4 Letter word
Using the results I’ve got so far I’m going to predict the total number of combinations for the letters in a 4 letter word. I predict that the answer is 1x2x3x4, because the number of arrangements for the 1st letter is 4, then the number of arrangements for the 2nd letter after assigning a letter as the 1st is 3. Then after assigning a letter as the 1st and 2nd letters there are only 2 arrangements for the 3rd letter and finally once you’ve assigned letters as the 1st, 2nd and 3rd letter there’s only 1 letter that can be assigned to the 4th letter.
Results prediction equation
= 1x2x3x4= 24 combinations in total.
I am now going to list all the numbers of combinations to see if my prediction is correct.
BSTI BSIT BTSI BTIS BIST BITS
ITSB ITBS IBST IBTS ISTB ISBT
STIB STBI SBIT SBTI SITB SIBT
TSIB TSBI TBIS TBSI TISB TIBS
As you can I see I was right.
Formula for no doubles:
My example word is BITE a 4 letter word.
ITBS BITS STBI TSBI = number of letters that can be assigned as the 1st letter
BSTI BTSI BIST= number of letters that can be assigned as the 2nd letter after the 1st 1 has been assigned.
IBST IBTS= number of letters that can be assigned to the 3rd letter once the 1st and 2nd letters have been chosen.
IBST= number of arrangements for the 4th letter once the 1st, 2nd and 3rd letters have been assigned.
1x2x3x4.
I have done some research by looking in an A-level text book called A Concise Course In A-Level Statistics written by J. Crawshaw and J. Chambers. In this book I found an expression called Factorial. To get the number of combinations for a word you have to get the product of all the whole numbers from 1 up to that number or in my example up to the number of letters there are in the word. This should give you the amount of combinations in a word with no doubles. Factorial can be represented as !.
For example:
4!= 1x2x3x4
C=n!
Where C= Total number of combinations.
n= number of letters in word. != Factorial
I predict that a 20 letter word with no doubles will be:
20!=2.432 x 10^18(in standard form).
part 2: Words with 1 double
2 letter word
I have found there is only one combination for a 2 letter word.
The only calculation that I think so far is the total number of arrangements is
1x2=2. 2÷2=1
3 letter word
MMS I have found there are 3 combination for a 3 letter word.
MSM 1x2x3= 6. 6÷2=3
SMM
4 letter word
I have noticed a patten which seems to be that if you divide the factorial of the numbers of letter by 2 you will get the total number of combinations.
This is because if you want to assign a letter as the 1st there’s only 4 different combinations that can be. Then once you have the 1st letter assigned theres only 3 different combinations of letters than can be place as the 2nd letter. Then once you assigned a letter as the 1st 2nd letters theres only a double letter that can be place as the 3rd and 4th letters.
1x3x4
EMMA
EMAM
EAMM
AMME
AMEM
AEMM
MMEA
MMAE
MEMA
MAME
MEAM
MAEM
Formula for 1 double:
The formula for finding the total number of combinations in a word with 1 double is in a 4 letter word for example: 1x3x4=12. or (1x2x3x4) ÷2.
C= n! ÷2
For a 20 letter word with 1 double I predict that the total number of combinations will be: 20!÷2= 1.216x10^18(in standard form).
Part 3: extension 1 triple
3 letter word
MMM There’s only one combination for a 3 letter word
4 letter word
I predict that the amount of combinations for a 3 letter word with 1 triple will be 4 because with a word with 1 double you divide factorial by 2 so with a triple I’m guessing you divide the amount of combinations you have with 1 double by 3.
You divide the factorial of 4 by 2 and 3 because there’s letters which are the same. Dividing the total number of letters by 2 and 3 gives the factorial of the total numbers of letters divided by the factorial of the letters repeated.
1x2x3x4÷2÷3=4 combinations.
Triple equation:
For example a 4 letter word:
1x2x3x4=24. 24÷2=12. 12÷3=4
C= n! ÷2÷3
For a 20 letter word with 1 triple I predict that the total number of combinations will be: 20!÷2÷3= 4.05x10^17(in standard form)
Part 4: Any word with one set of repeated letters
I think I’ve figured out an equation for all words with one set of doubles ,or triples or with any letter repeated.
With a 4 letter word with a triple to get the answer you do 1x2x3x4÷2÷3 and with only a double 1x2x3x4÷2. I think that the equation will be
C=n! ÷m!. n= number of letters in word, m= how many times the letter is repeated. E.g a double would be 2.
For example a 6 letter word with a triple would be:
6! ÷3!= 120 combinations.
To explain this formula im going to split it into two halves
1st halve being n! and 2nd halve being ÷m!.
The n! will be the total number of letters factorialised. The m! will be the number of letters repeated. Then you divide n! by m! and you will get the total number of combinations.
Part 5: words with more than 1 set of doubles:
C=n! ÷m! n= number of letters in word, m= how many times the letter is repeated. E.g a double would be 2.
4 letter word:
6 combinations
MMNN
MNMN
MNNM
NNMM
NMNM
NMMN
6 letter word:
MMNNAA
MMAANN
MMANAN
MMNANA
12 combinations
AANNMM
AAMMNN
AAMNMN
AANMNM
NNMMAA
NNAAMM
NNAMAM
NNMAMA
Final equation:
n! ÷m!( the equation to find the total number of combinations of a word with one set of repeated letters) can be evolved into a more general equation.
I have noticed that for the following equation above
n! is equal to the number of different letters add the amount of each letter, and then factorialed.
Then the m! part is equal to the amount of different letters times the amount of each different letter and then factorialed.
n! x m!
(n+m)! ÷n! x m!
for example a 4 letter word with 1 double:
AABC
Lets split the equation into 2 halves. The first halve being (n+m)!.
In this part n would = AA=2.
And the m part would in fact be =BC=2.
So you would do:
2+2=4
and then find the factorial of 4
4!=24.
Then to the 2nd part which is the answer that you got from the 1st part ÷ n! x m!.
In this case n would be 2 and m would be made into 2 m’s which will all be 1. for example the equation would be
(n+m)! ÷ n! x m! x m!.
2! x 1! x 1! =2
Then we combine the two halves together.
24÷2= 12 combinations.
AABC
AACB
ACAB
ABAC
ABCA
ACBA
CBAA
CABA
CAAB
BAAC
BCAA
BACA
