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Lucy's Dilemma

Extracts from this document...

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.

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:

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

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:

n!.

s1!  x   s2!

9!      .

4!   x   5!

=  362880  .

2880

=  126

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