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  • Level: GCSE
  • Subject: Maths
  • Word count: 1160

Emma’s Dilemma.

Extracts from this document...

Introduction

Abdul K Khan        Mathematics Coursework        20/04/2007

Emma’s Dilemma

Introduction

In this mathematics coursework, I am investigate and find out the different combinations in a word after I have done this will be trying to find out a formula that will allow me to find out the different combinations quicker without showing all the different outcomes, and what I am also going to do is find a formula which will enable combinations easier for a word with repeated letters.

Different arrangements in the word LUCY (four letter word)

LUCY

ULCY

CLUY

YLUC

LUYC

ULYC

CLYU

YLCU

LCUY

UCLY

CULY

YULC

LCYU

UCYL

CUYL

YUCL

LYUC

UYLC

CYLU

YCLU

LYCU

UYCL

CYUL

YCUL

Number of combinations= 24

What I did to find out all the outcomes in the word LUCY was keeping the first two letters in the same place then just swapping the last two letters around until I was enable to find any more

Letter(s)

Number of letters

Total combinations

L

1         image00.png

1

LU

2image05.png

2

LUC

3image05.pngimage03.png

6

LUCY

4image04.png

24

1×2×3×4=24 combinations for the word LUCY          

Different arrangements in the word EMMA (four letter word with same letters)

EMMA

MEMA

AMME

EAMM

MAME

AEMM

EMAM

MMEA

AMEM

MMAE

MEAM

MAEM

...read more.

Middle

0

1

SS

2

2image03.png

2

1

1 × 2 ÷ 1 ÷ 2 = 1 combination for the word SS

Formula meanings

n = number of letters in the word

y = number of repeated letters

! = factorial

My formula

My formula for word with different letters is

n! =

n × (n – 1) × (n – 2) . . . 2 × 1

This is because it is quicker to find the total combinations for a certain amount of letters without writing them all up.  

The formula for letters that have been repeated in the word is

n! ÷ y!=

n × (n – 1) × (n – 2) . . . 2 × 1

y × (y – 1) × (y – 2) . . . 2 × 1

This is because without dividing you will get some same combinations when written up.

Total combinations for words with letters up to ten

Number of letters

formula

Total combinations

1

1! =

1

2

2! =

2

3

3! =

6

4

4! =

24

5

5! =

120

6

6! =

720

7

7! =

5040

8

8! =

40320

9

9! =

362880

10

10! =

3628800

Total combinations for words with repeated letters up to ten

Z= total number of combinations with same letters

Number of letters

What you can get when different letters

Number of repeated letters

Total combinations

1

1

n!image06.png

z

2

2image07.png

n!

z

3

6

n!

z

4

24image07.png

n!image06.png

z

5

120image06.png

n!image06.png

z

6

720image07.png

n!image06.png

z

7

5040

n!image06.png

z

8

40320image07.png

n!image06.png

z

9

362880image06.png

n!image06.png

z

10

3628800image06.png

n!image08.png

z

Extension

What I am going to do now is find a formula for a word with different sets of same letters.

Different arrangements for the DEDE (four letter word with sets of same letters)

DEDE

EDED

DEED

EDDE

DDEE

EEDD

Number of combinations= 6

Letter(s)

Number of letters

What you can get when different letters

Number of repeated letters that are same

Number of repeated letters that are same

Total combinations

D

1

1image08.png

0image08.png

0image06.png

1

DE

2

2image09.png

0image08.png

0image08.png

2

DED

3

6image08.png

2image10.png

0image08.png

3

DEDE

4

24image08.png

2image09.png

2image09.png

6

...read more.

Conclusion

image12.png

This is because it will make it easier with things with more letters and you may also write down some same combinations.

This is a table showing more set of letters in a certain amount of letters up to ten

Number of letters

What you can get when different letters

sets of repeated letters

Total combinations

1

1image04.png

n! ÷ n! . . . n! ÷ n!image08.png

z

2

2image13.png

n! ÷ n! . . . n! ÷ n!image06.png

z

3

6image00.png

n! ÷ n! . . . n! ÷ n!image08.png

z

4

24image13.png

n! ÷ n! . . . n! ÷ n!image08.png

z

5

120

n! ÷ n! . . . n! ÷ n!image08.png

z

6

720image04.png

n! ÷ n! . . . n! ÷ n!image09.png

z

7

5040image00.png

n! ÷ n! . . . n! ÷ n!image09.png

z

8

40320

n! ÷ n! . . . n! ÷ n!image08.png

z

9

362880image04.png

n! ÷ n! . . . n! ÷ n!image09.png

z

10

3628800image13.png

n! ÷ n! . . . n! ÷ n!image08.png

z

Conclusion

In this coursework I have discovered that the letters in a word or a lot of letters put together can be re-changed to find different combinations. For with and without same letters I have a formula to make it easier to the total of combinations without the working out. I have also discovered it would work other situation e.g. when there is more sets of same letters.

        -  -

...read more.

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