• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  • Level: GCSE
  • Subject: Maths
  • Word count: 2265

Emma's Dilemma

Extracts from this document...

Introduction

GCSE Coursework –

Emma’s Dilemma!

I am going to investigate the number of different arrangements of the letters of Emma’s name. I will investigate this by using a system. My system is to find all the arrangements beginning with ‘E’, then all the arrangements beginning with ‘M’, and finally all the arrangements beginning with ‘A’.

EMMA

MMAE

AEMM

EMAM

MMEA

AMME

EAMM

MAME

AMEM

MEMA

MEAM

MAEM

I have found 12 different arrangements for the name Emma. I have also found a new systematic way, to find the arrangements. This system is to find everything beginning with the same letter, by fixing two, moving two and so on.

I will use this to investigate the number of different arrangements for the name Lucy.

LUCY

UCYL

CYLU

YLUC

LUYC

UCLY

CYUL

YLCU

LCYU

UYLC

CLUY

YUCL

LCUY

UYCL

CLYU

YULC

LYUC

ULCY

CUYL

YCLU

LYCU

ULYC

CULY

YCUL

I have found 24 different arrangements for the name Lucy.

So far I have used names with 4 letters but ‘Emma’ has a double letter and ‘Lucy’ does not. So this means a double letter equals half the number of arrangements of a name without single letters. You can use anything to work this out – symbols, numbers etc. and the number of combinations will still be the same.

But if you were to make the double letters different I predict that there will be the same amount of arrangements as Lucy’s name – 24. I will now try to prove this:

EMMA

MMAE

MMAE

AEMM

EMMA

MMEA

MMEA

AEMM

EMAM

MAEM

MAEM

AMME

EMAM

MAME

MAME

AMEM

EAMM

MEMA

MEMA

AMEM

EAMM

MEAM

MEAM

AMME

Each of these arrangements are in pairs.

...read more.

Middle

RSOIE

OISER

SEIRO

IREOS

EORSI

REOIS

ORSEI

SOIRE

ISEOR

EIRSO

REOSI

ORSIE

SOIER

ISERO

EIROS

REISO

OREIS

SOREI

ISORE

EISOR

REIOS

ORESI

SORIE

ISOER

EISRO

RESOI

ORISE

SOEIR

ISROE

EIORS

RESIO

ORIES

SOERI

ISREO

EIOSR

RISEO

OEIRS

SREOI

IORES

ESOIR

RISOE

OEISR

SREIO

IORSE

ESORI

RIEOS

OERSI

SROIE

IOESR

ESIRO

RIESO

OERIS

SROEI

IOERS

ESIOR

RIOSE

OESIR

SRIEO

IOSRE

ESROI

RIOES

OESRI

SRIOE

IOSER

ESRIO

I have found 120 different arrangements for the name Rosie.

I have also proved my prediction to be right.

You can also predict this in a different way. We know how many combinations are in a four-letter word, so you can also multiply the number of letters by the arrangements for a four-letter word. For example GEMMA, we know EMMA has 12 arrangements and if we multiply it by 5 we will get: 12*5=60, which is the correct number of arrangements.

I will now investigate the name Tasha.

TASHA

ASHAT

SHATA

HATAS

TASAH

ASHTA

SHAAT

HATSA

TAHAS

ASATH

SHTAA

HAAST

TAHSA

ASAHT

SATAH

HAATS

TAASH

ASTHA

SATHA

HASTA

TAAHS

ASTAH

SAAHT

HASAT

TSAHA

AHTAS

SAATH

HTSAA

TSHAA

AHTSA

SAHAT

HTASA

TSAAH

AHAST

SAHTA

HTAAS

THAAS

AHATS

STHAA

HSAAT

THASA

AHSTA

STAAH

HSATA

THSAA

AHSAT

STAHA

HSTAA

ATSAH

ATSHA

ATAHS

ATASH

ATHSA

ATHAS

AAHTS

AAHST

AATSH

AATHS

AASHT

AASTH

I have found 60 arrangements for the name Tasha.

If I made the make the ‘A’s different there would be the same amount of arrangements as Rosie.

Table of Results.

Number of arrangements with:

Number of letters

Double letters

Single letters

1

image00.png

1

2

1

2

3

3

6

4

12

24

5

60

120

I have now seen what the formula is. The formula that can be used is the factorial formula. To find out the number of single letters you use x! (X = the number of letters in the name), to find out the number of double letters you use (x!) / 2  (x = the number of letters in the name). This is because to get the number of arrangements for a three-letter name you have to multiply the one-letter arrangement by two, then you have to multiply it by three. These numbers are known as factorial numbers, the pattern goes like this:image01.png

1! = 1

2! = 1 x 2 =2

3! = 1 x 2 x 3 = 6

4! = 1 x 2 x 3 x 4 = 24 … etc. image02.png

We can also work out the nth term using this:

N! = 1*2*3*4*5……….(n–1)*n.

This works because (4-1)*4 = 12 and this is equal to 4! /2 is.

Using this formula I can find the number of arrangements for a six-letter name with a double letter and a six-letter name with single letters. To find out the double letter arrangement I would use (6!) / 2 which is 360, and the single letter arrangement is 6! which is 720.

I predict that if you have a triple letter in a name the number of arrangements will be a third of the double letter arrangement. Or to find the number of arrangements you could divide the single letter arrangements by 6.

This is what I think the results may be if a name had a triple letter:

Number of letters

No Repeats

Double letter

Triple letter

3

6

3

1

4

24

12

4

5

120

60

20

To seeif my predictions are correct, I will investigate names with a triple letter.

The first name I will use is a three-letter name –XXX

XXX

There is only one arrangement for XXX.

I will now investigate a four-letter word, with a triple letter; this name will be Lill.

LILL

ILLL

LLLI

LLIL

...read more.

Conclusion

Double letter, double letter = (4! / 2!) / 2! = 6

Double letter, triple letter = (5! / 3!) / 2! = 10

Triple letter, triple letter = (6! / 3!) / 3! = 20

These are the same as my results so the formula is correct.

I will now prove that N! works

                           X!Y!

XXAA has 6 arrangements.

4!    =        24    =        24   = 6

2!2!      2 x 2          4

XXXAA has 10 arrangements.

5!    =        120    =        120   = 10

3!2!      6 x 2          12

XXXAAA has 20 arrrangements.

6!    =        720    =        720   = 20

3!3!      6 x 6        36

In conclusion to my investigation, I have found out that a name with a double letter has half the arrangements of a name (with the same number of letters) with no repeated letters.  A name with 3-letters the same has a 6th of the arrangements of a name with no repeats and a 3rd of a name with a double letter. I have also found out that a name with four letters the same has a 24th of the arrangements of a name with no repeats and a 4th of a name with a triple letter.

I have seen how the factorial numbers make it easier to work out the amount of arrangements a word has. I have also found which formula works to find out the number of arrangements names with double letter-double letter, double letter-triple letter and triple letter-triple letter. This formula is

N!

X!Y!

I have also found out that a five-letter word with all the letters different has the same amount of arrangements as a four-letter word multiplied by five. The formula would be 4! * 5, the overall formula is (N-1)*N

...read more.

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

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Emma's Dilemma essays

  1. Arrangements for names.

    As each letter has its own number of arrangements i.e. there were 5 beginning with x, and 5 beginning with y, I think that factorial has to be used again. Also in a 5-letter word there are 120 arrangements and 24 arrangements (120 divided by 5)

  2. Emma's Dilemma

    Answer, One letter repeated Twice: Two letters: AA Total: 1 Three letters: AAB Total: ABA 3 BAA Four letters: AABC AACB ABAC ABCA Total: ACAB 12 ACBA BAAC BACA BCAA CAAB CABA CBAA With the four letter word above, I didn't need to retype this combination of letters ( four

  1. Maths GCSE Coursework: Emma's Dilemma

    A/A/B/B AABB ABAB ABBA BABA BAAB BBAA 6 combinations I will now look at the same case, but with a letter following it: 5 letters: A/A/B/B/C AABBC AABCB AACBB ABBAC ABBCA ABCAB ABCBA ABABC ABACB ACBBA ACBAB ACABB (12) (12)

  2. I have been given a problem entitled 'Emma's Dilemma' and I was given the ...

    I then looked at the word and realised that the number 4 in the equation may be produced by the fact that there are two sets of letters, both repeated twice. So it may be that as there are two letters you multiply these together to get the number you divided by in the equation.

  1. To investigate the combination of arrangement of letters in Jeans name and then for ...

    try 5 letters in a word to see if I can find a pattern from there. JAMES: I predict there will be 100 combinations to the name James. J A M E S J A M S E J A S M E J A S E M J A

  2. Emma's Dilemma

    EYYLY EYLYY ELYYY LYYYE LYYEY LYEYY LEYYY There are 20 different arrangements. TABLE OF RESULTS Number of different arrangements (a) Number of letters (n) No letter repeated One letter repeated 2? One letter repeated 3? 2 2 1 N/A 3 6 3 1 4 24 12 4 5 120 60

  1. Emma's Dilemma Question One: Investigate the number of different arrangements of the letters

    different combinations possible, from a selected number of letters ( using every letter only once ), can be achieved by using the following word formula: To find the total number of combinations, take the number of letters ( n ), and times that number by the number of letters minus

  2. I am doing an investigation into words and their number of combinations. I will ...

    2 letters the same and 3 letters the same: Now I will look at words with 2 of the same letter the same and 3 of another letter the same. (eg AABBB) 5 letters 6 letters 7 letters AABBB AABBBC=10 AABBBCD=60 ABABB AABBCB=10 AABBBDC=60 ABBBA AABCBB=10 AABBDBC=60 ABBAB AACBBB=10 AABDBBC=60

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work