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  • Level: GCSE
  • Subject: Maths
  • Word count: 2695

Emma's Dilemma

Extracts from this document...

Introduction

        Danny Soopen –“Emma’s Dilemma” - Page  of

Danny SoopenGCSE Maths Project – “Emma’s Dilemma”

I have been told to investigate the number of different arrangements of the letters of 'EMMA'. I was then told to investigate the number of different arrangements of the letters of 'LUCY'. After these, I must investigate into different arrangements of letters and find a rule to find the number of different arrangements of letters of any name.

Part 1:

        Ways of arranging EMMA’s name:

EMMA

MEMA

MMAE

AMME

EMAM

MEAM

MAEM

AMEM

EAMM

MMEA

MAME

AEMM

For each new beginning letter, there are 3 different possible combinations.

There are a total of 12 possible different combinations.

Part 2:

        Ways of arranging LUCY’s name:

LUCY

ULCY

CULY

YUCL

LUYC

ULYC

CUYL

YULC

LCUY

UCLY

CLUY

YCUL

LCYU

UCYL

CLYU

YCLU

LYUC

UYLC

CYLU

YLUC

LYCU

UYCL

CYUL

YLCU

For each new beginning letter, there are 6 different possible combinations.

There are a total of 24 possible different combinations.

This combination consists of all the letters being different. I will also try and find a formula for this arrangement, and number of letters. Once I have discovered these formulae I am going to investigate, other combinations of letters and different amounts of letters. I will then try to discover a link between the formulae to enable me to find a formula for the general case.

image03.png

There are 24 different possibilities in this arrangement of 4 letters all different. I have noticed that with Lucy beginning each different letter. For example there are 6 arrangements with LUCY beginning with L, and 6 beginning with u and so on 6*4(the amount of letters) gives 24.

...read more.

Middle

image02.png

           1          1          1          1

We can say that  tn = n!

For example:  

                For FRANK’s name,         tn = 5!

                                        tn = 5 x 4 x 3 x 2 x 1

= 120

The formula for the number of combinations for a word where all the letters are different can be obtained, (as with the LUCY example on page 2) by simply writing out all of the permutations. As no letters are repeated, it is a straightforward formula.

There are a total of four letters to be placed into four different positions each time. When the first letter is placed, there are four different places to put it in. When the second letter is being placed, one position is already being taken up by the first letter, and so there are only three places where it can be positioned. When the third letter is being placed, there are only two places in which it can be placed, and the final letter must fill the final space.

In the formula, we start with the number four as there are four letters in the word, and at each stage of the arrangement process, there is one less available place in which to put the letters, so the multiplying number decreases at each stage

Therefore, where there are four letters to be placed, we can summarise it as:

tn = 4 x 3 x 2 x 1

or        tn = 4!

(b).        Investigating the number of different arrangements of letters in names with varying numbers of letters, where one letter is repeated once.

...read more.

Conclusion

image14.pngimage15.pngwhere n = the number of letters in the word,

                   and R = the number of times the letter occurs.

Part 4:

Investigating the number of different arrangements of letters where more than one letter occurs more than once.

For example: words such as XXYY, XXYYXX, and XXXYYY

The above examples all have more than one letter that repeats more than one time.

Investigation:

Name: “XXYY”

Number of letters: 4

Arrangements:

XXYY

YYXX

XYXY

YXYX

XYYX

YXXY

Total Number of Arrangements: 6

Name: “XXXYYY”

Number of letters: 6

Arrangements:

XXXYYY

XYYXYX

YYYXXX

YXXYXY

XXYYYX

XYYXXY

YYXXXY

YXXYYX

XYYYXX

XYXXYY

YXXXYY

YXYYXX

XXYXYY

XYXYXY

YYXYXX

YXYXYX

XXYYXY

XYXYYX

YYXXYX

YXYXXY

Total Number of Arrangements: 20

Name: “XXYYY”

Number of letters: 5

Arrangements:

XXYYY

XYXYY

XYYYX

YYXXY

YYXYX

XYYYX

XYYXY

YYYXX

YXYYX

YXXYY

Total Number of Arrangements: 10

Summary of Part 4:

When there is only one letter repeated, the formula obtained was image16.png

Therefore, for words where more than one letter is repeated, I will try the formula image17.png

Where n = the total number of letters

And   R1 = the number of times the first letter occurs

And   R2 = the number of times the second letter occurs

Examples:

Name: “XXXYYY”

Number of letters: 6

image18.png

This formula does not work. I think that the formula could perhaps be image19.png

Examples:

Name: “XXXYYY”

Number of letters: 6

image20.png

Name: “XXYYY”

Number of letters: 5

image21.png

This equation appears to be correct.

  • We have therefore arrived at the formula image19.png
  • The n! Is there as the total number of letters, as before. We previously decided that where one letter is repeated a certain number of times, the equation would be image16.png
  • It is therefore logical to assume that where more than one letter is repeated, n! Should be divided by the total number of the first repeated letters multiplied by the total number of the second repeated letters.  

...read more.

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