier to see what is happening if I used larger words.
What if there was a five-letter word? How many different arrangements would there be for that?
As I have found that there were 24 arrangements for a 4 letter word with all different letters and that there were 6 combinations beginning with each of the letters in the word I predict that there will be 120 arrangements for qlucy, 24 for q, 24 for l, 24 for u; and so on. 120 divided by 6 (number of letters) equals 24. Previously in the 4-letter word, 24 divided by 4 equals 6, the number of possibilities there were for each letter.
For the sake of convenience I have used lucy and put a q in front of it to show that there are 24 different possibilities with each letter of a 5 lettered name being all different.
qlucy qucyl qcylu qycul
qluyc qucly qcyul qyclu
qlcuy qulcy qculy qyulc
qlcyu qulyc qcuyl qyucl
qlyuc quycl qclyu qylcu
qlycu quylc qcluy qyluc
I can see that the numbers of different arrangements are going to dramatically increase as more different letters are used. So as a general formula for names with x number of letters all different I have come up with a formula. With Lucy's name; 1x2x3x4 = 24. With qlucy; 1x2x3x4x5 = 120. This is called a factorial. There is a button on most scientific calculators which basically does this formula for you according to the number of letters in the word. If I key in (assuming all the letters are different) factorial 6, I get it gives me 720, which makes sense because 720 divided by 6 equals 120 which was the number of arrangements for a 5 letter word, and it continues to follow that pattern.
Total Letters (all different) Number of Arrangements
1-1
2-2
3-6
4-24
5-120
6-720
So now that I've explained the pattern of general x lettered words, what do I do if any letters are repeated? Like in Emma; it has 4 letters but 2 of which are the same. 4 factorial equals 24, but I could only find 12, which means that there must be more to it that just factorial in that way.
To make it a bit easier instead of using letters as such I will use x's and y's (any letter). I will start with xxyy:
Arrangements for xxyy:
x=a, y=b
Arrangements for aabb:
This is a 4-letter word with 2 different. I have done this with;
aabb abab baab
aaba baba bbaa
There are 6 arrangements. What if I had xxxyy?
x=a, y=b
aaabb aabab aabba ababa abaab
abbaa bbaaa baaab babaa baaba
There are 10 different arrangements for this sequence.
What if I had an arrangement of aaaab?
aaaab aaaba aabaa
abaaa baaaa
There are 5 different arrangements for this sequence.
If I go back to aaabb; there are 3 a's and 2 b's in a total of 5 unknowns. As each letter has its own number of arrangements i.e. there were 5 beginning with a, and 5 beginning with b, I think that factorial has to be used again. Also in a 5-letter word there are 120 arrangements with 24 arrangements (120 divided by 5) for each letter. As that seemed to work, I had a go at trying to work out a logical universal formula. I came up with; The total number of letters factorial, divided by the number of a's, b's ect factorised and multiplied
Formula for emma=
4!/1!x1!x2!=48 /4=12
For example:
A five letter word like aaabb; this has 3 a's and 2 b's. So: 1x2x3x4x5 / 1x2x3 x 1x2 = 120 / 12 = 10
A four letter word like aabb; this has 2 a's and 2 b's
So: 1x2x3x4 / 1x2 x 1x2 = 24 / 4 = 6
A five letter word like aaaab; this has 4 a's and 1 b
So: 1x2x3x4x5 / 1x2x3x4 x 1 = 120 / 24 = 5
Five letter words like abcde; this has 1 of each letter (no letters the same)
So : 1x2x3x4 / 1x1x1x1x1x1 = 24 / 1 = 24
All these have been proved in previous arrangements. This shows that my formula works.
