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Lucy's Dilemma

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Introduction

        Lucy was playing with her name. She sees the there are a number of different combinations. These are:

  1. LUCY
  2. LUYC
  3. LCUY
  4. LCYU
  5. LYUC
  6. LYCU
  7. ULCY
  8. ULYC
  9. UCLY
  10. UCYL
  11. UYLC
  12. UYCL
  13. CLUY
  14. CLYU
  15. CULY
  16. CUYL
  17. CYLU
  18. CYUL
  19. YLUC
  20. YLCU
  21. YULC
  22. YUCL
  23. YCLU
  24. YCUL

I noticed that the amount of combinations was 4 factorial (i.e. 4x3x2x1), written 4!.

I tried with other names and found there to be a pattern. I used Ben, Henry, and Thomas. The numbers were: For Ben, 6 or 3!; For Henry, 120 or 5!; and for Thomas, 720 or 6!.

I realised the formula for this was:

n!

This is because, e.g.

...read more.

Middle

Then I tried Sally and that also had half of the factorial so I get the equation:

n!

2

I then tried Bobby and wrote down the different combinations:

  1. BOBBY
  2. BOBYB
  3. BOYBB
  4. BBOBY
  5. BBOYB
  6. BBBOY
  7. BBBYO
  8. BBYOB
  9. BBYBO
  10. BYOBB
  11. BYBOB
  12. BYBBO
  13. OBBBY
  14. OBBYB
  15. OBYBB
  16. OYBBB
  17. YBOBB
  18. YBBOB
  19. YBBBO
  20. YOBBB

I noticed that this is 5! divided by 3!. I, therefore get the equation, when s=number of letters the same:

n!

s!


I then tried with more than one than one combination of more than one letter. Instead of thinking of names I will use letters. I will do AABBB:image01.png

  1. AABBB
  2. ABABB
  3. ABBAB
  4. ABBBA
  5. BAABB
  6. BABAB
  7. BABBA
  8. BBAAB
  9. BBABA
  10. BBBAA

...read more.

Conclusion

This is because it is the factorial but the first letter there is n-s-1, in the second there is a majority of n-s-2, and in the third there is a majority of n-s-3. This keeps on going until the penultimate number, which, along with the last, stays the same as normal.


Then we were asked to find out how many possibilities there are for a combination of 4 Xs and 5 Ys. I used the formula to figure this out in this way:image02.png

       n!.

s1!  x   s2!

      9!      .

4!   x   5!

=  362880  .

     2880

=  126


image07.pngimage05.pngimage06.pngimage03.pngimage04.png

...read more.

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