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  • Level: GCSE
  • Subject: Maths
  • Word count: 2380

Emma's Dilemma

Extracts from this document...

Introduction

Emma’s Dilemma

Emma and Lucy are playing with arrangements of the letters of their names.

Part 1

Investigate the number of different arrangements of the letters of Lucy’s name.

Part 2

Investigate the number of different arrangements of the letters of Emma’s name.

Part 3

Investigate the number of different arrangements of various groups of letters.

Emma’s Dilemma

Emma and Lucy were playing with arrangements of the letters of their names.

Part 1

My first task was to investigate the number of different arrangements for the letters in Lucy’s name. I began by listing the arrangements, one by one, following a certain structure (a structure which will hereafter be used in similar situations throughout the rest of this investigation). This started with me dividing the name “Lucy” up into its individual parts i.e. the letters L, U, C and Y. By re-arranging these, I was then able to find new arrangements. Starting with LUCY (obviously), I moved on to find LUYC (by exchanging the two letters at the end). When I found that there were no longer any arrangements starting with LU, I simply replaced the second letter with the subsequent letter and started to look for arrangements starting with LC. When the arrangements starting with L had run out altogether, I took the next letter i.e. U and started to look for arrangements starting with that. In the end, I found that there were 24 different arrangements for the letters in Lucy’s name:

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

EAMM

MMEA

AMME

MMAE

MAEM

MAME

I discovered 12 different arrangements in which the two M’s in Emma’s name could be exchanged and 24 in which they were individual. This tells me that the prior was half of the latter. Looking at my results now, this is quite understandable. I can see that for every two arrangements in which the M’s were separate, there was one of the opposite. For example, EM1M2Aand EM2M1A could just as easily have been EMMA.

However, to verify the results, I decided to have a look at a similar name “Zeek”. Again, I found 24 different combinations where the M’s were identifiable and 12 where they weren’t so there shouldn’t be a problem:

ZE1E2K

E1ZE2K

E2ZE1K

KZE1E2

ZE1KE2

E1ZKE2

E2ZKE1

KZE2E1

ZE2E1K

E1E2ZK

E2E1ZK

KE1ZE2

ZE2KE1

E1E2KZ

E2E1KZ

KE1E2Z

ZKE1E2

E1KZE2

E2KZE1

KE2ZE1

ZKE2E1

E1KE2Z

E2KE1Z

KE2E1Z

ZEEK

EZEK

KZEE

ZEKE

EZKE

KEZE

ZKEE

EEZK

KEEZ

EEKZ

EKZE

EKEZ

I then decided to see if there was a trend between the number of repetitions there were in a word, and the number of different arrangements, in which those repetitions were interchangeable, that the letters in that word could form. In order to do this, I kept the length of the words constant (i.e. at four characters per word) so as it would not interfere with the results I got.

I decided to look at the word “baaa”. I found that, as with the names “Emma” and “Zeek” before it, the letters in it could appear in 24 different arrangements where the repeated letters were separable. However, there were only 4 different arrangements in which they were not:

BA1A2A3

A1BA2A3

A2BA1A3

A3BA1A2

BA1A3A2

A1BA3A2

A2BA3A1

A3BA2A1

BA2A1A3

A1A2BA3

A2A1BA3

A3A1BA2

BA2A3A1

A1A2A3B

A2A1A3B

A3A1A2B

BA3A1A2

A1A3BA2

A2A3BA1

A3A2BA1

BA3A2A1

A1A3A2B

A2A3A1B

A3A2A1B

BAAA

ABAA

AABA

AAAB

Again, to credit my findings, I investigated a similar word “mooo”. And, again, I found that the letters it is spelt of could fit in 24 different arrangements where the repetitions of the letter O could be identified and 4 where they couldn’t:

MO1O2O3

O1MO2O3

O2MO1O3

O3MO1O2

MO1O3O2

O1MO3O2

O2MO3O1

O3MO2O1

MO2O1O3

O1O2MO3

O2O1MO3

O3O1MO2

MO2O3O1

O1O2O3M

O2O1O3M

O3O1O2M

MO3O1O2

O1O3MO2

O2O3MO1

O3O2MO1

MO3O2O1

O1O3O2M

O2O3O1M

O3O2O1M

MOOO

OMOO

OOMO

OOOM

...read more.

Conclusion

D1E1E2D2

E1D1E2D2

E2D1E1D2

D2D1E1E2

D1E1D2E2

E1D1D2E2

E2D1D2E1

D2D1E2E1

D1E2E1D2

E1E2D1D2

E2E1D1D2

D2E1D1E2

D1E2D2E1

E1E2D2D1

E2E1D2D1

D2E1E2D1

D1D2E1E2

E1D2D1E2

E2D2D1E1

D2E2D1E1

D1D2E2E1

E1D2E2D1

E2D2E1D1

D2E2E1D1

DEED

EDED

DEDE

EDDE

DEED

EEDD

The formula “the number of different arrangements for the letters in a name = n!/a!” obviously didn’t apply in this case as there were two completely unrelated repetitions. Therefore:

The number of different arrangements for the letters in a name = n!/a!×b!

N.B. b referring to the number of repetitions of the second variety.

The b! works on the same principle as the a! before it; that, just as if they were words themselves, the duplicate letters can be arranged in a number of different arrangements, taking into account that they are distinguishable, according to the number of them there were. Hence, the number of different arrangements for the letters in a name is equal to n! divided by a! b! or a! multiplied by b!. If there were more repeated letters, this could be expanded. c would be used to represent the number of times the third letter is repeated and d for the next and so on.

...read more.

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