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# Emma&amp;#146;s Dilemma.

Extracts from this document...

Introduction

Emma’s Dilemma

Part 1

Introduction

In this investigation, I am going to find out the different arrangements of the letters of the name LUCY.

Here are the possibilities for the name LUCY

LUCY UYCL YCLU CULY

LCYU UCLY YLCU CYLU

LYCU ULUC YULC CLYC

LCUY ULUY YLUC CUYL

LUYC UCLY YCUL CYUL

LYUC UYLC YULC CLUY

First of all I took the letter L from the name LUCY then wrote the second letter being U and after the two other possibilities being C and Y. Then after I used the same letter L, wrote the second letter down being C and wrote the two possibilities again being U and Y. I took another letter which I used that to be the second letter, keeping the first letter the same in till I have found the rest of the possibilities in the letter.

Middle

I am now going to try to find a pattern in the possibilities of combinations of different letters.

a) Firstly with one letter.

A

There is one permutation with one letter.

b) With two letters.

AB BA

There are two possibilities with two different letters.

c) With three letters.

ABC BAC CBA

ACB BCA CAB

There are six permutations with three different letters.

d) With four letters.

ABCD BACD CABD DABC

Conclusion

pan>

1

2

6

24

?

1! = 1 x 1 =1

2! = 1 x 2 =2

3! = 1 x 2 x 3 =6

4! =1 x 2 x 3 x 4 =24

From this table I predict that with five letters there will be 120 permutations.

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

e) With five letters.

ABCDE BACDE CABDE DABCE EABCD

ABCED BACED CABED DABEC EABDC

ACDEB BCDEA CBDEA DBCEA EBCDA

ACEDB BCEDA CBEDA DBECA EBDCA

AEBCD BEACD CEABD DEABC EDABC

AECDB BECDA CEBDA DEBCA EDBCA

AEDBC BEDAC CEDAB DECAB EDCAB

AEDCB BEDCA CEDBA DECBA EDCBA

Ali Khizar Set 3

Maths Coursework

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