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GCSE Mathematics Coursework - Emma's Dilemma

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Introduction

 GCSE Mathematics Coursework - Emma's Dilemma

Method:

First, I tested different arrangements of the name 'EMMA', by systematically rearranging the letters in the name such as:

1-EMMA        4-AEMM        7-MMEA        10-MEAM

2-EAMM        5-AMME        8-MEMA        11-MAEM

3-EMAM        6-AMEM        9-MMAE        12-MAME

        I found that there were twelve arrangements for the name 'EMMA'. I then investigated different arrangements of letters in the name 'LUCY', using the same method:

        1-LUCY        7-UCYL        13-CYLU        19-YCLU

        2-LUYC        8-UCLY        14-CYUL        20-YCUL

3-LCUY        9-UYLC        15-CUYL        21-YUCL

4-LCYU        10-UYCL        16-CULY        22-YULC

5-LYCU        11-ULCY        17-CLYU        23-YLCU

6-LYUC        12-ULYC        18-CLUY        24-YLUC

        I found that there were many more arrangements in the name 'LUCY' than in the name 'EMMA'. This is because the name 'EMMA' contains repeated letters and the name 'LUCY' does not. I then investigated how many different arrangements of letters there were in other names of different lengths. The lengths of names that I used were 2, 3, and 4 letter names. I recorded my results in a table.

...read more.

Middle

10

3628800

         I then investigated more names, which had repeated letters, and found that there were fewer arrangements of letters. There also appeared to be a pattern involving factorials in this case, although it was not the same pattern as was found in names with no repeated letters. I recorded my results in a table.

Table showing the number of different arrangements of letters possible in words of different lengths with no repeated letters

Number of letters

Number of different arrangements

2

1

3

3

4

12

        I found that the numbers in these results were in the same pattern as the results for the names with no repeating letters, but the numbers of arrangements of the names with no repeating letters was halved for example:

Number of letters

Arrangements (letter repeated once)

Arrangements (no repeating letters)

1

-

1

2

1

2

3

3

6

4

12

24

        So, the formula for the number of arrangements for words containing two repeated letters (for example, sArAh) is:

(Number of letters in word)!

2!

        I then investigated the number of different arrangements of letters in a word with tree repeated letters, and, once again found a similar pattern. By testing the name ‘PIPPA’, I found this set of results:

Number of letters

Arrangements (letter repeated once)

1

-

2

-

3

1

4

4

5

20

...read more.

Conclusion

Number of letters

Number of sets of repeated letters

Number of repeated letters

Number of possible arrangements

4

2

4

6

5

2

4

30

I continued to investigate different formulae which could be used to find the number of arrangements in a word with two sets of repeated letters.

        A four-letter word such as XXYY has 2 X's and 2 Y's. Therefore 1x2x3x4 (as there are 4 letters) or (1x2)x(1x2) will give the number of different arrangements when:

Number of letters in word (n) factorial

(number of X's)!x(number of Y's)!

or

n!

(Number of x!)x(number of y!)

         I then used this formula to calculate the number of arrangements possible in words of length 6,7 and 8 letters containing two sets of repeated letters. I also tested words which contained different numbers of X's and Y's, for example: XXXYY or XXYYY. These results are also shown in the table below. I also calculated some values for other words with greater numbers of letters.

Number of letters

Number of sets of repeated letters

Number of repeated letters

Number of possible arrangements

X

Y

4

2

2

2

6

5

2

2

2

30

6

2

2

2

180

7

2

2

2

1260

8

2

2

2

10080

5

2

3

2

6

2

3

2

7

2

3

2

7

2

4

3

...read more.

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