• 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
  13. 13
    13
  14. 14
    14
  • Level: GCSE
  • Subject: Maths
  • Word count: 1772

Emma’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.

...read more.

Middle

  1. MHDAE
  1. MHEAD
  1. MHEDA
  1. MEHDA
  1. MEHAD
  1. MEDAH
  1. MEDHA
  1. MEADH
  1. MEAHD
  1. MDEAH
  1. MDEHA
  1. MDHEA
  1.  MDHAE
  1. MDAHE
  1. MDAEH
  1. EAHMD
  1. EAHDM
  1. EADHM
  1. EADMH
  1. EAMDH
  1. EAMHD
  1. EMAHD
  1. EMADH
  1. EMDAH
  1. EMDHA
  1. EMHDA
  1. EMHAD
  1. EDMAH
  1. EDMHA
  1. EDHAM
  1. EDHMA
  1. EDAMH
  1. EDAHM
  1. EHMAD
  1. EHMDA
  1. EHDMA
  1. EHDAM
  1. EHAMD
  1. EHADM
  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

...read more.

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        image02.pngimage03.png

x!                x=4                4!=   24image04.png

  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!         image05.png

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        

...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. Arrangements for names.

    Like in Emma; it has 4 letters but 2 of which are the same. 4 factorial equals 24, but I could only find 12, which means that there more to it that just factorial in that way. To make it a bit easier instead of using letters as such I will use x's and y's (any letter).

  2. Emma's Dilemma

    X ( No. of letters - 2 ) X etc... combinations = n X (n-1) X (n-2) X (n-3) X ... ... X 3 X 2 X 1 For Example: To find the number of combinations which can be made from using 10 letters can be found by using the following formula: No.

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

    the same arrangements occur just with one 'A' being replaced with a 'B' in the letters that are moving around. The following 20 permutations can be produced for this word: BAAABB BAABAB BAABBA BABABA BABAAB BABBAA BBAAAB BBAABA BBABAA BBBAAA BAAABB BAABAB BAABBA BABABA BABAAB BABBAA BBAAAB BBAABA BBABAA BBBAAA

  2. EMMA's Dilemma Emma and Lucy

    arrrangements 1322 2213 ---- 6 arrangements 3221 2231 2312 2321 total arrangement is 12 Try 5 fig: 42213 12234 42231 12243 42123 12324 42132 12342 42321 12423 42312 -------- 12 arrangements 12432 -------- 12 arrangements 41223 13224 41232 13242 41322 13422 43122 14223 43212 14232 43221 14322 21234 23124 31224

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

    3 X 2 This is due to the fact that the six different letters can be arranged in any order, and because all of them are different, it dose matter which order they go in. However, with one letter repeated three times, the letter which is repeated can be swapped with itself, with-out making a new arrangement.

  2. To investigate the various arrangements that could be made with different names or words. ...

    I will now create a table showing the results of this investigation and also the results from my second investigation, to compare the arrangements with all different letters to the arrangements with three of the same. Numbers of letters in the name Arrangements made from words with all different letters

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