# Emma's Dilemma

Extracts from this document...

Introduction

Emma’s Dilemma

In my investigation I am going to investigate the number of different arrangements of letters in a word, e.g.

Tim Is one arrangement

Mit Is another

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

LUCY

LUYC

LYCU

LYUC

LCYU

LCUY

ULCY

UCLY

UYLC

ULYC

UCYL

CYUL

CYLU

CULY

CUYL

CLUY

CLYU

YCUL

YCLU

YLUC

YLCU

YUCL

YULC

There are 4 different letters and there are 24 different arrangements.

SAM

SMA

MSA

MAS

ASM

AMS

There are 3 different letters in this name and 6 different arrangements.

JO

OJ

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

With a two letter name with no repeats I had 2 possible arrangements

With a three letter name with no repeats I had 6 possible arrangements

With a four letter name with no repeats I had 24 possible arrangements

From this Information I have noticed that I can find out the number of possible arrangements without writing them all out. This is because I have noticed a way to work out the answer using sufficient information. I have noticed that the answer to 2L, for example, is the previous answer multiplied by the number of letters in that name. I.E. The possible arrangements for a 5 letter name with no repeats is (24x5)

I shall now test my hypothesis:

JAMES

JAEMS

JASEM

JASME

JAMSE

JAESM

JSAME

JSMEA

JSMAE

JSAEM

JSEAM

JSEMA

JMASE

JMAES

JMESA

JMEAS

JMSAE

JMSEA

JEAMS

JEASM

JESAM

JESMA As there is five letters in the name James there will obviously be 120 arrangements as there is 24 for each letter

JEMAS

JEMSA Just as I had predicted there are a possible 120 arrangements for a five letter name with no repeats.

Middle

For 4 letters with one letter repeating once: ½ x (4 x 3 x 2 x 1) = 12

The formula works!

I shall now look at names where two letters repeat once e.g Lulu

I will have to start at 4 letter names as you cant possibly have 2 letters repeating once in any smaller character name.

ANNA

NANA

NNAA

NAAN

ANAN

AANN

There are 6 possible arrangements for four lettered names with two letters repeating once

EDDIE

EDIDE

EDEID

EDDEI

EDIED

EDEDI

EEDDI

EEDID

EEIDD

EIDED

EIEDD

EIDDE

DDIEE

DDEIE

DDEEI

DIDEE

DIEDE

DIEED

DEEID

DEEDI

DEIDE

DEIED

DEDIE

DEDEI

IEEDD

IEDED

IEDDE

IDEDE

IDDEE

IDEED

There are 30 possible arrangements for a 5 letter name with two letters repeating once

Here are all my results I have found so far in a table:

In this table 1 repeat means there is one letter that was repeated. I.E Aandy However 2 repeats means there were two letters that each repeated once I.E. EEDDYA

Once again I have noticed that the arrangements for the names with two letters each repeating once are ¼ of the names with no repeats, and ½ of the names with one letter repeating once. Therefore the formula for a name with two letters repeating once is ¼ x N!

From this I can predict that the possible arrangements for a 5 letter name in which 2 letters each repeat once is 90. A 6 letter name in which 2 letters each repeat once has a possible 1760 arrangements.

I Can therefore work out a formula to suit all names in which a certain amount of letters repeat once.

Conclusion

* The numbers with question marks next to them are predictions based on other results I have found. They are correct though as they fit with all the other pieces of information*

It is clear that the answers to names with two letters repeating N times will always be half of the answers where the name has 1 letter repeating N times

E.g. where a 6 letter name with 3 of one letter in it will have twice as many possible arrangements than a 6 letter name with 3 of two letters in it. This will help in finding a suitable formula as 2L is always going to be half the answer of 1L i.e. 2L = ½ N when 1L = N

Formula: l! % x! = possible arrangements (A)

(a) represents the total arrangements

(l) represent the number of letters in the name

(x) represent the numbers of the same letter in the name

(I) represents the recipicle

For example:

for 2 of the same letter in the name the formula is a=li/xixi

for 2 pairs of different letters in name the formula is a=li/x1ix2i

for 3 of the same letter in the name the formula is a=li/xixixi

for 3 pairs of different letters in the name the formula is a=li/x1ix2ix3i.

For example:

xxyy

the arrangement for this is a=(4*3*2*1)/(2*1*2*1)=6

xxyyy

the arrangement for this is a=(5*4*3*2*1)/(3*2*1*2*1)=10

xxxxxxyyyyyyyyyy

the arrangement for this is:

a=ni/x1ix2i=(16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1)/(10*9*8*7*6*5*4*3*3*2*1*6*5*4*3*2*1)=8008

Use this formula, we can find out the total arrangements of all numbers and letters.

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