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• Level: GCSE
• Subject: Maths
• Word count: 1848

# Emma's Dilemma.

Extracts from this document...

Introduction

Emma’s Dilemma Coursework

Aim:        To investigate the number of different arrangements of the letters in the name Emma and several other names I have chosen and find a formula for the names with different amounts of letters. I will also investigate the number of different arrangements of the letters X and Y and find a formula for the words with different amounts of X’s and Y’s.

Prediction:        I think that the more letters in the name the more arrangements of that name there will be. I also think that there will be more arrangements of Lucy than of Emma because all the letters in Lucy are different whilst two letters in the name Emma are the same.

Method:   To work out the number of different arrangement of names I will write down the first letter of the name, then systematically work out all the arrangement of the name that begin with that letter by using the branching method. The branching method looks like this

Y        C                        LUYC

## Y        C        U                        LYCU

As you can see there is six different arrangement of the name Lucy that begin with L.

Middle

MDYAN
###### Y        N        A        MDYNA

A        Y        MDNAY

D        N        Y        A        MDNYA

N        Y        MDANY

A        Y        N        MDAYN

###### D        N        A        MYDNA

D        A        MYNDA

## A        N        D        MYAND

D        Y        ANMDY

M        Y        D        ANMYD

N                M        Y        ANDMY

D        Y        M        ANDYM

D        M        ANYDM

Y        M        D        ANYMD

###### D        N        M        AYDNM

A        M        AYNAM

Y        N        M        A        AYNMA

N        D        AYMND

M        D        N        AYMDN

###### D        N        Y        AMDNY

D        Y        AMNDY

## M        N        Y        D        AMNYD

D        N        AMYDN

Y        N        D        AMYND

D        Y        NAMDY

M        Y        D        NAMYD

D        M        NAYDM

Y        M        D        NAYMD

A        M        NDYAM

Y        M        A        NDYMA

M        Y        NDAMY

A        Y        M        NDAYM

A        Y        NDMAY

D        M        Y        A        NDMYA

###### D        A        M        NYDAM

Y        A        M        D        NYAMD

M        D        A        NYMDA

## M        A        Y        D        NMAYD

D        A        NMYDA

N        Y        DAMNY

M        Y        N        DAMYN

A                M        Y        DANMY

N        Y        M        DANYM

N        M        DAYNM

Conclusion

So: 1x2x3x4 / 1x2 x 1x2 = 24 / 4 = 6

Five letter words like vwxyz; this has 1 of each letter (no letters the same)
So: 1x2x3x4 / 1x1x1x1x1x1 = 24 / 1 = 24

A= ni/6

Formula

The formula is easier to show using numbers

Formula for all different number:

a=ni

formula when 2 number are the same

a=ni/2

formula when 3 numbers are the same

a=ni/6

Let’s put them is this way:

 n 1 2 3 x 1 2 6

n represent the number of figures of a number

x represents the divided number in the formula

X is equal the last x*n, so I predict that the formula for 4 same number of a number is:

A= ni/24

4 fig, one arrangement.

a=n/24=(1*2*3*4)/24=1so the formula works

(4)

if a figure has two pairs of three same numbers the formula is (1*2*3*4*5*6*7*8)/1*2*3*4*1*2*3*4=70

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

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)

From this I have worked out the formula to find out the number of different arrangements:

n! = the number of letters in the word

p! = the number of letters the same

a=n!/p!(formula)

In conclusion: I have learnt how to systematically rearrange different numbers and letters in a sequence.

I have also developed my factorial skills.

Natalie Young 10CP

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