Emma's dilemma.

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Emma’s dilemma

Part 1

I am investigating the different combinations that are possible to create by rearranging the letters of the word Lucy. I will mix up the letters to find as many different possibilities of arrangements as possible, for example LUCY rearranged could be ULCY or LYUC. I will try a systematic method of rearrangement that should show that I haven’t missed out any potential arrangements. Later I will go on and investigate bigger words that include repeated letters like successfully.

Methodology

I am abut to write out all the possible rearrangements of the name Lucy starting with all the combinations starting with the letter L then U, C and Y

Hopefully I will notice a pattern that will be able to show how many amalgamations can be made with a word and then apply it to others to predict the rearrangements of any word no matter how big without the laborious process of writing all of the combinations out.

LUCY                          ULCY                           CLUY                         YLCU

Once I had written down the first letter (orange) the other three letters (blue) could only be used twice each as the second letter of the combination, since the remaining two letters could only be rearranged twice to give a new combination .There are twenty four different arrangements of LUCY and four letters in the name, there are six possible combinations that can be made starting with each letter.

E.g.

It is impossible to rearrange these remaining two black letters more than twice to create a new combination.

My conclusion is that two letters can be rearranged twice, three-six times and four-24 times:

4 letters                                3 letters                     2 letters

Now I will investigate the same problem but using the name EMMA

Part 2

Similar to what I did with Lucy, I will use the same systematic method and write out all of the different combinations that are possible to create by rearranging the letters of the word using colour coded letters

I have encountered a problem with this table as there are only twelve viable combinations of EMMA despite it having the same amount of letters as Lucy, there are only half the number of combinations .I have crossed out and highlighted in blue the combinations that are not acceptable. They are not acceptable because they look exactly the same as the other twelve combinations and the only way to distinguish them is to colour code them. This is because one of the letters is repeated and it means that despite having twenty four combinations using different letters, twelve of them look exactly the same as the other twelve. E.G.

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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 ...

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