Hannah
If a word contains six letters – all of which are doubles e.g. Hannah, it would be able to be rearranged 90 times (according to my theory, which would be (1x2x3x4x5x6)
(1x2)(1x2)(1x2) = 90
Why do we halve?
Proof
If you treat the double letters as individual letters then you don’t needto halve in the formula. You halve because of the double letter, the example above proves it.
Factorial – A shorthand algebraic method
A much faster way of factorising (1x2x3x4x5…) on a scientific calculator would be the x! button.
For example: 4! Would equal 24. (1x2x3x4)
2! Would equal 2. (1x2)
6! Would equal 720. (1x2x3x4x5x6x7)
What would happen if you had a repeat? Well all you then need to do is x!
2
What if you have more than one repeat? Well then you would have to do (in the example of Hannah) _
_6!__ Number of Letters
(2!)(2!)(2!) = 90
Answer
Repeated Letters
This proves that the formula for Hannah is right and it equals 90.
Letters with a triple.
I predict that the formula for a five letter word with a triple in it will be 5!
3! = 20
Proof
The method is correct.
If a word has a double and a triple in it like Anana the formula might be 5!
(3!)(2!) =10
Or 5
(3!)+(2!) =132
Investigation
Formula number one was correct.
(4) Investigate the arrangements of X and Y.
XX – 1 arrangement Formula= 1! =1
XY
YX – 2 arrangements Formula = 2! =2
XYY
YXY
YYX – 3 arrangements Formula = 3!
2! =3
XYYY
YXYY
YYXY
YYYX – 4 arrangements Formula = 4!
3!
XYYYY
YXYYY
YYXYY
YYYXY
YYYYX – 5 arrangements Formula = 5!
4! = 5
XYYYYY
YXYYYY
YYXYYY
YYYXYY
YYYYXY
YYYYYX – 6 arrangements Formula = 6!
5! = 6
Three X’s and Two Y’s
In theory this should equal 10 because 5!
(3!)(2!) = 10
Proof
Four X’s and Four Y’s
A double quad should add up to 70 because 8!
(4!)(4!)
Proof
General Equation
To find a general equation I will make a table of results:
As you can see by the table there is an undisputed relationship between the number of repeted letters, the number of letters and the formula.
General Formula
Using the table of results I can make a formula which will fit every word :
(X+Y)!
(X!)+(Y!)
Conclusion
To conclude, any word no matter how long or haw many repeats it has can be solved using the above formula for example Computation has eleven letters in it and the letter o is repeted twice, t is repeted twice, c is repeted twice and n is repeted twice so the formula for it would be:
11!
(2!)(2!)(2!)(2!)
That would equal 2494800 combinations.
