I am highlighting one M in orange so it is possible to differentiate between the two M’s:
EMMA and EMMA are exactly the same word although they have the swapped the letters of the M’s around it is the word so one is cancelled and because the M’s are repeated twice each combination is going to be repeated twice, there is going to be half the amount of combinations
Part 3
In this section I will research and test as many different methods to work out the different combinations of a word. I will investigate the combinations of various varying words. I will work out an equation to make the task of finding a word’s combinations, simpler as opposed to writing out all combinations by hand.(a very tedious task with words of many letters. And finally to find one that can be used to predict any word’s combinations.
Methodology
I will write out all the combinations for words with one letter, two letters, three letters and four letters. And try to find a pattern and/or equation, using a systematic method as shown in part one.
One letter
C
This has 1 combination
Two letters
2 combinations
Three letters four letters
This has 6 combinations
This is another pattern explanation to show how to prove prediction of combinations:
To every word that you add another letter to, you can place the new letter between each letter of the combinations of the old word to make new combinations
E.G:
1 2 3 4 5 6
E D A , E A D , A E D , A D E , D A E , D E A
A new letter could have been placed between any of these letters and it would have produced a new combination. Since there are four possible places for the new letter to be placed in, you can multiply four by the number of combinations from the previous word, in this case it would be six. Four × six equals twenty four which is the number of combinations for a four letter word with no repetitions.
This table shows a pattern that can be generalised as the number of combinations from the previous word ×the number of letters in the word you are investigating gives you the number of combinations for the word you are investigating. This is because each word has a certain amount of places for each letter to go into e.g.
Has one possible place for the letter to be placed in
Have two places for each letter to be placed in
If you focus on just the A’s you can see there is a space
That the A can be placed into .this is because once the A
Is in place, the other two letters can be rearranged to make
New combinations. It is because once the A is in place
It is like having a two letter word again, you rearrange the
B and C exactly like you did in the previous word.
Now I will investigate repeated letters with one different letter in each word:
Three letters
MEE
EME
EEM
This has three combinations
Four letters
MEEE
EMEE
EMEE
EEEM
This has four combinations
I have observed that in all words that only have one different letter than the rest, the number of letters they have is proportional to the number of combinations they have .this is because you are only capable of moving round the M to make a new combination since moving around the Es will have no effect and will result in a repeated combination.
E.G.:
MEEE is the same as MEEE, as previously explained in part two.
Factorial
This is a method that can be incorporated into working out the number of combinations in a word. Number of letters multi plied by factorial gives you the number of combinations in a word, factorial can be written as N!
The equation can be written out as:
N! = 1×2×3×4……………..….N
The equation is worked out by multiplying all of the numbers up to and including N (which is the number of letters in the word you are trying to work out.)for example if the word had seven letters in it, then you would multiply one by two by three by four by five by six by seven and it would give you the number of combinations in the word. Factorial can be abbreviated to the number of letters in the word followed by an exclamation mark.
The logic behind factorial is that it is a way of working out the amount of possibilities for different combinations there is going to be.
E.G.
If you start off with one letter, A, then you
Add another letter, B, it has two places it
can be placed in to create new combinations.
then you add another letter ,C, and it has 4
three places it can be placed in to make
new combinations. Finally you add, D, and 3
it has four places it can be placed in to
make new combinations. 2
1
Another way to describe it with the name Lucy:
_ _ _ _
when no letters have been introduced, there are four possible letters 4
that can be used ,L, U,C,Y.
L _ _ _
Now there are three possible spaces for any of the remaining letters 3
to be used.
LU_ _
Now there are two possible spaces for any of the remaining letters 2
To be used
LUC_
There is only one space left and Y has only one possible space to go.
I will now investigate a word with repeated letters and see if the factorial equation works with the combinations.
Two letters
AA
One combination
Three letters
AAB
ABA
BAA
Three combinations
Four letters
AABC
AACB
ABCA
ACBA
CBAA
BCAA
BACA
BAAC
CABA
CAAB
ACAB
ABAC
Twelve combinations
These words all have exactly half the amount of combinations than those without repeated letters; this is because they all have two letters the same, whereas the others have one of each letter. So this means that if you were to apply factorial to words with repeated letters, you would need to divide the answer in half.
E.G.
AABC 4×3×2×1 24 24 combinations
- 2
Now I will see if the same rules apply to a word with three letters the same:
Three letters
AAA
One combination
Four letters
AAAB
AABA
ABAA
BAAA
Four combinations
Five letters
AAABC
AACBA
AABCA
AAACB
ABCAA
ACBAA
CBAAA
BCAAA
BACAA
BAACA
BAAAC
CABAA
CAABA
CAAAB
ABACA
ACABA
ACAAB
ABAAC
AACAB
AABAC
Twenty combinations
If you divide the factorial of five by 2 it equals sixty, but that is not the right amount of combinations. To get twenty you have to divide 120 by 6.
If you think of factorial in terms of repeated letters, you can see there is a pattern, by dividing the amount of letters in a word by the amount of repeated letters; you are left with the number of combinations in the word. There are three letters the same so I will divide the number of letters in the word, five factorial by three which is twenty, the right number of combinations.
Now I need to see if the new improved formula works on the names LUCY and EMMA:
LUCY: 4! 24 combinations, which is right
1
EMMA: 4! 12 combinations, which is right
2!
The reason that this formula works is that every time you add another repeated letter, you get a smaller amount of combinations that are valid, not repeated .it works like an opposite to normal factorial ,the more repeated letters you have, the more the more repeated combinations you are going to have. When you add a repeated letter you have to add a number onto the factorial of both the nominator and the denominator because the denominator factorial is how many times less the number of combinations is going to be than if there were no repeated letters at all
E.G.
With AB adding the letter C:
It can be placed any where to make a new combination: A B
3×2×1=6
B A
But adding a letter that is already there like B: A B
B A
3×2×1=6
2×1=2
In this case you have to divide the second word by two which gives you three, half of what the number wouldn’t have been if it wasn’t for the repeated letters.
However this would not work with words such as AABBB or AABB
But if I divided the number of letters in the word by two factorial numbers, one for each set of repeated letters
E.G.
AABB 4! = 24 = 6
2! ×2! 4
AABB
ABBA
BBAA
ABAB
BABA
BAAB
Six combinations
This has the same principles as only one set of repeated letters except that you need to add another factorial for a new set of repetitions.
Now I will apply this to a word with unequal numbers of repeats to show that it works.
BBAAA
ABBAA
AABBA
AAABB
BABAA
BAABA
BAAAB
ABABA
ABAAB
AABAB
Ten combinations
5! = 120 = 10 combinations which is right so my adapted theory is
2! ×3! 12 SUCCESSFUL
I will now use my formula on SUCCESSFULLY which has four sets of repeated letters to tell you how many times it can be rearranged
11! = 39916800 = 831600 combinations
3!×2!×2!×2! 48