# Emma&#146;s Dilemma

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Introduction

Emma’s Dilemma

In my investigation I am going to investigate of the different arrangements of letters in a word. My aim is to be able to generalise a formula for all my findings.

E.g. Tim Is one arrangement

Mit Is another

- First I am going to investigate how many different arrangements in the name “Emma”, which has no letters the same.

EMMA AMME AMEM EMAM AEMM EAMM MMEA

MMAE MEMA MAME MEAM MAEM

4-letter word, 2 letters repeated, 12 different arrangements.

Now I am going to investigate the different arrangement of Emma’s friends name “Lucy”

LUCY LCUY LYUC LUYC LCYU LYCU

UCLY UYCL UYLC ULYC UCYL ULCY

CYUL CYLU CUYL CULY CLUY CLYU

YCUL YUCL YULC YLCU YLUC YCUL

- There are 4 different letters and there are 24different arrangements.
- In the name “SAM” (Investigating the different number of arrangement in a word with 3 different letters)

SAM SMA MSA MAS ASM AMS

- There are 3 different letters in this name and 6 different arrangements.
- Investigating the number of arrangements in a two-letter word with 2 different letters.

JO OJ

There are 2 different letters in this name and there are 2 different arrangements.

Table of Results

Number of Letters | Number of Different Arrangements |

2 | 2 |

3 | 6 |

4 | 24 |

5 | 120 |

6 | 720 |

7 | 5040 |

- From the table of results I have found out that a 2-letters word with individual letters has 2 arrangements, and a 3-letter word with individual letters has 6.
- Taking for example a 3-letter word, I have worked out that if we do 3 (the length of the word) x 2 = 6, the number of different arrangements.

In a 4 letter word, to work out the amount of different arrangements you can do 4 x 3 x 2 = 24, or you can do 4! Which is called 4 factorial that is the same as 4 x 3 x 2.

- So, by using factorial (!) I can predict that there will be 40320 different arrangements for an 8-letter word.

The formula for this is: n! = a

- Where n = the number of letters in the word and

a = the number of different arrangements.

Now I am going to investigate the number of different arrangements in a word with 2 letters repeated, 3 letters repeated and 4 letters repeated.

- This is a 3-letter word with 2 letters repeated

Middle

PQMMM PMQMM PMMQM PMMMQ

QPMMM QMPMM QMMPM QMMMP

MPQMM MPMQM MPMMQ MQPMM

MQMPM MQMMP MMPQM MMQMP

MMMPQ MMMQP MMPQM MMMQP

- 5-letter word, 3 letters repeated, 20 different arrangements.
- This a 4- letter word with 3 letters repeated

SMMM MSMM MMSM MMMS

4-letter word, 3 letters repeated, 4 different arrangements.

- This is a 5-letter word with 4 letters repeated

RRRRK RRRKR RRKRR RKRRR KRRRR

5-letter word, 4 letters repeated, 5 different arrangements.

## Table of Results

No of letters (n) | No letters repeated | Same letter repeated 2x (p) | Same letter repeated 3x (p) | Same letter repeated 4x (p) | Same letter repeated 5x (p) |

3 | 6 | 3 | 1 | 0 | 0 |

4 | 24 | 12 | 4 | 1 | 0 |

5 | 120 | 60 | 20 | 5 | 1 |

6 | 720 | 360 | 120 | 30 | 6 |

7 | 5040 | 2520 | 840 | 210 | 42 |

I have worked out that if I say 5! = 120, to find out how many different arrangements in a 3 letter word (3x2 x1=6) it would be 5! Divided by 6= 20, so, a 6-letter word with 4 letters (4x3x2x1=24) repeated would be 6! Divided by 24 = 30, as you can see in the “No letters repeated” column these are the numbers we are dividing by:

- 2 letters the same = n!

(2x1 = 2)

- 3 letters the same = n!

( 3x2x1 = 6)

Conclusion

A five letter word like aaaab; this has 4 a's and 1 b (4 x's and 1 y)

So: 1x2x3x4x5 (5 letter words) = 120 = 5 different arrangements

(1x2x3x4) x (1) (4a’s x 1b) 24

A five letter words like abcde; this has 1 of each letter (no letters the same)

So: 1x2x3x4 = 24 = 24 different arrangements

(1x1x1x1x1x1) 1

A five letter word like aaabb; this has 3 a's and 2 b's (3 x's and 2 y's).

So: 1x2x3x4x5 = 120 =10 different arrangements

(1x2x3) x (1x2) 12

This shows that my formula works:

n! = The number of letters in the word

x!y! = The number of repeated letters the same

### Conclusion

#### I was able to find these formulas

- The formula for this is: n! = a

Where n = the number of letters in the word and

a = the number of different arrangements

- This formula is used to find out the different arrangements and possibilities from individual letters in a word.

- 2 letters the same = n!

(2x1 = 2)

- 3 letters the same = n!

( 3x2x1 = 6)

- 4 letters the same = n!

(4x3x2x1 = 24)

- 5 letters the same = n!

(5x4x3x2x1 = 120)

- 6 letters the same = n!

(6x5x4x3x2x1 = 720)

- Finding similarities in letters the formula I found are shown above.

Deborah Taiwo

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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