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• 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’.

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

 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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