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• Level: GCSE
• Subject: Maths
• Word count: 1298

# Emma&amp;#146;s Dilemma

Extracts from this document...

Introduction

By Thomas Hackford

Emma’s Dilemma

Emma is playing with arrangements of the letter of her name. One arrangement is

E M M A

Another arrangement can be:

## E M AM

In this investigation I am going to investigate the different number of letters in a word, and see if there are any patterns.

The first two names I will investigate are EMMA and LUCY.

## EMMA                EMAM                EAMM

AEMM                AMEM                AMME

MMEA                MMAE                MEAM

MAEM                MEMA                MAME

There are 12 different arrangements for the name EMMA.

LUCY                LUYC                LYUC

LYCU                LCYU                LCUY

ULCY                ULYC                UYCL

UYLC                UCLY                UCYL

CULY                CUYL                CYLU

CYUL                CLUY                CLYU

YCUL                YCLU                YLUC

YLCU                YUCL                YULC

There are 24 different arrangements for the name LUCY.

Despite both names having the same amount of letters, there are a different number of arrangements. This is because in EMMA there are 3 different letters, and in LUCY there are 4 different letters.

First I will investigate words in which all the letters are different.

A 1 letter word, such as A only has 1 arrangement:

## A

A 2 letter name such as JO has 2 arrangements:

## A 3 letter name such as TOM has 6 arrangements:

Middle

N!

On my scientific calculator, I can factorise numbers by first putting in the number I want to factorise (N), and then I press:

I will now make a few predictions. If your name is CHRIS, which has 5 different letters, how many arrangements will there be? I will do the same for THOMAS (6 letters), PAULINE (7 letters), REGINALD (8 letters) and GERALDINE (9 letters).

CHRIS: 5! = 1 x 2 x 3 x 4 x 5 = 120arrangements

## THOMAS: 6! = 1 x 2 x 3 x 4 x 5 x 6 = 720 arrangements

PAULINE: 7! = 1 x 2 x 3 x 4 x 5 x 6 x 7 = 5040 arrangements

REGINALD: 8! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 = 40320 arrangements

GERALDINE: 9! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 = 363880 arrangements

This formula makes sence. If you have LUCY:

LUCY                LUYC                LYUC

LYCU                LCYU                LCUY

ULCY                ULYC                UYCL

UYLC                UCLY                UCYL

CULY                CUYL                CYLU

CYUL                CLUY                CLYU

YCUL                YCLU                YLUC

YLCU                YUCL                YULC

And you add a new letter such as A, you get:

ALUCY                ALUYC                ALYUC

ALYCU                ALCYU                ALCUY

AULCY                AULYC                AUYCL

AUYLC                AUCLY                AUCYL

ACULY                ACUYL                ACYLU

ACYUL                ACLUY                ACLYU

AYCUL                AYCLU                AYLUC

AYLCU                AYUCL                AYULC

Because you add another letter, in this case, A, you will have to times this by 5.

Conclusion

24

2 x 1 x 1 = 12 is correct!

24

2 x 2 = 6 is correct!

In this case, the formula is:

N! / (A! x B! x C! x X!) = number of arrangements

WithN! being the total number of letters,

A! being the number of times A was used in the word

B! being the number of times B was used in the word

X! being the number of times X was used in the word

To make absolutely sure this formula is correct, I am  going to test it one more time on a 5 letter word such as AAAAB:

AAAAB        AAABA        AABAA        ABAAA        BAAAA

In this word N is 5, A is 4 and B is 1

5!                             5 x 4 x 3 x 2 x 1                          120

4! x 1! =                   (4 x 3 x 2 x 1) (1)  =                     24 =        5

Now that I have a formula, I am going to made a prediction. I am going to work out, using my formula how many arrangements there would be for the following sequence of letters:

AAAAAAABBBCCD

A = 7                B = 3                C = 2                 D = 1

N = 13

Because this is a difficult calculation to make, I will do it by hand:

My prediction is that for the word:

AAAAAAABBBCCD

There will be                                                                       arrangements

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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