The total rearrangements for any word with 5 letters, 2 of them being the same letter = 60
The total arrangements for any word with 5 letters, 3 of them being the same letter = 20
A word that has 5 letters, 4 of them being the same, has a total number of rearrangements of 5
6x4=24 3x4=12 1x4=4
24x5=120 5x12=60 4x5=20 1x5=5
FOR WORDS WITH 1 LETTERS THE SAME.
3x2x1=6
4x3x2x1=24
5x4x3x2x1=120
For words with 0 letters the same there is a formula:
‘n(n-1)(n-2)(n-3)…(n-(n-1))’
I have done some research, using the internet, on this formula. I found a web site, http://homepages.which.net/~gk.sherman/mea.htm. This page is shown on the next page. It explains that this formula is normally expressed as ‘factorial’, that has the symbol ‘!’.
So the overall formula for words with 0 letters which are the same is:
‘n!’
n=total number of letters in a word
Factorials
The factorial of a number, n, is denoted by n!
One definition:
n! = 1 * 2 * 3 * ... * (n-1) * n
Another definition:
I'm trying to find out about the history of factorials. Please e-mail me if you have any info on their discovery.
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Created 26/3/00
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FOR WORDS WITH 2 LETTERS THE SAME.
3x2x1/2=3
4x3x2x1/2=12
5x4x3x2x1/2=60
The formula is:
‘n!/2’
n=total number of letters in a word
FOR WORDS WITH 3 LETTERS THE SAME.
3x2x1/6=1
4x3x2x1/6=4
5x4x3x2x1/6=20
The formula is:
‘n!/6’
n=total number of letters in a word
FOR WORDS WITH 4 LETTERS THE SAME.
4x3x2x1/24
5x4x3x2x1/24=5
The formula is:
‘n!/24’
n=total number of letters in a word
THE OVERALL FORMULA!
1 letters the same 2 letters the same 3 letters the same 4 letters the same
n! n!/2 n!/6 n!/24
n!/2x1 n!/3x2x1 n!/4x3x2x1
Notice that the ‘n!’ is always divided by the factorial of the number letters that are the same.
From this information I have made a new formula:
n!/s!
n=the total number of letters in a word.
s=the number of letters that are the same.
TESTING THE FORMULA
To test this formula, I am going to calculate the number of rearrangements for a word with 6 letters, 5 of them being the same letter.
6x5x4x3x2x1/5x4x3x2x1=
=720/120=6
There are no more rearrangements. This shows that my formula is correct!
Oh NO!
The formula ‘n!/s!’ does not work with words such as the above. This is because there are two groups that contain letters which are the same.
2/5 letters the same and 3/5 letters the same in one word has 10 rearrangements
2/4 letters the same and 2/4 letters the same in one word has 6 rearrangements
5x4x3x2x1/12=10 4x3x2x1/4=6
n!/12 n!/6
n!/d!xs!
n=total number of letters in the word
d=The number of letters in the first group that contains the same letters
s=The number of letters in the second group that contains the same letters
I believe that for every different letter in a word, there should be another number which the ‘n!’ should be divided by, e.g:
abbccc
This word has a total number of 6 letters, but only 3 of the letters are different. So I believe that the formula for this would be:
n!/a!xb!xc!
n= total number of letter is the word
a=number of same letter(1) in a word
b=number of same letter(2) in a word
c=number of same letter(3) in a word
So the total number of rearrangements for the word would be:
6!/1!x2!x3!=
=720/12=60 rearrangements
IS IT CORRECT?
My formula does is correct! There are 60 rearrangements here.
So the new overall formula is
n!/a!xb!xc!
n= total number of letter is the word
a=number of same letter(1) in a word
b=number of same letter(2) in a word
c=number of same letter(3) in a word
From this information I can make some new formulas for different words:
Aabbcc This has the formula, ‘6!/2!x2!x2!’ which equals 90 rearrangements
Aaabbbcc This has the formula, ‘11!/3!x3!x2!’ which equals 554400 rearrangements
Conclusion
My prediction was incorrect! What I did to begin with was:
A1A2BB
A2A1BB
I did not know that this was not permitted until I was told so by person X.
Evaluation
To find my first formula, ‘n(n-1)(n-2)(n-3)…(n-(n-1))’, I used the fact that the number of letters in that word was multiplied by the number of rearrangements for the previous number;
2x1=2
3x2=6
4x6=24
Using the explanation in these paragraphs
I made the formula.
The reason I always put a ‘1’ at the end of each equation (such as 3x2x1) is because I found it difficult to put an ending to my first formula. Without a 1, my formula would be ongoing, i.e.
N(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)…
I made a mistake, forgetting about words that contain two groups of letters. This put me behind. If I had remembered, I could have saved myself some time.
