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Emma’s Dilemma

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Introduction

Emma’s Dilemma

In this investigation I will try and work out how many combinations I can make out of any type of word.

4 letters

The name EMMA has 4 letters, with 2 letters the same.  I will experiment how many combinations I can make with this name.

I find it easier to try and make each combination using one letter, then another etc.

EMMA        MEMA     MEAM    AMME

EMAM        MMEA     MAEM    AMEM

EAMM        MMAE     AEMM

The name LUCY has 4 letters, which are all different.

LUCY    LYUC    ULCY    CULY    CUYL    YLUC

LCUY    UCLY    UYLC    CYUL    YCLU    YULC

LYUC    UCYL    UYCL    CLYU    YLCU    YUCL

LCYU    ULCY    CYLU    CLYU    YCUL    LUYC

The name SASS has 4 letters, 3 of which are the same.

SASS    SSSA

SSAS    ASSS

The name AAAA has 4 letters, which are all the same.

AAAA has no variations.

3 letters

The name LOU has three letters, all different.

LOU    ULO    OUL

...read more.

Middle

image04.png I shall multiply 5 X 24 to get the combinations for 5 letters. I will then half the answer I get from this to find out the combinations for 5 letters with 2 the same. I’ll divide this answer to find the combination for a 5 letter word with 3 the same, and so on.

No. of letters

All different

2 the same

3 the same

4 the same

5 the same

2

2

1

3

6

3

1

4

24

12

4

1

5

5X24= 120

120image03.png 2=60

60 image03.png 3=20

20 image03.png 4=5

1

From these results I have now found out how this works. The number of combinations is narrowed down each time there is more that 1 letter the same, or 1 letter is chosen to begin.

This is what I mean:

E.g. 2 letter word

2 choices for letter, multiplied by one choice for letter = 2

3 letter word

1

...read more.

Conclusion

E.g. 10 ! = 10 X 9 X 8 X 7 X 6 X 5 X 4 X 3 X 2 X 1

        10!  = 3, 628, 800

I can therefore say that a 10 letter word with all of the letters different could make 1, 628, 800 combinations of itself.

I have come up for this equation for finding the combination of an n letter word, with all letters different.

  N! = n x (n - 1) x (n - 2) x (n - 3) x (n - 4) x (n - 5) x (n - 6) x (n - 7) … etc.

If 2 letters were the same, the equation would be:

n! image03.png2

And if 3 letters were the same, it would be:

n! image03.png 2 image03.png 3                                                                                                         …etc.

This concludes my coursework, as I have now found how many combinations you can get from an n letter word, even when the word has 2 letters the same, 3 letters the same, 4, or 5 …etc.

- Louise Manly

...read more.

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