• 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
  10. 10
    10
  • Level: GCSE
  • Subject: Maths
  • Word count: 2900

Emma's Dilemma

Extracts from this document...

Introduction

Maths Coursework                                                                                

Emma’s Dilemma

I will be investigating the different arrangements of letters of words, which don’t have any identical letters in them and those that do. Then I will try and find a formula that can calculate the total arrangements of letters of any word, which does not have any identical letters in it. I will also try and find a formula to find the total arrangements of words, which have some identical letters in them.

Firstly I will look at those words, which have no identical letters in them and then work out a formula that can find the total arrangements of letters in them. So I will begin by finding the total arrangements of a one-lettered word to finding the total arrangements of a four-lettered word. Thereafter I will try and predict the total arrangements of a five-lettered word and then I will check if I was correct by finding the total arrangements of it’s letters

Part 1

One-lettered word
A               =1
There is only one arrangement for a one-lettered word.
Two-lettered word
WE              
EW              =2

There are two different arrangements for a two-lettered word whose letters are all different.

Three-lettered word

CAR CRA    

ACR ARC       =6

RCA RAC

There are six different arrangement for a three-lettered word whose letters are all different.

Four-lettered word

LUCY LUYC LYUC LYCU LCYU LCUY  

ULCY ULYC UCLY UCYL UYCL UYLC    

CLUY CLYU CULY CUYL CYUL CYLU    

YUCL YULC YCUL YCLU YLCU YLUC      =24

There are 24 different arrangement for a four-lettered word whose letters are all different.

...read more.

Middle

1

1

2

2

3

6

4

24

5

120

6

720

It is apparent that as the number of letters in a word increase the number of different arrangements increase. When there is a two-lettered word the total arrangement is 2, which is worked out by doing 2 x 1 and 1 is also the total number of arrangement for a one-lettered. So I think that to find the total arrangements of a word you have to multiply the total number of arrangements of the word before it by the number of letters in that word. This theory can be confirmed by multiplying 24 (the total number of arrangements of a four-lettered word) by 5 (the number of letters in that word) which equals 120. But to find the total arrangements of a four lettered-word you have to multiply 1, which is the total arrangement of a one-lettered word by 2 and then multiply that by 3, which gives you 6. Then you multiply 6 by 4 to find the total arrangements of a four-lettered word. You can put all this together like this: 1x2x3x4x5 = 120.

So to find the total arrangements of any letter you have to start by multiplying from 1 to the number of letters of that word. So to find the total arrangements of a seven-lettered word you do this: 1x2x3x4x5x6x7, which equals 5040. This is called factorial.

Factorial is known by ! which can be found on a scientific calculator.

So the formula to find  the total arrangement of a word, which has no repeated letters in it is n!.

Part 2

I will now investigate the total arrangements of words, which have some repeated letters in them. Then I will try and find a formula for finding the total arrangement.

First I will investigate the total arrangements of words, which have a letter repeated in it twice.

Two-lettered word with a letter repeated twice          

AA      = 1

There is only one arrangement for a two-lettered word, which has a letter repeated in it twice.

Three-lettered word with a letter repeated twice

RAA ARA AAR      = 3  

There are only three different arrangements for a three-lettered word, which has a letter repeated in it twice.

Four-lettered word with a letter repeated twice

EMMA   EAMM   EMAM

MEAM   MAEM   MAME

MMEA   MMAE   MEMA        

AEMM   AMEM   AMME       = 12

There are 12 different arrangements for a four-lettered word, which a letter repeated in it twice.

Five-lettered word with a letter repeated twice

SLEEP   SLEPE   SLPEE   SPLEE   SPELE   SPEEL

SEEPL   SEPLE   SELPE   SEPEL   SEELP   SELEP

LSEEP   LSEPE   LSPEE   LPEES   LPESE   LPSEE

LEEPS   LEESP   LESPE   LESEP   LEPES   LEPSE

PSLEE   PSELE   PSEEL   PLSEE   PLESE   PLEES

PEELS   PEESL   PESEL   PESLE   PELSE   PELES

ESPEL   ESPLE   ESEPL   ESELP   ESLEP   ESLPE

EPSEL   EPSLE   EPLSE   EPLES   EPELS   EPESL

ELSPE   ELSEP   ELPSE   ELPES   ELEPS   ELESP

EELSP   EELPS   EEPLS   EEPSL   EESPL   EESLP    = 60

There are 60 different arrangements for a five-lettered word, which has a letter repeated

Here is a table showing the total arrangements of words, which have a letter repeated in it twice.

 Number of Letters

Total arrangement

2

1

3

3

4

12

5

60

If you compare this with the total arrangements of words, which have no letters repeated in them you will notice that the total arrangement of a word, which has a letter repeated twice is half of the arrangement of a word, which has the same number of letters but has no letter repeated twice.

So to find the formula I would have to use the formula for a word, which has no letter repeated which is n! and then divide it by 2. So when you divide 24 (which is the total arrangement of LUCY) you get 12 (which is the total arrangement of EMMA).

So the formula to find the total arrangement of a word, which has a letter repeated in it twice is (n!)/2.

Now I will investigate the total arrangements of words, which have a letter repeated three times.

Three-lettered word which has a letter repeated three times

TTT    = 1

There is only one arrangement for a three-lettered word, which has a letter repeated three times.

Four-lettered word which has a letter repeated three times

STTT   TSTT   TTST   TTTS   = 4

There are four different arrangements for a four-lettered word, which has a letter repeated in it three times.

Five-lettered word which has a letter repeated three times

ASTTT   ATSTT   ATTST   ATTTS

SATTT   STATT   STTAT   STTTA

TASTT   TATST   TATTS   TSATT

TSTAT   TSTTA   TTSAT   TTSTA

TTAST   TTATS   TTTAS   TTTSA         = 20

There are 2o different arrangements for a five-lettered word, which has a letter repeated three times.

Here is a table showing the total arrangements of words, which have a letter repeated three times.

Number of Letters

Total arrangement

3

1

4

4

5

20

...read more.

Conclusion

Four-lettered word which has two types of letters each repeated twice

RRAA   RARA   RAAR   ARRA   ARAR   AARR      = 6

There are 6 different arrangements of a four-lettered word, which has two different types of letters each repeated twice.

The formula for a four-lettered word, which has only two types of letters is n!/4.

You get 4 when you multiply 2! by 2! So I think that the formula to find the total arrangement of a word, which has two types of letters repeated two times or more is (n!)/(r!)(x!) (x stands for the other letter which is also repeated)

So to find the total arrangement of a four-lettered word, which has two types of letters repeated twice, you would do this:      4!

                                                          (2!)(2!)  

                                                =          24

                                                             4

                                               =      6

But if there are more than two types of letters repeated all you do is add more factorials according the number of types of different letters that are repeated.

So to find the total arrangement of RRHHAA you will do this:            6!  

                                                                                                        (2!)(2!)(2!)

You have three factorials because there are three different types of letters repeated twice

                                                                                                 =             720

                                                                                                                  8

                                                                                                =  90

So the total arrangement of RRHHAA is 90.

So the formula to find the total arrangement of a word, which has more than one type of letter repeated is (n!)/(Factorials) (the number of factorials depend on the number of different types of letters repeated)

Here are all the formulas that I have found from this investigation.

n!                         When there are no letters repeated.

(n!)/(r!)            When there is a letter repeated 2 or more times.

(n)/(Factorials) When there are more than one type of letter repeated.

From this investigation I have found out many formulas, which will help me to find the total arrangements of different types of words. Finding the formulas weren’t too difficult but trying to put them into words was a bit hard.  

 

...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. Emma's Dilemma

    This means that the total number of arrangements possible is reduced by a factor of 4, or "2 ! X 2!" ( See later ). For example For 6 letters, all different, there are "6 X 5 X 4 X 3 X 2 X 1"arrangements.

  2. GCSE maths coursework: Emma's dilemma

    212211 total arrangement is 20 let's see the formula: a=(1*2*3*4*5*6)/1*2*3 =120 no, it doesn't work but if we divide it by another 6 which is (1*2*3) Can you see the pattern, the formula still work if we times the mutiply again.

  1. I am investigating the number of different arrangements of letters

    Let's confirm this formula, if a nuber has 1fig it has 1 arrangements formular: 1*1=1 It works 2fig it has 2 arrangements formular: 1*2=2 It works 3fig it has 6 arrangements formular: 1*2*3=6 It works 4fig it has 24 arrangements formular: 1*2*3*4=24 It works Formula is confirmed What about if

  2. Emma's Dilemma

    I have now come to find the correct answer which is shown below: 5! / 3! = 20 NUMBER OF POSSIBILITIES After I got this far I thought to myself what if there are repeats of different letters, for example 'AABB.'

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

    with two of the chosen letters being the same, can be achieved by using the following formula: To find the total number of combinations, take the number of letters ), and times that number by the number of letters minus 1, timed by the number of letters minus 2, and

  2. EMMA'S DILEMMA

    =6 The formula works. To use the formula for a 6 lettered word, which 1 letter is repeated twice and the other letter is also repeated twice and the final letter is also repeated twice. To find the arrangement, if I triple the repeated letters arrangement.

  1. We are investigating the number of different arrangements of letters.

    The formula is a=ni/xixi but we need to change the formula, because there are 2 pairs of same numbers with different number of figures. so we change the formula to a=ni/x1i*x2i Let's try 5 figures with 3 same number, and 2 same number.

  2. Emma's Dilemma

    2 60 Otojo 5 3 20 XX......XXYY.......Y 20 10 X's and 10 Y's 184756 I observed that to find the maximum number of arrangements of a word with 2 or more identical letters you have to follow the same sequence as to find the arrangements as if there were no identical letters.

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