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• Level: GCSE
• Subject: Maths
• Word count: 1772

# Emma&amp;#146;s Dilemma

Extracts from this document...

Introduction

A.M.D.GJune 2001

## Emma’s Dilemma

I firstly found the different number of arrangements for Emma’s name.

1. EMMA
1. EMAM
1. EAMM
1. AEMM
1. AMEM
1. AMME
1. MAME
1. MAEM
1. MEAM
1. MEMA
1. MMAE

I then found the different number of arrangements for Lucy’s name.

1. LUCY
1. LUYC
1. LYUC
1. LYCU
1. LCUY
1. LCYU
1. ULCY
1. ULYC
1. UYLC
1. UYCL
1. UCYL
1. UCLY
1. CULY
1. CUYL
1. CYUL
1. CYLU
1. CLYU
1. CLUY
1. YLUC
1. YLCU
1. YCLU
1. YCUL
1. YUCL
1. YULC

The method I used to investigate the number of different arrangements of Emma’s name was by keeping the first two letters the same and rearranging the last two letters. I kept on doing this until I had the maximum number of arrangements, which was 12. I repeated this procedure with Lucy’s name and found 24 arrangements.

Next I found the arrangements of some different names to see if I could spot a pattern.

1. TOM
1. TMO
1. MTO
1. MOT
1. OMT
1. OTM

My strategy for finding the number of arrangements for the name Tom, was by this time, keeping only the first letter the same and rearranging the last two letters. I kept on doing this until I had found the maximum number of arrangements, which was 6.

In order to get an orderly pattern, I then decided to find the arrangements of a name with two different letters and one letter.

1. J

One letter gives just the one arrangement.

1. JO
1. OJ

Two different letters give two arrangements.

Middle

1. MHDAE
1. MHEDA
1. MEHDA
1. MEDAH
1. MEDHA
1. MEAHD
1. MDEAH
1. MDEHA
1. MDHEA
1.  MDHAE
1. MDAHE
1. MDAEH
1. EAHMD
1. EAHDM
1. EAMDH
1. EAMHD
1. EMAHD
1. EMDAH
1. EMDHA
1. EMHDA
1. EDMAH
1. EDMHA
1. EDHAM
1. EDHMA
1. EDAMH
1. EDAHM
1. EHMDA
1. EHDMA
1. EHDAM
1. EHAMD
1. DAHME
1. DAHEM
1. DAMEH

100.DAMHE

101. DAEMH

102. DAEHM

103. DEAMH

104. DEAHM

105. DEHMA

106. DEHAM

107. DEMAH

108. DEMHA

109. DMEAH

110. DMEHA

111. DMHEA

112. DMHAE

113. DMAHE

114. DMAEH

115. DHEAM

116. DHEMA

117. DHAEM

118. DHAME

120. DHMEA

From this, we can see that the formula, lp=a works because the 5-lettered name Ahmed, gives 120 arrangements, as I predicted. However, the letter ‘p’ may not be very useful because you do not know the number of arrangements from the previous name in all cases. Now I must find a simpler formula for this. I wrote the arrangements out again.

J-1

JO-2

TOM-6

LUCY-24

I noticed that ‘l’ was fine, as it is in the formula but ‘p’ needed to be broken down a little. I then wrote down a formula for all the names substituting the values in for ‘l’, ‘p’ and ‘a'

JO                1*2=2

TOM        2*3=6

LUCY        6*4=24

From this, I could see that in the formula for the name Tom, ‘p’ which is 2, is the same as the calculation for Jo; therefore, the calculation for Tom’s name is 1*2*3. This works for Lucy’s name as well because ‘p’ in

Conclusion

x. This is because 120/3=40 and not 20 which is the value of ‘a’. I then looked at other names in which the same letter occurs 3 times.
1. EMMM                                 1. MMM
1. MEMM
1. MMEM
1. MMME

I then arranged the names in size order and wrote ‘l’ factorial for each one.

MMM-1                3!=6

EMMM-4                4!=24

SUSIS-20                5!=120

I noticed that all these arrangements with 1 letter repeated 3 times were the quotient of l! and 6 because 6 is the factorial of 3 and 3 is the number of same letters. This gives us a new formula of a=l!/x!. I shall test this with a 5-lettered name with 4 same letters. Firstly, I will predict my answer, then I will write all the different arrangements.

l!   =a                l=5                5!=  120

x!                x=4                4!=   24

1. MMMME
1. MMMEM
1. MMEMM
1. MEMMM
1. EMMMM

From this we can see that the formula l!/x!=a works for names which contain one letter which is repeated.

This formula works because the number of arrangements for a name that repeats a letter will obviously be less.

Now I will investigate a general formula by seeing AABBBCD

b= another pair of same letters or more.     C=number of different letters i.e. ‘c’

We know that the formula        a=l!

x!b!c!

EMMA        4! / 2!*1!*1!=12

AHMED        5! / 1!*1!*1!=120

LUCY                4! / 1!*1!*1!=24

SUSIE        5! / 2!*1!*1!=60

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

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