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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

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

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.

This student written piece of work is one of many that can be found in our GCSE Emma's Dilemma section.

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