Three letters different
SAM
SMA
AMS
ASM
MAS
MSA
6 different combinations
Lucy
See above
24 different combinations
5 Letters different
ANDIE
ANDEI
ANEDI
ANEID
ANIDE
ANIED
ADNIE
ADNEI
ADENI
ADEIN
ADIEN
ADINE
AIDNE
AIDEN
AIEDN
AIEND
AINED
AINDE
AEIDN
AEIND
AENID
AENDI
AEDNI
AEDIN
So far I have 24 combinations. All combinations begin with ‘a’ . As previous results have shown the combinations beginning with the following letters. Either e d n or I will each give 24. Therefore for the total number of combinations all I need to do is multiply 24 by 5.
This gives a total of 120 different combinations for the five-letter name, with all letters different 120.
I am now going to put the results I have so far into a table. This way I can see if a pattern has emerged
I can see a pattern
The number of letters in the current name multiplied by the number of arrangements in the previous.
4 *6 =24
5*24 = 120
Therefore the arrangements for a name with 6 different letters will be. 6* 120 which gives you 720. And so on for seven different arrangements.
Or better still you can do 5!
Which is called 5 factorial. Which is the same as 5 * 4*3*2*1
Therefore by using factorial (!) I can predict that for an 8-letter world there will be 40320 different arrangements.
This is a lot simpler
The formula for this is : N! = A
Part 2
Now I am going to investigate the number of different arrangements in a word with 2 letters repeated 3 letters repeated and 4 letters repeated.
(See separate sheet for combinations for the combinations which led me to this table.)
Table of Results
To work out the formula for the amount of letters repeated in a word. I took a look at the table which a put together.
I am most concerned with the no letters repeated column and the number of letters repeated. For a 5-letter word, with all five letters different, you get an optimum of 120 combinations. When you repeat a letter you subtracting all the combinations that that would have made, if the letter were different from the other four words.
For example Rabbi gives a total of 60 combinations.
But if I replaced a b with a different letter such as e to get rabei the combinations would double.
I therefore came to the conclusion that to see how many combinations a certain word would get. You factorial the amount of letters in that word to get the optimum number of combinations. And to subtract the appropriate amount of combinations that the repeated letter takes the place of you must divide it by the amount of letters repeated factorial!
n! = the number of letters in the word
x! = the number of letters the same
Part 3
I am now looking to find a formula for a word with two or more letters the same.
Firstly I ma going to see how many combinations are made with the word aabb
Aabb
Bbaa
Abba
Abab
Baba
Baab
This gives 6 combinations
xxxyy
xxxyy xxyxy xxyxx xyxyx xyxxy
xyyxx yyxxx yxxxy yxyxx yxxyx
this gives 10 combinations
In this example each letter has its own amount of combinations. When you begin the word with x you get 6 combinations, and when you begin the word with y you get 4 combinations. With this in mind I think factorial(!) must be used again, As in the original formula n!/x!
I believe that the new formula will be n!/ something.
To work out what that something was I looked to the X’s and Y’s. And figured that when you multiply together, the amount of Y's and the amount of X's in the letter, and factorial them both. You get the total combinations that the word in question gives.
I am now going to test the formula to prove it works
A four letter word like aabb; has 2 A’s and 2 B’s
So: 1x2x3x4
(1x2) * (1x2)
= 24
4
= 6 different arrangements
A five letter word like aaaab; this has 4 A’s and 1 b
So: 1x2x3x4x5
(1x2x3x4) * (1)
= 120
24
= 5 different arrangements
This shows that my formula works
N! = The number of letters in the word
x!y! = the number of repeated letters the same
