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• 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.

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