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• Level: GCSE
• Subject: Maths
• Word count: 1064

# Emma's Dilemma

Extracts from this document...

Introduction

Chris Huggins                                                                                                        Page  of

EMMA is investigating the amount of different arrangements of letters in her name; she does the same with her friend LUCY.  LUCY has twice as many arrangements as EMMA, they are curious as to why this is and decide to investigate other names and find reasons for their answers.

EMMA –

emma,

eamm,

emam,

aemm,

amme,

amem,

meam,

maem,

mame,

mema,

mmea,

mmea,

LUCY -
lucy,
luyc,
lycu,
lyuc,
lcyu,
lcuy,
ulcy,
ulyc,
uylc,
uycl,
ucly,
ucyl,

cluy,
clyu,
culy,
cuyl,
cyul,
cylu,
yluc,
ylcu,
yulc,
yucl,
yclu,
ycul,

Emma has 12 combinations and Lucy has 24.       ½ of 24 = 12 so Emma has half the amount that Lucy has;  this may be because Emma has 2 letters the same.  I will be investigating whther this happens to other names as well.

I will Investigate –

• The amount of combinations for names with 2-10 letters
• What happens when those names have 2 letters the same
• What happens when they have 2,3,4,5 etc. letters the same
• Whether 3 letters the same means 1/3 of the combinations it would have if no letters were the same
• Whether 4 letters the same means 1/4 of the combinations it would have if no letters were the same (and 5, 6 ,7 etc.)
• Whether there are any patterns or rules to follow when estimating amounts of combinations
• What happens when words have more than 1 letter twice (e.g. LIANNA)

Middle

IYJMM

IYMMJ
IMMYJ
IMMJY
IMJYM

IMYJM
IMYMJ
IMJMY

MJMIY

MJMYI

MJYIM

MJYMI
MJIMY
MJIYM
MIJMY
MIJYM
MIYJM
MIYMJ
MIMYJ
MIMJY
MMJIY
MMJYI
MMIYJ
MMIJY
MMJYI
MMYIJ
MYMJI
MYMIJ
MYJIM
MYJMI
MYIMJ
MYIJM

YJMMI
YJMIM

YJIMM

YMJMI
YMJIM
YMIMJ
YMIJM
YMMJI
YMMIJ
YIMMJ
YIMJM

YIJMM

 Name Letters, letters the same Combinations Jo 2, 0 2 Jim 3, 0 6 Lucy 4, 0 24 James 5, 0 120

Looking at my table of results I can see that a 3 letter word (with none the same) has 6 combinations, 3 x 2

• A 4 letter word (none same) has 24, 4 x 3 x 2
• A  5 letter word (none same) has 120, 5 x 4 x 3 x 2

I estimate that this pattern continues for all names with no letters the same, e.g. I predict that a name with 6 different letters will be 6 x 5 x 4 x 3 x 2 which equals 720. To work out larger numbers you can use the ‘!’ on a scientific calculator, it is called the factorial button and does the process automatically

• If you key in 7! The calculator will perform the sum 7 x 6 x 5 x 4 x 3 x 2, which equals  5040.

The formula for this sum is     n! = y

where n is the number of letters and y is the number of combinations

 Name Letters, letter the same combinations Dd 2, 2 1 Bob 3, 2 3 Emma 4, 2 12 Bobb 4, 3 4 Jimmy 5, 2 60 Jimmm 5, 3 20

Conclusion

there are 10 combinations

I think factorials need to be used again because, with xxxyy, there are 6 combinations beginning with x and 4 with y, because of this I will need the old formula -

n!

a!    =   y

through investigation I have decided that I will need to factorise and multiply the factorials of the amounts of different letters

e.g. if there is 4 a’s and 2b’s (aaaabb)

I will do           6!       (because there are 6 letters in total)

4! X 2!

This is
6 x 5 x 4 x 3 x 2720
(4 x 3 x 2) X (2)          which equals           48       =   15 combinations

This shows that the formula is
n!                                            *

a! x b!

If you include another letter (e,g, xxyyzz) the formaul continues

n!                                                                                *
a! x b! x c!

for mmtookaanooba(2 x m, 3 x a, 4 x o) the formula would be

13!                                                                                *
2! X 3! X 4!              =
21621600

This formula is correct for all words and can be any length as long as you have the factorial of the amount of letters divided by the factorials of the letters which appear more than once multiplied together.

98382.doc

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# Related GCSE Emma's Dilemma essays

1. ## Arrangements for names.

Formula = n!/x!y! For example: A five letter word like aaabb; this has 3 a's and 2 b's (3 x's and 2 y's). So : 1x2x3x4x5 / 1x2x3 x 1x2 = 120 / 12 = 10 !!! A four letter word like aabb; this has 2 a's and 2 b's (2 x's and 2 y's)

2. ## Emma's Dilemma

This means that the total number of arrangements possible is reduced by a factor of 2, or "2 !" ( See later ). For example For 6 letters, all different, there are "6 X 5 X 4 X 3 X 2 X 1"arrangements.

1. ## Emma&amp;amp;#146;s Dilemma

XYXXYYY L! / SL! = A 7! / 3! X 4! = 35 I will now list all the arrangements. XXXYYYY XXYYYYX XYYXYYX XXYYYXY XYYYXYX XYXYYYX XXYYXYY XYYYXXY XYXYYXY XXYXYYY XYYXYXY XYXYXYY XYYYYXX XYYXXYY XYXXYYY These are all the combinations beginning with X.

2. ## Emma's Dilemma.

XXXYZ 2) XXXZY 3) XXYZX 4) XXZYX 5) XYZXX 6) XZYXX 7) ZXXXY 8) ZXXYX 9) ZXYXX 10) ZYXXX 11) YXXXZ 12) YXXZX 13) YXZXX 14) YZXXX 15) XZXYX 16) XZXXY 17) XYXZX 18) XXZXY 19) My prediction was right I shall now compare all my results for words having 3 letters the same with words having 2

1. ## Emma's Dilemma

If we look at EMMA (a 4-letter word) for example, it has the same letter (M) repeated twice, and has 12 different arrangements. LUCY (a 4-letter word with no letters repeated) has 24 arrangements. If we take each M in EMMA to be a different letter, by changing the colour of one of the Ms, we would get 24 arrangements.

2. ## Emma's Dilemma Question One: Investigate the number of different arrangements of the letters

This formula is shown below: No. of = No. of letters ! combinations ( no. of times a letter X ( no. of times a letter has been repeated ) ! different has been repeated ) ! As you can see, the factorial symbol has been added to the ends of each part of this equation.

1. ## I am doing an investigation into words and their number of combinations. I will ...

Nth term 4 letters 1 24 3! /1x2x3=1 5 letters 5 120 4! /1x2x3=4 6 letters 30 720 5! /1x2x3=20 n letters n n!(4! I believe the formula is n!/ 4!. I will now test this and predict the combination for 7 letters.

2. ## Emma&amp;amp;#146;s Dilemma.

Part Two: Moving onto the second task on the list, I have decided to adopt the method used in the first part in order to find all the possible combinations of the letters in the name 'Lucy' - i.e. by numbering the letters to avoid errors.

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