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

# Emma&amp;#146;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 1 LU 2 2 LUC 3 6 LUCY 4 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

Middle

0

1

SS

2

2

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! z 2 2 n! z 3 6 n! z 4 24 n! z 5 120 n! z 6 720 n! z 7 5040 n! z 8 40320 n! z 9 362880 n! z 10 3628800 n! 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 1 0 0 1 DE 2 2 0 0 2 DED 3 6 2 0 3 DEDE 4 24 2 2 6

Conclusion

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 1 n! ÷ n! . . . n! ÷ n! z 2 2 n! ÷ n! . . . n! ÷ n! z 3 6 n! ÷ n! . . . n! ÷ n! z 4 24 n! ÷ n! . . . n! ÷ n! z 5 120 n! ÷ n! . . . n! ÷ n! z 6 720 n! ÷ n! . . . n! ÷ n! z 7 5040 n! ÷ n! . . . n! ÷ n! z 8 40320 n! ÷ n! . . . n! ÷ n! z 9 362880 n! ÷ n! . . . n! ÷ n! z 10 3628800 n! ÷ n! . . . n! ÷ n! 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.

-  -

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