• 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
  11. 11
    11
  12. 12
    12
  • Level: GCSE
  • Subject: Maths
  • Word count: 3131

Emma's Dilemma.

Extracts from this document...

Introduction

Emma’s Dilemma

In this investigation I am going to investigate the number of different arrangements of letters in a word.  

I will firstly see the number of arrangements of the letters of Emma’s name.

ARRANGEMENTS FOR EMMA

1) EMMA 2) EMAM 3) EAMM 4) AEMM 5) AMEM 6) AMME 7) MMAE 8)MMEA 9) MEAM 10) MAEM 11) MAME 12) MEMA

I have found that there are 12 different arrangements. In each word there are total of 4 letters and 2 of them are the same, ’M’ which is repeated.

Now I shall see how many different arrangements there are in the name Lucy

ARRANGEMENTS FOR LUCY

1) LUCY 2) LUYC 3) LYUC 4) LYCU 5) LCUY 6) LCYU 7) ULYC 8) ULCY  9) UYCL 10) UYLC 11) UCLY 12) UCYL 13) CLUY 14) CLYU 15) CYUL 16) CYLU 17) CULY 18) CUYL 19) YUCL 20) YULC 21) YLCU 22) YLUC       23) YCUL 24)YCLU

There are 24 different arrangements possibilities in this arrangement of 4 letters that are all different, which is twice as many arrangements than EMMA.

I shall now investigate on different words, which have different number of letters. I will start by using the word ‘JO’

ARRANGEMENTS FOR JO

  1. JO 2) OJ  

There are two arrangements for this 2-lettered name. I will now investigate the number of arrangements for ‘JIM’ which is a 3 lettered word

ARRNAGEMENTS FOR JIM

  1. JIM 2) JMI 3) IJM 4) IMJ 5) MJI 6) MIJ

There are 6 arrangements for this 3-lettered name, which is triple the number of arrangements of the ‘JO’. I already know the number of arrangements for a four –lettered name ‘LUCY’ that is 24 arrangements. This is quadruple the number of arrangements than ‘JIM’.

...read more.

Middle

1*2*3*4*5

120

6

1*2*3*4*5*6

720

7

1*2*3*4*5*6*7

5040

8

1*2*3*4*5*6*7*8

40320

9

1*2*3*4*5*6*7*8*9

362880

10

1*2*3*4*5*6*7*8*9*10

3628800

WHY FORMULA WORKS

Using ‘Lucy’ as an example n=4

Using the formula n!=a

4!= 4*3*2*1=a

a=24

4 is in the formula because there are 4 letters and so there are 4 sets of arrangements for which each letter can begin with:

1)LUCY  LUYC  LCUY  LCYU  LYCU  LYUC

2)ULCY  ULYC  UCLY  UCYL  UYLC  UYCL

3)CLUY  CLYU  CULY   CUYL CYLU  CYUL

4)YLUC  YLCU  YCLY YCUL   YULC  YUCL

3 is in the formula because with in each set of arrangements there are 3 letters left which can follow:

1)LUCY LUYC 2)LCUY LCYU 3)LYUC  LYCU

2 is in the formula because then there are only the last two letters which can be swapped

1)LUCY 2) LUYC

1 is in the formula because there is only one place to put the last letter

(1 arrangement)

1)LUCY

I will now investigate words that have repeated letters. Like in Emma it has 4 letters but 2 of letters are the same. I shall start of looking at words, which have 2 of the letters the same.

ARRANGEMENTS FOR YY

1)YY

There is one arrangement for this word, which is half the number of arrangements for when n=2 but all its letters different.

JO=2 arrangements YY=1 arrangement

n

r

a

2=YY

YES

1

2=JO

NO

2

ARRANGEMENTS FOR BBC

1)BBC 2)BCB 3)CBB

There are 3 different arrangements for this word, which is half the number of arrangements for when n=3 with all different letters.

BBC=3 arrangements           JIM=6 arrangements

n

r

a

3=BBC

YES

3

3=JIM

NO

6

Having already investigated the number of arrangements for the word Emma, I have compared the number of arrangements with the name Lucy. I have found that Emma has half the number of arrangements than LUCY, which has all different letters.

EMMA=12 Arrangements   LUCY=24 Arrangements

n

r

a

4=EMMA

YES

12

4=LUCY

NO

24

Prediction

Now I shall investigate the number of arrangements for when n=5 with 2 of the letters the same, I shall use the name DAVID.

Judging from the above results I can see that the number of arrangements for a word with 2 letters the same, will have half the number of arrangements than the word with the same total number of letters but with all different letters. Therefore as the name RALPH has 120 arrangements so as DAVID has 2 letters the same a shall be half of RALPH. Therefore as Ralph has 120 arrangements the name DAVID will have 60 arrangements. 120/2=60

ARRANGEMENTS FOR DAVID

1)DAVID 2)DAVDI 3)DADVI 4)DADIV 5)DAIVD 6)DAIDV 7)DDVIA 8)DDVAI 9)DDAIV 10)DDAVI 11)DDIVA 12)DDIAV 13)DVDIA 14)DVDAI 15)DVADI 16)DVAID 17)DVIDA 18)DVIAD 19)DIDVA 20)DIDAV 21)DIVAD 22)DIVDA 23)DIAVD 24)DIADV

25)AVDID 26)AVDDI 27)AVIDD 28)ADDVI 29)ADDIV 30)ADIVD 31)ADIDV 32)ADVID 33)ADVDI 34)AIDDV 35)AIDVD 36)AIVDD

37)VDDAI 38)VDDIA 39)VDIAD 40)VDIDA 41)VDADI 42)VDAID 43)VADDI 44)VADID 45)VAIDD 46)VIDDA 47)VIDAD 48)VIADD

49)IDDVA 50)IDDAV 51)IDVAD 52)IDVDA 53)IDAVD 54)IDADV 55) IVDDA 56)IVDAD 57)IVADD 58)IADVD 59)IADDV 60)IAVDD

n

r

a

5=DAVID

YES

60

5=RALPH

NO

120

As you shown, my prediction for being 60 arrangements for the 5-lettered name is right. Now I shall compare the words with 2 letters the same and words with all different letters.

n

r

a

2-JO

NO

2

2-YY

YES

1

3-JIM

NO

6

3-BBC

YES

3

4-LUCY

NO

24

4-EMMA

YES

12

5-RALPH

NO

120

5-DAVID

YES

60

OBSERVATIONS

The main comparison I noticed between a word with all different letters and a word with 2 of its letters the same, is that a for a word with all different letters is double the amount of arrangements for a word with the same amount of letters but 2 of its letters the same. So a general formula for a word with 2 letters the same is

n!/2=a

n= number of letters in word.

a= number arrangements.

Now I shall look at words, which have 3 of its letters the same.

ARRANGEMENTS FOR XXX

1) XXX

There is only 1 arrangement for this 3-lettered word, which has 3 letters the same. This is a third of the number of arrangements than a word with the same amount of letters but only 2 of its letters the same, and 1 sixth of the number of arrangements for a word the same number of letters but with all different letters.

n

r

r 2x

a

3=XXX

NO

YES

1

3=BBC

YES

NO

3

3=JIM

NO

NO

6

...read more.

Conclusion

I will now gather the information that I have to do with words with 4 of the same letters and compare them to words that have all different letters, words with 2 of its letters the same, and words that have 3 letters of its letters the same.

n

r

r 2x

r 3x

a

4=LUCY

NO

NO

NO

24

4=EMMA

YES

NO

NO

12

4=XXXY

NO

YES

NO

4

4=AAAA

NO

NO

YES

1

5=RALPH

NO

NO

NO

120

5=DAVID

YES

NO

NO

60

5=XXXYZ

NO

YES

NO

20

5=MMMMA

NO

NO

YES

5

Observations

The main comparison between a word with 4 letters the same, word with 3 of its letters the same, a word with 2 of its letters the same and a word with all different letters which has the same total of letters, is that the words with 4 letters the same has one twenty fourth the amount of arrangements than a word with all different letters, it has 1 twelfth the amount of arrangements for a word with the same amount of letters but 2 of its letters the same. Also it is a quarter the amount of arrangements than a word with the same amount of letters but 3 of its letters the same. Therefore I can show a formula for a word with 3 of its letters the same. I know this because the formula for a word with all different letters is n! and as the number of arrnagments for a word with all different letters has 24 times the amount of arrangements for a word with the same number of letters but 4 of its letters the same, so the formula must be for a word with 4 letters the same

n!/24

...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 Emma and Lucy

    For example: 1*2 = 2i 1*2*3 = 3i 1*2*3*4 = 4i so on So if n represent the number of figures of a number, then it has arrangements of ni. The formula: NI NI: Can be caculated on caculator. Process: pres key N (the number of figure), then press key I, then you would get the arrangements.

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

    I found out that my analysis was correct, and that the changing of letters didn't affect the number of different arrangements for four letters, all different. Results: Number of Letters: Number of different combinations: 1 1 2 2 3 6 4 24 Rule: To find the number of all the

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

    So firstly, I listed out all the different arrangements for the name EMMA, there were: EMMA EMAM EAMM MEAM MAEM MEMA MAME AEMM AMME AMEM MMAE MMEA There are 12 different arrangements for the name Emma. This is interesting as this is half the number of arrangements of the name

  2. EMMA'S DILEMMA

    This arrangement also matches with the formula. So the formula worked and gave me the right arrangement. What if a word had three letters same and the other 2 was also same. Lets work out the arrangement and find the right arrangement with the formula. 5-lettered word with 1 letter repeating 3 times and the other letter repeating twice: 1.

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

    it works Formular is confirmed What about if 3 numbers are the same let's try 333 only on arrangement Try 3331 3331 3313 ------ 4 arrangements 3133 1333 Try 33312 33312 31233 12333 33321 31323 13233 ---4 arrangements 33123 31332 ----12 arrangements 13323 33132 32331 13332 33231 32313 33213 32133

  2. Emma's dilemma

    =A For example: a three letter word with all letters different Prediction: for the word with six letters different. Using pattern noticed 1) N*(N-1) =A 6*5! =6*120 =720 Using the formula 2) A=N! A=6! A=720 The calculations using formula and the pattern noticed both agree.

  1. Emma's Dilemma

    YUCL LUYC ULYC CUYL YULC LCUY UCLY CLUY YCUL LCYU UCYL CLYU YCLU LYUC UYLC CYLU YLUC LYCU UYCL CYUL YLCU Total Number of Arrangements: 24 Name: "FRANK" Number of letters: 5 Arrangements: FRANK RFANK ARFNK NRAFK KRANF FRAKN RFAKN ARFKN NRAKF KRAFN FRNAK RFNAK ARNFK NRFAK KRNAF FRNKA RFNKA

  2. Emma's Dilemma

    UPAIEL UPAEIL UAPELI UPLAEI UPLAIE UPLIAE UPLIEA UPLEIA UPLEAI UPILAE UPILEA UPIELA UPIEAL UPIAEL UPIALE UPELAI UPELIA UPEAIL UPEALI UPEIAL UPEILA UAPLIE UAPLEI UAPELI UAPEIL UAPIEL UAPILE UALPIE UALPEI UALEPI UALEIP UALIPE UALIEP UAIPLE UAIPEL UAIELP UAIEPL UAILEP UAILPE UAEPLI UAEPIL UAELPI UAELIP UAEIPL UAEILP ULPAIE ULPAEI ULPEAI ULPEIA ULPIEA

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