A 3 letter name such as TOM has 6 arrangements:
TOM TMO
MOT MTO
OTM OMT
A 4 letter name such as LUCY has 24 arrangements:
LUCY LUYC LYUC
LYCU LCYU LCUY
ULCY ULYC UYCL
UYLC UCLY UCYL
CULY CUYL CYLU
CYUL CLUY CLYU
YCUL YCLU YLUC
YLCU YUCL YULC
I am now going to put this into a table of results:
I have found a formula for this. I found that if you factorise the number of letters in a word, you get the number of arrangements. If you times 1 x 1, you get 1 arrangement. If you time 1 x 2, you get 2 arrangement.
Test:
1 x 1 = 1
1 x 2 = 2
1 x 2 x 3 = 6
1 x 2 x 3 x 4 = 24
So the formula is N factorised = A, with N being the number of letters in a word, and A being the number of arrangements. N factorised can be shown as N!
On my scientific calculator, I can factorise numbers by first putting in the number I want to factorise (N), and then I press:
I will now make a few predictions. If your name is CHRIS, which has 5 different letters, how many arrangements will there be? I will do the same for THOMAS (6 letters), PAULINE (7 letters), REGINALD (8 letters) and GERALDINE (9 letters).
CHRIS: 5! = 1 x 2 x 3 x 4 x 5 = 120 arrangements
THOMAS: 6! = 1 x 2 x 3 x 4 x 5 x 6 = 720 arrangements
PAULINE: 7! = 1 x 2 x 3 x 4 x 5 x 6 x 7 = 5040 arrangements
REGINALD: 8! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 = 40320 arrangements
GERALDINE: 9! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 = 363880 arrangements
This formula makes sence. If you have LUCY:
LUCY LUYC LYUC
LYCU LCYU LCUY
ULCY ULYC UYCL
UYLC UCLY UCYL
CULY CUYL CYLU
CYUL CLUY CLYU
YCUL YCLU YLUC
YLCU YUCL YULC
And you add a new letter such as A, you get:
ALUCY ALUYC ALYUC
ALYCU ALCYU ALCUY
AULCY AULYC AUYCL
AUYLC AUCLY AUCYL
ACULY ACUYL ACYLU
ACYUL ACLUY ACLYU
AYCUL AYCLU AYLUC
AYLCU AYUCL AYULC
Because you add another letter, in this case, A, you will have to times this by 5. This is because the above arrangements all start with A, but if you had all the arrangements, you would 24 arrangements for each of the 5 letters at the beginning.
5 x 24 = 120 which is the same as the formula gives.
Having found a formula for words in which all letters are different, I will now attempt to find a formula for words in which some letters may be repeated.
To make it easier to me, I will use A, B and C etc. to mark the letters.
In a word like EMMA, which I will now translate into AABC, there are 12 different arrangements:
AABC AACB ABAC ABCA ACBA ACAB
BCAA BACA BAAC CBAA CAAB CABA
There are 4 different letters. A is used twice, B and C are both used once.
If it was just AABB, there would be only 6 arrangements:
AABB ABAB ABBA
BBAA BABA BAAB
If it was just AABB, there would be only 6 arrangements.
I will now do the same for a three letter word:
AAB ABA BAA
There are only 3 arrangements. There are 3 different letters, one of which has been used twice, the other has been used only once.
I will now put this into a table:
I then started playing around with this to see if I could find a formula. First I tried N times A times B times C, but this didn’t work:
3 x 2 x 1 = 3 is incorrect!
Then I tried doing N over (A times B times C), but this also did not work:
3
(2 x 1 x o) = 3 is incorrect!
Because my formula for words with 1 letter was X!, I thought I would re-do the above table but make all the numbers factorised, and see if there are any patterns:
Using the numbers from this table, I started playing around with the numbers again.
First I tried multiplying N! times A! times B! times C!, but this was, again incorrect:
6 x 2 x 1 x 0 = 3 is incorrect!
Then I tried N! over (A! times B! times C!):
6
2 x 1 = 3 is correct!
To test if this applies to all words in which letters are repeated I did the same for the other two words I investigated:
24
2 x 1 x 1 = 12 is correct!
24
2 x 2 = 6 is correct!
In this case, the formula is:
N! / (A! x B! x C! x X!) = number of arrangements
With N! being the total number of letters,
A! being the number of times A was used in the word
B! being the number of times B was used in the word
X! being the number of times X was used in the word
To make absolutely sure this formula is correct, I am going to test it one more time on a 5 letter word such as AAAAB:
AAAAB AAABA AABAA ABAAA BAAAA
In this word N is 5, A is 4 and B is 1
5! 5 x 4 x 3 x 2 x 1 120
4! x 1! = (4 x 3 x 2 x 1) (1) = 24 = 5
This is the correct answer!
Now that I have a formula, I am going to made a prediction. I am going to work out, using my formula how many arrangements there would be for the following sequence of letters:
AAAAAAABBBCCD
A = 7 B = 3 C = 2 D = 1
N = 13
Because this is a difficult calculation to make, I will do it by hand:
My prediction is that for the word:
AAAAAAABBBCCD
There will be arrangements
