# Emma's Dilemma.

Extracts from this document...

Introduction

Sean Jones

Maths coursework

Set 1; Mrs Loney

June2003

## Emma’s Dilemma

Emma is playing with the arrangements of the letters in her name.

One arrangement is EMMA

A Different arrangement is Meam

The following is a list of all combinations of the letters in the name EMMA.

Emma has 2 letters the same.

EMMA

AMME

AMEM

EMAM

AEMM

EAMM

MMEA

MMAE

MEMA

MAME

MEAM

MAEM

Emma has a friend named Lucy

Lucy has all 4 letters different.

LUCY

LUYC

LYCU

LYUC

LCYU

LCUY

ULCY

ULYC

UCYL

UCLY

UYCL

UYLC

CLUY

CLYU

CUYL

CULY

CYLU

CYUL

YLUC

YLCU

YUCL

YULC

YCLU

YCUL

The arrangements for Lucy has came out with 24 different combinations. This is double that of the combinations for the name Emma.

This is because of EMMA having two letters the dame. This illiminate's the possibility of creating more different combinations. No matter how much you jumble up MM, they will stay MM.

I ma now going to look to see how combinations for other names with all letters different figure out.

Two letters different

JO

OJ

Two combinations

Three letters different

SAM

SMA

AMS

ASM

MAS

MSA

6 different combinations

Lucy

See above

24 different combinations

5 Letters different

ANDIE

ANDEI

ANEDI

ANEID

ANIDE

ANIED

ADNIE

ADNEI

ADENI

ADEIN

ADIEN

ADINE

AIDNE

AIDEN

AIEDN

AIEND

AINED

AINDE

AEIDN

AEIND

AENID

AENDI

AEDNI

AEDIN

So far I have 24 combinations. All combinations begin with ‘a’ .

Middle

2

2

3

6

4

24

5

120

6

720

7

5040

I can see a pattern

The number of letters in the current name multiplied by the number of arrangements in the previous.

4 *6 =24

5*24 = 120

Therefore the arrangements for a name with 6 different letters will be. 6* 120 which gives you 720. And so on for seven different arrangements.

Or better still you can do 5!

Which is called 5 factorial. Which is the same as 5 * 4*3*2*1

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

This is a lot simpler

The formula for this is : N! = A

Part 2

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.

(

Conclusion

I believe that the new formula will be n!/ something.

To work out what that something was I looked to the X’s and Y’s. And figured that when you multiply together, the amount of Y's and the amount of X's in the letter, and factorial them both. You get the total combinations that the word in question gives.

I am now going to test the formula to prove it works

A four letter word like aabb; has 2 A’s and 2 B’s

So: 1x2x3x4

(1x2) * (1x2)

= 24

4

= 6 different arrangements

A five letter word like aaaab; this has 4 A’s and 1 b

So: 1x2x3x4x5

(1x2x3x4) * (1)

= 120

24

= 5 different arrangements

This shows that my formula works

N! = The number of letters in the word

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

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