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Emma’S Dilemma

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Introduction

EMMA’S DILEMMA

We are to find out a formula to predict the number of combinations in a word. This can be affected by the amount of words, the amount of different words and if the word is a palindrome (A word which can be read front-to-back and back-to-front, e.g. ANNA). I will first list the combinations of one name, EMMA and look at the amount of combinations and try to find a formula, which receives the same results.

  1. EMMA – The name Emma has four words so there can not be too many results. There are two letters the same, which should
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Middle

-Lcyu

-Lyuc

-Luyc

-Ulcy

-Ulyc

-Uycl

-Uylc

-Ucly

-Ucyl

-Clyu

-Cluy

-Cyul

-Cylu

-Culy

-Cuyl

-Ylcu

-Yluc

-Yucl

-Yulc

-Ycul

-Yclu                                                                                                    24 Combinations

The name LUCY proves that a word with two letters the same has fewer combinations because it has 24 combinations, where as EMMA had only 12 combinations. I have decided to use a “Shriek” (also known as a ‘Factorial’), which is resembled as a ‘!’. This causes a number (e.g. 5) to be multiplied my its predecessors (e.g. 5x4x3x2x1). This number is the amount of letters in the word (N). This gives the same answer to the number of combinations as the name (e.g. a four letter word, LUCY, has four letters, so N! = 4

...read more.

Conclusion

So the new formula is (N!/D!)/M.

This solves all my other problems, but to make sure, I will try it out with other words.

  1. OTEETO (this is not a name, but just an example, the word being a proper  

                         name does not change the outcome)

(N!/D!)/M

= (6!/3!)/2 = 240 Combinations, this is correct to the manual way, meaning my formula works.

...read more.

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