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Emma's Dilemma

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Introduction

Emma’s Dilemma

        For our maths coursework we had to investigate the number of different word arrangements possible for names. The names (or letter arrangements) I started off with had letters that were only used once n each name and were not repeated, like “Lucy”. As I continued experimenting with the different arrangements, I began to use words with letters repeating, such as “Emma”. My goal was to find formulas that worked for the different arrangements.

        I started off by working out the number of different arrangements in the name “Emma”. I shall now display the different arrangements of “Emma” that I found:

EMMA

EMAM

EAMM

MMEA

MMAE

MAME

MAEM

MEAM

MEMA

AMME

AEMM

AMEM

        Here are the different letter arrangements for “Lucy”:

LUCY          ULCY          CLUY          YLUC

LUYC          ULYC          CLYU          YLCU

LCUY          UCLY          CULY          YULC

LCYU          UCYL          CUYL          YUCL

LYUC          UYLC          CYLU          YCLU

LYCU          UYCL          CYUL          YCUL

...read more.

Middle

NMTYI          NITYM          NTIYM          NYITM

NMYIT          NIYMT          NTYMN         NYTMN

NMYTI          NIYTM          NTYNM         NYTNM

TMINY          TIMNY          TNMIY          TYMIN

TMIYN          TIMYN          TNMYI          TYMNI

TMNIY          TINMY          TNIMY          TYIMN

TMNYI          TINYM          TNIYM          TYINM

TMYIN          TIYMN          TNYMI          TYNMI

TMYNI          TIYNM          TNYIM          TYNIM

YMINT          YIMNT          YNMIT          YTMIN

YMITN          YIMTN          YNMTI          YTMNI

YMNIT          YINMT          YNIMT          YTIMN

YMNTI          YINTM          YNITM          YTINM

YMTIN          YITMN          YNTMI          YTNMI

YMTNI          YITNM          YNTIM          YTNIM

I displayed each number of letters in the way of a tree diagram, each time changing the last letter of the word, then working my way back along the word, until I eventually had the letters in reverse order. I then recorded the results in this table:

NUMBER OF LETTERS

DIFFERENT ARRANGEMENTS

1

1

2

2

3

6

4

24

5

120

6

720

7

5040

8

40320

9

362880

10

3628800

        Here is how I found the rule of working out the number of arrangements for a word with no repeated letters. If you multiply the number

...read more.

Conclusion

        I used the same method that I used before and discovered that words with one set of repeated letters were exactly the same as the previous results, except that the number of arrangements was halved when the letters were repeated. Therefore the algebraic formula would be n!/2. I then had to find a rule that could be applied to words with no repeated letters, two repeated letters and then three repeated letters. According to my previous formulas, an arrangement of words with three repeated letters should have an algebraic formula of n!/3. I decided to display my results in the following table:

NUMBER OF LETTERS

NO REPEATS

TWO REPEATED LETTERS

THREE REPEATED LETTERS

1

1

-

-

2

2

1

-

3

6

3

1

4

24

12

4

5

120

60

20

6

720

360

120

7

5040

2520

840

8

40320

20160

6720

9

362880

181440

60480

10

3268800

1814400

604800

        Whilst looking for different rules I found that you can also get the same answer by using the algebraic expression n!/r!, which means that the number of letters factorized divided by the number of repeated letters.

...read more.

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