Table showing the number of different arrangements of letters possible in words of different lengths with no repeated letters
After recording the results in a table, I found that there was a pattern in my results. The number of arrangements of letters in a 3-letter name without repeated letters is 6, and the number of arrangements in a 2-letter name without repeated letters is 2. There is only one possible arrangement of letters in a 1-letter word.
The number of arrangements of letters in each name increases like this:
1x1=1
1x2=2
1x2x3=6
1x2x3x4=24
Or
1 2 6 24
X2 x3 x4
This can also be written as factorials,
e.g. 2 factorial (2!)=1x2=2.
3 factorial (3!)=1x2x3=6.
4! =1x2x3x4=24.
So, to find the number of arrangements for a word with no repeated letters, the formula used would be:
(Number of letters)!=number of arrangements.
Or
n!=a
(Where n=the number of letters in the word and a=the number of arrangements)
I was then able to calculate the numbers of arrangements in names of other lengths, and so extend the amount of results which I had.
Table showing the number of different arrangements of letters possible in words of different lengths with no repeated letters
I then investigated more names, which had repeated letters, and found that there were fewer arrangements of letters. There also appeared to be a pattern involving factorials in this case, although it was not the same pattern as was found in names with no repeated letters. I recorded my results in a table.
Table showing the number of different arrangements of letters possible in words of different lengths with no repeated letters
I found that the numbers in these results were in the same pattern as the results for the names with no repeating letters, but the numbers of arrangements of the names with no repeating letters was halved for example:
So, the formula for the number of arrangements for words containing two repeated letters (for example, sArAh) is:
(Number of letters in word)!
2!
I then investigated the number of different arrangements of letters in a word with tree repeated letters, and, once again found a similar pattern. By testing the name ‘PIPPA’, I found this set of results:
I found that the number of arrangements which can be made when a word has 3 repeated letters is:
(Number of letters in word)!
3!
I found a pattern in the formulae which I had used to find the number of arrangements in each case.
The formulae which I have so far are:
(Number of letters)! (Number of letters)! (Number of letters)!
1! 2! 3!
I was then able to predict the formulae for other words, such as words with 4 and 5 repeated letters. I tested these formulae which I had predicted by writing out the arrangements of letters in the smaller words. I found that these formulae were correct. I then recorded all of my results in a table.
The formulae I found for the number of arrangements of letters in words containing different numbers of repeated letters were:
For a word with 4 repeated letters:
(Number of letters)!
4!
For a word containing 5 repeated letters:
(Number of letters)!
5!
I then used these formulae to find a general formula giving the number of arrangements which can be made when a word has repeated letters. The formula which I found was:
(Number of letters)!
(Number of repeated letters)!
After investigating the number of arrangements made in a word with only one set of repeated letters, I investigated the number of arrangements in a word with two sets of repeated letters, for example ‘XXYY’. I tested the number of arrangements of letters which were possible within this word.
XXYY XYXY YYXX
YXYX XYYX YXXY
I tested more words of different lengths containing two and three sets of repeated letters, and found a pattern in the results that was similar to those which I had found in the work which I had previously done, using words with only one set of repeated letters. I found this set of results:
I continued to investigate different formulae which could be used to find the number of arrangements in a word with two sets of repeated letters.
A four-letter word such as XXYY has 2 X's and 2 Y's. Therefore 1x2x3x4 (as there are 4 letters) or (1x2)x(1x2) will give the number of different arrangements when:
Number of letters in word (n) factorial
(number of X's)!x(number of Y's)!
or
n!
(Number of x!)x(number of y!)
I then used this formula to calculate the number of arrangements possible in words of length 6,7 and 8 letters containing two sets of repeated letters. I also tested words which contained different numbers of X's and Y's, for example: XXXYY or XXYYY. These results are also shown in the table below. I also calculated some values for other words with greater numbers of letters.