# Emma's Dilemma.

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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

- 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

- 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’.

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 |

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

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