# Emma&#146;s Dilemma.

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Introduction

Emma’s Dilemma

Question 1

There are 12 ways of writing EMMA, they are:

- EMMA
- EMAM
- EAMM
- MEMA
- MMEA
- MAEM
- MMAE
- MEAM
- MAME
- AMEM
- AMME
- AEMM

It takes a long time to think of and write all of the different combinations EMMA but there is a way of shortening the task. If you look at all of the different combinations of EMMA all starting with E then you will find that there are only 3.

- EMMA
- EMAM
- EAMM

When you have done this you can then multiply 3 by the number of letters in the word, in this case 4. So the sum is 3 x 4= 12.

## Question 2

I will now use the same rule for the name LUCY to see if this rule works constantly.

All of the combinations of LUCY beginning with L

- LUCY
- LCUY
- LCYU
- LUYC
- LYUC
- LYCU

As we can see there are 6 ways of writing LUCY beginning with L so we times this by the number of letters in the word and we get the sum 6 X 4 = 24.

- LUCY
- LCUY
- LCYU
- LUYC
- LYUC
- LYCU
- CULY
- CLUY
- CYLU
- CYUL
- CUYL
- CLYU
- UCLY
- ULCY
- UYCL
- UYLC
- ULYC
- UCYL
- YCLU
- YCUL
- YLUC
- YLCU
- YULC
- YUCL

The reason that there are more ways of writing LUCY than EMMA although they both have the same amount of letters is that EMMA has two letters that are the same.

Middle

12 (number of ways of writing PIPPA beginning with P) X 5 (number of letters in the word)

3 (number of letters the same in the word).

## OR MORE SIMPLY:

12 X 5

As you can see there are, indeed 20 ways of writing PIPPA.

- PPPIA
- PPPAI
- PPAIP
- PPIAP
- PPIPA
- PPAPI
- PAIPP
- PIAPP
- PAPPI
- PIPPA
- PIPAP
- PAPIP
- IPPAP
- IAPPP
- IPAPP
- IPPPA
- AIPPP
- APIPP
- APPIP
- APPPI

I am now going to look at a number of different words all with letters repeated. I predict that there will be 10 ways of writing XXXYY. I know this because there are six ways of writing it with X at the beginning so the equation that I get is 6 x 5 = 10.

3

- XXXYY 1. YYXXX
- XYYXX 2. YXYXX
- XXYYX 3. YXXYX
- XYXXY 4. YXXXY
- XYXYX
- XXYXY

3 letter word with 2 letters repeated.

- XXY
- XYX
- YXX

4 letter word with 3 repeated.

- XXXY
- XXYX
- XYXX
- YXXX

5 letter word with 4 repeated.

- XXXXY
- XXXYX
- XXYXX
- XYXXX
- YXXXX

6 letter word with 5 repeated.

- XXXXXY
- XXXXYX
- XXXYXX
- XXYXXX
- XYXXXX
- YXXXXX

So once we have worked out how many ways there are of writing a word all beginning with a certain letter then we can work out the total number of combinations of that word even if it has letters repeated.

I have

Conclusion

No of letters (n) | No letters repeated | Same letter repeated 2x (r) | Same letter repeated 3x (r) | Same letter repeated 4x (r) | Same letter repeated 5x (r) |

3 | 6 | 3 | 1 | 0 | 0 |

4 | 24 | 12 | 4 | 1 | 0 |

5 | 120 | 60 | 20 | 5 | 1 |

6 | 720 | 360 | 120 | 30 | 6 |

7 | 5040 | 2520 | 840 | 210 | 42 |

Half of 5040 A Third 2520 A Quarter of 840

We can simply see from this table that each time that you add a letter, to be repeated, then the number of possible combinations of that word is reduced by a factor of firstly 2 then by 3 and then by 5.

All of the combinations that are in Bold Italic I have worked out using my rule.

I will now try to prove my rule.

Lets take the combination of letters XXYYZZ, I will now predict and prove both of my rules with this combination.

First of all:

- XXYYZZ
- XXZZYY
- XXYZYZ
- XZYZXY
- XZXYYZ
- XZYYZX
- XZYZYX
- XZZXYY X 6 = 90
- XZZYXY
- XYZZYX
- XYZYXZ
- XYXZYZ
- XYZZYX
- XYZYZX
- XYZZXY

Then second:

6! (number of letters in the word)

2! X 2! X 2! (number of letters the same)

=

720

2 X 2 X 2

=

720

8

=

90

Well both of the rules seem to work so, my final equation for my second rule turn out to be

N!

R! X R!

N = number of letters in the word.

R = number of letters repeated in the word.

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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