● 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. As it is easier to see what is happening with more difficult arrangements I will do a table for more letters and try and look for a more meaningful explanation..
-I can ask a question, what if there was a five-letter word? How many different arrangements would there be for that? And how many words will be identical?
Prediction:
As I have found that there were 24 arrangements for a 4 letter word with all different letters and that there were 6 ways begging with one of the letter I predict that there will be 120 arrangements for “esamz”, 24 for “e”, 24 for “s”, 24 for “a",24 for “m”,24 for “z”, so120 ( arrangements) 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. 9pYc3uWc
For the sake of convenience I have used “esam” and put a “z” in front of it to show that there are 24 different possibilities with each letter of a 5 lettered name being all different.
zesam, zsame, zames, zeasa, zesma, zsaem, zamse, zmaes, zeasm, zseam, zasem, zmsea, zeame, zsema, zasme, zmsae, zemsa, zsame, zaems, zmeas, zemas, zsmea, zaesm and zmesa.
- I can observe that the numbers of possibilities for different arrangements are going to 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 Esam’s name; 1x2x3x4 = 24. With zesam; 1x2x3x4x5 = 120. If I key in (lets say the number of letters all different) factorial 6 it gives me 720, with makes sense because 720 divided by 6 equals 120 which was the number of arrangements for a 5 letter word and it continues to fall in that pattern.
Total Letters (all different) Number of Arrangements
1 letter gives us an arrangement of 1
2 letters will give us an arrangement of 2
3 letters will give us an arrangement of 6
4 letters will give us an arramgement of 24
5 letters will give us an arrangement of 120
6 letters will give us an arrangement of 720
-So now that I've explained the pattern of general x lettered words, what do I do if there are repeated letters? 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 are more to it than just factorial in that way.
-To make it a bit easier instead of using letters I will use “x”'s and “y”'s (or any other letter any letter). I will start with xxyy:
-The possible arrangements for xxyy:
I will make x=a, y=b
This is a 4-letter word with 2 different. I have done this with;
aabb, abab, baab,aaba, baba and bbaa
-There are 6 possible arrangements arrangements, but what if I had 3 “x” ’s and 2 “y” ’s?
■ I will alsomake x=a, y=b and the arrangements will be as follows:
aaabb, aabab, aabba, ababa, abaab, abbaa, bbaaa, baaab, babaa and baaba.
-There were 10 different arrangements.
What if I had an arrangement of 4 “x”’s and 1 “y” ?
I will make x=a, y=b:
aaaab, aaaba, aabaa, abaaa and baaaa:
-There were 5 different arrangements
If I go back to xxxyy; there are 3 x's and 2 y's in a total of 5 unknowns. As each letter has its own number of arrangements i.e. there were 5 beginning with x, and 5 beginning with y, I think that factorial has to be used again. Also in a 5-letter word there are 120 arrangements and 24 arrangements (120 divided by 5) for each letter. As there I a divide issue involved I had a go at trying to work out a logical universal formula. I came up with; The number of total letters factorial, divided by the number of x's, y's e.c.t factorised and multiplied.
-Then the formula is:
N over X times Y:
- I will give an example:
A five letter word like aaabb; this has 3 a's and 2 b's (3 x's and 2 y's). So : 1x2x3x4x5 / 1x2x3 x 1x2 = 120 / 12 = 10
A four letter word like aabb; this has 2 a's and 2 b's (2 x's and 2 y's) So : 1x2x3x4 / 1x2 x 1x2 = 24 / 4 = 6
A five letter word like aaaab; this has 4 a's and 1 b (4 x's and 1 y) 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 results have been very successful and were proved in previous arrangements and this shows that my formula works.
