# Emma's Dilemma

Extracts from this document...

Introduction

David Leiwy

EMMA'S DILEMMA

4 LETTER NAME WITH 2 LETTER THE SAME

EMMA

- EMMA
- EMAM
- EAMM
- MEMA
- MEAM
- MMEA
- MMAE
- MAME
- MAEM
- AEMM
- AMEM

12) AMME

4 LETTER NAME

LUCY

- LUCY
- LCUY
- LCYU
- LUYC
- LYUC
- LYCU
- ULCY
- ULYC
- UCLY
- UCYL
- UYCL
- UYLC
- CYLU
- CYUL
- CUYL
- CULY
- CLUY
- CLYU
- YLCU
- YLUC
- YULC
- YUCL
- YCLU
- YCUL

I am now going to test this out on different names and see if I can find a pattern.

2 LETTER NAME

MO

1) MO

2) OM

3 LETTER NAME

JOE

1) JOE

2) JEO

3) EJO

4) EOJ

5) OEJ

6) OJE

5 LETTER NAME

RICKY

- RICKY
- RICYK
- RIKCY
- RIKYC
- RIYCK
- RIYKC
- RCIKY
- RCIYK
- RCYIK
- RCYKI
- RCKIY
- RCKYI
- RKYCI
- RKYIC
- RKICY
- RKIYC
- RKCYI
- RKCIY
- RYICK
- RYIKC
- RYKCI
- RYKIC
- RYCKI
- RYCIK
- ICKYR
- ICKRY
- ICRYK
- ICRKY
- ICYRK
- ICYKR
- IKYCR
- IKYRC
- IKCYR
- IKCRY
- IKRCY
- IKRYC
- IYRKC
- IYRCK
- IYCRK
- IYCKR
- IYKRC
- IYKCR
- IRCKY
- IRCYK
- IRKCY
- IRKYC
- IRYCK
- IRYKC
- CKYRI
- CKYIR
- CKRIY
- CKRYI
- CKIRY
- CKIYR
- CYRIK
- CYRKI
- CYIRK
- CYIKR
- CYKRI
- CYKIR
- CRYKI
- CRYIK
- CRKYI
- CRKIY
- CRIKY
- CRIYK
- CIRKY
- CIRYK
- CIKYR
- CIKRY
- CIYKR
- CIYRK
- KYRIC
- KYRCI
- KYICR
- KYIRC
- KYCRI
- KYCIR
- KRCIY
- KRCYI
- KRIYC
- KRICY
- KRYIC
- KRYCI
- KICYR
- KICRY
- KIYCR
- KIYRC
- KIRYC
- KIRCY
- KCYIR
- KCYRI
- KCIYR
- KCIRY
- KCRYI
- KCRIY
- YRICK
- YRIKC
- YRCIK
- YRCKI
- YRKCI
- YRKIC
- YICKR
- YICRK
- YIKCR
- YIKRC
- YIRCK
- YIRKC
- YCKRI
- YCKIR
- YCRIK
- YCRKI
- YCIRK
- YCIKR
- YKRIC
- YKRCI
- YKIRC
- YKICR
- YKCRI
- YKCIR

Name | No. Of letters | No. Of permutations |

MO | 2 | 2 |

JOE | 3 | 6 |

LUCY | 4 | 24 |

RICKY | 5 | 120 |

I found that when you multiply the no. of letters by the previous amount of permutations, you will find the correct amount of permutations.

To find out how many permutations there are in a 3-letter word: 3*2 = 6 (which is 3!)

To find out how many permutations there are in a 4-letter word: 4*6= 24 (which is 4!)

I also noticed that each no.

Middle

NO. OF LETTERS IN NAME | NO. OF COMBINATIONS |

1 | 1 |

2 | 2 |

3 | 6 |

4 | 24 |

Therefore,

NAME | NO. OF LETTERS | NO. OF PERMUTATIONS |

TIM | 3 | 6 |

FRED | 4 | 24 |

CRAIG | 5 | 120 |

CLARKE | 6 | 720 |

CHARLIE | 7 | 5040 |

CHARLTON | 8 | 40320 |

I am now going to test weather duplicates of letters will make a difference in the no. of permutations.

3 LETTER NAME

WITH 2 LETTERS THE REPEATED

BOB

- BOB
- BBO
- OBB

4 LETTER NAME

WITH 2 LETTER THE REPEATED

EMMA

- EMMA
- EMAM
- EAMM
- MEMA
- MEAM
- MMEA
- MMAE
- MAME
- MAEM
- AEMM
- AMEM
- AMME

4 LETTER NAME

WITH 3 LETTERS THE REPEATED

LILL

- LILL
- ILLL
- LLIL
- LLLI

5 LETTER NAME

WITH 2 LETTERS THE REPEATED TWICE

EDDIE

- EDDIE
- EDDEI
- EDIDE
- EDIED
- EDEID
- EDEDI
- EIEDD
- EIDDE
- EIDED
- EEIDD
- EEDID
- EEDDI
- DDIEE
- DDEIE
- DDEEI
- DEDIE
- DEDEI
- DEIDE
- DEIED
- DIEED
- DIEDE
- DIDEE
- DEEID
- DEEDI
- IEEDD
- IDDEE
- IDEDE
- IDEED
- IEDED
- IEDDE

5 LETTER WORD WITH 3 LETTERS REPEATED

DADDY

- DADDY
- DDDAY
- DDDYA
- DDADY
- DDAYD
- DADYD
- ADDDY
- ADDYD
- ADYDD
- AYDDD
- YDDDA
- YDDAD
- YDADD
- YADDD
- DDYAD
- DDYDA
- DAYDD
- DYADD
- DYDAD
- DYDDA

NAME | NO. OF LETTERS | NO. OF REPEATED LETTERS | NO. OF PERMUTATIONS | |||

BOB | 3 | 2(B) | 3 | |||

EMMA | 4 | 2(M) | 12 | |||

LILL | 4 | 3(L) | 4 | |||

EDDIE | 5 | 2(E) + 2(D) | 30 | |||

DADDY | 5 | 3(D) | 20 |

Conclusion

X!

permutations was divided by 2 (which is 2!) to find the number of permutations.

For 5 letters with 2x’s & 3y’s after using the formula N! , that no. of

X!

permutations was divided by 6 (which is 3!) to find the number of permutations

This shows another pattern of factorial.

Therefore you need to divide the formula by Y!. This now becomes:

N!

X! *Y!

Now using this formula I will be able to determine the no. Of permutations in any name or letter combination.

Using my formula I can prove how many permutations are in these names

## NAME | ## NO. OF LETTERS | NO. OF PERMUTATIONS |

ROBERT | 6 | 360 |

## RICHARD | 7 | 2520 |

## FREDDIE | 7 | 1260 |

## HARRISON | 8 | 20160 |

## THANCANAMOOTOO | 14 | 151351200 |

## Now I will find out the no. Of permutations using x’s & y’s

## PERMUTATION | NO. OF LETTERS | ## NO. OF X’S | ## NO. OF Y’S | NO. OF PERMUTATIONS |

## XXYYYYY | 7 | 2 | 5 | 21 |

## XXXXXYYY | 8 | 5 | 3 | 56 |

## XXXXXXXXY | 9 | 8 | 1 | 9 |

XXXXXYYYYY | 10 | 5 | 5 | 252 |

XXXXXXXXXXX YYYYYYYYY | 20 | 11 | 9 | 167960 |

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