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Level: GCSE
Subject: Maths
Word count: 2380

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Emma's Dilemma

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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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Related GCSE Emma's Dilemma essays

Emma's Dilemma

M A EMMA MEMA MEMA AEMM EMAM MEAM MEAM AEMM EAMM MMYE MMEA AMME EAMM MMEA MMAE AMEM EMMA MAME MAEM AMEM EMAM MAEM MAME AMME Now that we have crossed out the duplicates we can create a new table: E M A EMAM MEMA AEMM EMMA MEAM AMME EAMM

Emma's Dilemma.

DANNY 5 2 60 12*5=60 I predict the name Danny will have 60 different arrangements DANNY DANNY YDANN NYDAN NNYDA ANNYD DANYN YDNAN NYDNA NNYAD ANNDY DAYNN YDNNA NYAND NNADY ANDYN DYNNA YANND NYADN NNAYD ANDNY DYNAN YANDN NYNDA NNDYA ANYDN DYANN YADNN NYNAD NNDAY ANYND DNANY YNAND NANYD AYNND

To investigate the various arrangements that could be made with different names or words. ...

Letters Arrangements 4 24 5 120 6 720 7 5040 I have succeeded in finding an algebraic rule for this investigation. It is quicker and more efficient to find the number of arrangements. Investigation Three: I am now going to be investigating words with two letters in it, the same.

Emma's dilemma.

So the new formula for the whole table would be L! / S!. Here is a list of formula for the different number of letters the same: 2 letters the same = . L! . (2x1 = 2) 3 letters the same = L!

Emma&amp;#146;s Dilemma.

Both names contain four letters, however, the only difference is that Emma's name has got the letter 'M' twice and Lucy's name has all four letters different. So if the name contains a repeating letter then it will have fewer combinations than a name with all different letters, but with the same number of letters.

Emma&amp;#146;s Dilemma

When that letter has been placed there are now only 4 places for the next letter to go into. G _ _ _ _ And then 3 for the next and so on until the word is full. But when you have two letters the same and there are only two spaces then there is only 1 arrangement possible.

Emma&amp;#146;s Dilemma

/ r! = 5! / 3! = 120 / 6 = 20 this is also the right amount. This also works for when there are two repetitions: 4th = x! / r! = 4! / 2! = 24 / 2 = 12 and for when there is only one repetition of each letter (i.e.

Emma's Dilemma

MEMA MMEA AEMM AMEM AMME The letters in italics have already been written out but using the other m, so therefore there is half the amount of possible combinations.

Emma's Dilemma

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