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• Level: GCSE
• Subject: Maths
• Word count: 1493

# Emmas dilemma.

Extracts from this document...

Introduction

Emma’s Dilemma

Arrangements for EMMA:

EMMA        EMAM        EAMM        MMAE

MMEA        MEAM        MEMA        MAME

MAEM        AMME        AMEM        AEMM

Arrangements for MIKE

MIKE                 MIEK                MKEI                MKIE

MEIK                MEKI                IKEM                IKME

IEMK                IEKM                IMEK                IMKE

KEMI                KEIM                KIME                KIEM

KMEI                KMIE                EMIK                EMKI

EKMI                EKIM                EIMK                EIKM

If I choose a word where all of the letters are different there will be more combinations.

There were 24 different possibilities in the arrangement of 4 letters that are all different. That is twice as many as EMMA, which has four letters and 2 the same. I have noticed that with MIKE there were 6 possibilities beginning with each different letter. For instance there are 6 arrangements with MIKE beginning with M, and 6 beginning with I and so on. 6 X 4 (the amount of letters) gives 24, the number 6 may have come from 1 x 2 x 3, the number of letters, and multiplied by four because that is how many numbers there are all together.

The difference between how many combinations of the names EMMA and MIKE is the double M in EMMA. There are 24 possible combinations for MIKE and 12 for EMMA.

I will now investigate further words with all letters different.

I will now test a word with 3 letters:

DOG                OGD                GDO

DGO                ODG                GOD

There are 6 possible combinations, 2 for each letter.

Now I will investigate a 2 letter word.

IT        TI

There are two possible arrangements, one for each letter.

Middle

MEIK                YMEKI                YIKEM                YIKME

YIEMK                YIEKM                YIMEK                YIMKE

YKEMI                YKEIM                YKIME                YKIEM

YKMEI                YKMIE                YEMIK                YEMKI

YEKMI                YEKIM                YEIMK                YEIKM

This proves my prediction is correct, as there are 24 arrangements that can be made from a four letter word, when this is multiplied by 5 I have the total number of arrangements made from a 5 letter word.

These is the arrangements for MIKE but with a letter ‘Y’ in front of them, therefore the ‘Y’ can move into five different positions throughout the word, therefore that is why we multiply 24 by 5.

I will now investigate more words with one set of repeated letters:

AA:

AA, only one combination, this is half as many as a normal 2 lettered word.

EGG:

EGG                GEG                GGE

There are 3 possible combinations; this is half as many as a normal 3 lettered word, such as DOG. This is because of the double letter.

I will now put this data into a table so it is easier to see al of the information.

 Number of letters Arrangements when all letters are different Arrangements with one set of repeated letters 1 1 - 2 2 1 3 6 3 4 24 12

From this table I can see that a word with one set of repeats has half as many, or is divided by two, arrangements as a word with the same number of letters and no repeats.

Conclusion

1 x 2 x 3 x 4        or          4!

1 x 2                      2!

A word with 5 letters and 3 repeated letters:

1 x 2 x 3 x 4 x 5          or     5!

1 x 2 x 3                             3!

I now know that a word with:

n letters and 2 repeated letters:

n!

2!

n letters and 3 repeats:

n!

3!

n letters and 4 repeats:

n!

4!

n letters and 5 repeats:

n!

5!

Therefore with n letters and x repeats:

1 x 2 x 3 … n             or     n!

1 x 2 x 3… x                                  x!

Now I will explain what happens when two sets of letters are repeated twice.

AABB - 24 possible combinations

The formula for AABB is 4! / 2! x 2! = 6 possible combinations.

To make it easier instead of using letters as such I will use a’s (any letter) and b’s (any other letter). I will start with AABBB:

AABBB – 120 combinations

This is because the formula is      5!

2!3!

n = the number of letters in the word.

a,b,c,d….x = the number of times a single letter is repeated.

So another version of this formula is

(a + b + c + … x)!

Because of this formula I can improve the formula for my original words:

MIKE changes from 4! to          4!

1!1!1!1!

And EMMA changes from        4!         to           4!

2!                    2!1!1!

Because I must show that there are numbers there, but because 1! equals one it has no effect on the answer to the problem.

I will now finish my investigation with a long word with many repeated letters, MISSISSIPPI:

11!                Because of the number of letters in the word

2!4!4!1!

There are 19958500 combinations

for the word MISSISSIPPI

-  -

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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