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  • Level: GCSE
  • Subject: Maths
  • Word count: 2493

Emma's dilemma.

Extracts from this document...

Introduction

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.

L

U

C

Y

L

U

C

Y

U

L

C

Y

C

L

U

Y

Y

L

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U

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U

LUCYULCYCLUY   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.

L

U

C

Y

L

U

Y

C

...read more.

Middle

E

A

M

M

E

A

M

A

M

M

E

Eimage03.png

A

M

M

M

A

E

M

M

A

E

M

Aimage09.png

M

M

E

E

A

M

M

M

A

M

E

M

A

M

E

A

M

E

M

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.

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

C

Y

Y

C

2 combinations

Three letters                          four letters

L

U

C

Y

L

U

Y

C

L

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Y

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C

image19.png

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   Aimage01.pngimage01.pngimage01.pngimage01.pngimage01.pngimage01.pngimage01.pngimage01.pngimage01.pngimage01.pngimage01.pngimage01.pngimage01.pngimage01.pngimage01.pngimage01.pngimage01.pngimage01.pngimage01.pngimage01.png

image04.png

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.

Number of combinations

Number of letters in a word

1        ×image05.png

1

image06.png

2        ×image05.pngimage07.png

2

6        ×image08.pngimage05.png

3

image10.png

24

4

...read more.

Conclusion

E.G.image33.pngimage33.png

With AB adding the letter C:

It can be placed any where to make a new combination:    A   B

3×2×1=6image33.pngimage33.pngimage33.png

                                                                                           B   Aimage33.png

But adding a letter that is already there like B:  A   Bimage33.png

                                                                           B   A

3×2×1=6                                                 image34.png

  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

...read more.

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