# Emma&#146;s Dilemma.

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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:

- LU
- UL

LUC has six arrangements:

- LUC 3) ULC 5) CLU
- 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:

- JJO
- JOJ
- 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 letters | ## 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:

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

29) XYYXYXYX

30) XYXYYYXX These are all the arrangements which start with X.

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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