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Investigate all the possible combinations there are of a person's name, using only the letters in their name.

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Introduction

Emma’s Dilemma

Introduction

This piece of course work was set to investigate all the possible combinations there are of a person’s name, using only the letters in their name.

For example Joe:

The combinations are

Joe

Jeo

Eoj

Ejo

Oej

Oje

Part 1

Lucy

In this section, I will be exploring all the combinations of the name Lucy and if there is any correlation between the number of combinations and the number of letters in her name.

Possible Combinations:

  1. lucy
  2. luyc
  3. lcuy
  4. lcyu
  5. lyuc
  6. lycu
  7. ulcy
  8. ulyc
  9. uylc
  10. uycl
  11. ucly
  12. ucyl
  13. cluy
  14. clyu
  15. culy
  16. cuyl
  17. cylu
  18. cyul
  19. yluc
  20. ylcu
  21. yucl
  22. yulc
  23. yclu
  24. ycul

image00.png

Method

Firstly you choose one letters to start with, then change the second letter, for the example of Lucy we would have

Lucy

Lucy ← The U stays the same but the last two letters change, this pattern is repeated.

From

...read more.

Middle

In this section I will be exploring the possible combinations of a name that includes a double letter.

Possible Combinations:

  1. emma
  2. emam
  3. eamm
  4. meam
  5. maem
  6. mame
  7. mema
  8. mmea
  9. mmae
  10. amme
  11. amem
  12. aemm

From this we can see that there are 12 possible combinations of the name Emma. This could mean that to find the number of combinations in a name with repeated letters, you take the number of cominations in that amount of letters without repeated letters and divide it by to.

For example

24/2 = 12

As Lucy had 24 combination and when adding a double letter the number of combinations were 12.

I will investigate into this further in Part 3

Part 3 – Further Research

Rohan

I know from investigation that Rohan has 120 combinations of his name, and has 5 letters in his name.

There for it is 24 x 5.

...read more.

Conclusion

The formula for a name with repeated letters is

X!/R! where R= the amount a letter is repeated.

Using this, the word ABCBDB would be

6!/3!

If a name has 2 sets of repeated letters then you simply divide it by the amount of repeated letters. So a name like Hannah would be

6!/2!2! as there are 2 A’s and N’s.

This is true of any word. For example ABACABCBACABA would be

14!/6!4!3!

Also from using ! we can easily work out all the combinations of a name without having to write them all out:

Number of letters in the word

Amount of possible permutations

1

1

2

2

3

6

4

24

5

120

6

720

7

5040

8

40320

9

362880

To test this table we can take 9!

Which is 1x2x 3x4x5x6x7x8x9 which is 362880

Tom Edwards 10CHMY – Emmas Dilemma      Page  of

...read more.

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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