Lucy's Dilemma

I realised the formula for this was:

n!

This is because, e.g. for 5, there are 5 possibilities for the first column, 4 for the second, 3 for the third, 2 for the forth, and 1 for the fifth.

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        Then Lucy played around with her friend, Emma’s, name. She writes down the different combinations:

1. EMMA
2. EMAM
3. EAMM
4. MEMA
5. MEAM
6. MMEA
7. MMAE
8. MAEM
9. MAME
10. AEMM
11. AMEM
12. AMME

I notice that there is only half the amount of combinations that there are for Lucy even though they have the same number of letters.

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!

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        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

I noticed that this is 5! divided by 2! x 3!. I also did AABBCC and that came out as 6!divided by 2! x 2! x 2!. Therefore I found the equation:

          n!           .

s1!s2!s3!(etc.)

This equation works on any word.

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.

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        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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