• 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
• Level: GCSE
• Subject: Maths
• Word count: 1322

# Emma&amp;#146;s Dilemma.

Extracts from this document...

Introduction

Ross Holden

Emma’s Dilemma

This task involves playing with different arrangements of different names. The names will vary in size and with letters being the same or not the same.

Different arrangements for the name EMMA are:

1) EMMA        4) AEMM        7) MAME        10) MEAM

2) EMAM        5) AMEM        8) MMAE        11) MAEM

3) EAMM        6) AMME        9) MEMA        12) MMEA

Different arrangements for the name LUCY are:

1) LUCY        7) ULCY         13) CULY        19) YCLU

2) LUYC        8) ULYC         14) CUYL        20) YCUL

3) LYUC        9) UYLC         15) CYLU        21) YULC

4) LYCU        10) UYCL         16) CYUL        22) YUCL

5) LCUY        11) UCYL         17) CLUY        23) YLUC

6) LCYU        12) UCLY         18) CLYU        24) YLCU

I am going to find out if there is a relationship between number of letters and number of arrangements when all the letters in the combination are different.

L has only one arrangement which is L.

LU has two arrangements:

1. LU
2. UL

LUC has six arrangements:

1. LUC        3) ULC        5) CLU
2. LCU        4) UCL        6) CUL

LUCY, as we know has 24 different arrangements.

Here is a table showing the number of letters and the number of arrangements when all the letters are different:

 Number of Letters Number of Arrangements 1 1 2 2 3 6 4 24

Middle

To do this, you can use the method of factorial notation which multiplies a number by all the previous consecutive numbers. On a calculator the factorial button looks like this:

!

If you press a number and then !, then it multiplies that number by all the previous consecutive numbers.

Therefore, the formula when all the letters are different is

A = x!

Where X is the number of letters and A is the number of arrangements.

Now I am going to find out the relationship between the number of letters and the number of arrangements when there are two letters the same.

JO has one arrangement:

1) JJ

ANN has 3 different arrangements:

1. JJO
2. JOJ
3. OJJ

EMMA, as we know has 12 different arrangements.

Here is a table showing the number of letters and the number of arrangements where 2 letters are the same.

## Number of arrangements

2

1

3

3

4

6

Even though there are 2 letters the same, the relationship between the number of letters and the number of arrangements is very similar to when there are no letters the same.

Conclusion

Here is a table showing the number of letters, the number of X’s, Y’s and the number of arrangements.

 Number of letters Number of X’s Number of Y’s Number of Arrangements 4 3 1 4 5 3 2 10 6 3 3 20

Here, there is definitely a relationship. If you use where there and 4 letter with 3 X’s and 1 Y you do      4 x 3 x 2 x 1          which makes  24  which equals 4 which

(3 x 2 x 1)x(1 x 1)                               4

agrees with the table.

Therefore with XXXXYYYY, there should be 70 different arrangements because

8!       = 70

4! X 4!

Here are the arrangements for XXXXYYYY:

1. XXXXYYYY        8) XXYYXXYY        15) XYXYXXYY        22) XYYXXYYX
2. XXXYXYYY        9) XXYYXYXY        16) XYXYXYXY        23) XYYYYXXX
3. XXXYYXYY        10) XXYYXYYX        17) XYXYXYYX        24) XYYYXXXY
4. XXXYYYXY        11) XXYYYXXY        18) XYXYYXXY        25) XYYYXXYX
5. XXXYYYYX        12) XXYYYXYX        19) XYXYYXYX        26) XYYYXYXX
6. XXYXXYYY        13) XXYYYYXX        20) XYXXYYXY        27) XYYXYXXY
7. XXYXYXYY        14) XYXXXYYY        21) XYXXYYYX        28) XYYYYXXX

29) XYYXYXYX

31) XYXXXXYY               As there are an equal number of X’s and Y’s, you

32) XYYYXXYX               multiply the number of arrangements so far by 2

33) XYYXXYXY               to get the total number of arrangements.

34) XYXYYYXX

35) XYYXXXYY

Therefore the overall formula is

A =   X!

S!L!

Where A is the number of arrangements, x is the number of letters, s and l are the number of letters the same.

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

# Related GCSE Emma's Dilemma essays

1. ## Emma's Dilemma

However, with two letters repeated twice, the letters which are repeated can be swapped with their opposites, with-out making a new arrangement. Answer, Two letters repeated Three Times: Six letters: AAABBB BAAABB AABABB BAABAB AABBAB BAABBA AABBBA BABAAB ABAABB BABABA Total: ABABAB BABBAA 20 ABABBA BBAAAB ABBAAB BBAABA ABBABA BBABAA

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

AAABB: This word has two different letters. A is repeated three times and B is repeated twice. I managed to work out the number of permutations, using help from the above word AABB. As in the word AAABB, there is one extra A, I could then add an A onto the beginning of 'AABB's permutations' giving me 6 permutations.

1. ## To investigate the combination of arrangement of letters in Jeans name and then for ...

I B L I L I B I L PAGE 4 My prediction was correct. So if I tried the combination ozzzo there would b less then 30 combinations. I predict there will be I predict there will be 20 combinations because 120 / 3 / 2 = 20.

2. ## Emma's Dilemma

First, I will investigate a 4-letter word. LLLL There is only 1 arrangement. Now I will investigate a 5-letter word. HHHHU HHHUH HHUHH HUHHH UHHHH There are 5 arrangements. Now I am going to investigate a 6-letter word. SSSSEF SSSSFE SSSESF SSSEFS SSSFSE SSSFES SSESSF SSESFS SSEFSS SSFSSE SSFSES SSFESS

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

This is due to the fact that the six letters in the previous number of arrangements can be rearranged with another letter. This increases the total number of arrangements , for this number of arrangements, by a factor of 7.

2. ## Emma&amp;amp;#146;s Dilemma.

Four Letters: 'LUCY' I have already found the number of possible permutations for a name with four letters (Lucy) in part two. There are a total of 24 permutations, which shows the pattern widening out even more. Five Letters: 'KACIE' K = 1 A =2 C = 3 I =

1. ## Emma is playing with arrangements of the letters of her name to see how ...

I will continue with my method to see if my prediction is correct. MILYE MLYEI MYEIL MEILY MILEY MLYIE MYELI MEIYL MIYEL MLIYE MYLIE MELIY MIYLE MLIEY MYLEI MELYI MIEYL MLEYI MYIEL MEYLI MIELY MLEIY MYILE MEYIL I have found a further 24 arrangements beginning with M.

2. ## Emma's Dilemma

For example, the number 4 can indicate the word 'Lucy.' The numbers on the bottom row indicate the various outcomes I obtained by using the method of listing. For example, the number 24 indicates that there are 24 different possible outcomes for a 4 lettered word, as it is placed under the number 4.

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