Emma's dilemma.

So the formula is A = L!

This means that whatever L is, it is multiplied by every number smaller than itself appart from 0 (zero). So another example would be if L = 10, then A = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 = 3628800. Although the number 1 in the equation, does not make a difference whether it is in or out of the equation, it is still added in, because it is in the formula, L!.

If there were 3 letters, which make 6 arrangments and I were to add 1 more letter, then there would be 4 letters. The reason why we multiply by 4, is because the 3 letters can still have six different arrangments with the same fourth letter starting. E.g LUCY

LUYC

LCUY

LCYU

LYCU

LYUC  

All of these arrangments start with the same letter ‘L’. so not including the letter L, there would be 6 arrangments 1 x 2 x 3; but as there is the L making 4 letters, you can start with the same letter and end with 6 arrangments; but as there is 4 letters, this mean you can having 4 different starting letters which all will each form 6 arrangments, so therefore if L = 4, then the number of arrangments = 4 x 3 x 2 x 1 =  4 x 6 = 24

The number of arrangments for words with the number of letters from 1 – 5, and 1 pair of letters so there is 2 same.

1. JJ        Number of arrangments = 1

1. MUM
2. MMU
3. UMM                Number of arrangments = 3

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

Number of arrangments         = 24 + 36

        = 60

The number of arrangments for words with the number of letters from 1 – 5, and 1  3 letters the same, so 3 same.

1. JJJ                Number of arrangments = 1

1. NUNN
2. NNUN
3. NNNU
4. UNNN        Number of arrangments = 4

The number of arrangments for words with the number of letters from 1 – 5, and 1 pair of letters so there is 2 same.

1. JJJJ                Number of arrangments = 1

1. ASSSS
2. SASSS
3. SSASS
4. SSSAS
5. SSSSA        Number of arrangments = 5

Table to show all of the number of arrangments with different number of letters, and different numbers of letters the same.

From the table above I have seen a pattern and have worked out the formula to answer all results. Say 5! = 120 as seen from the first formula; to find out how many different arrangments in a 5 letter word with 3 letters the same, it would be 5! / 6 = 20. I got the 6 from 1 x 2 x 3 because there is 3 the same. So the new formula for the whole table would be L! / S!.

Here is a list of formula for the different number of letters the same:

2 letters the same =   .     L!    .

                                 (2x1 = 2)    

3 letters the same =            L!      .

                                  ( 3x2x1 = 6)

4 letters the same =                     L!        

                                   (4x3x2x1 = 24)

5 letters the same =                  L!            .

                                  (5x4x3x2x1 = 120)

6 letters the same =                  L!    _        

                                   (6x5x4x3x2x1 = 720)

Formula = L! / S!   or   L!

                 S!        

To prove this formula again, I will answer the 6 letter words.

If L = 6 and S = 1, then A = 6 x 5 x 4 x 3 x 2 x 1 / 1 = 720 / 1 = 720

If L = 6 and S = 2, then A = 6 x 5 x 4 x 3 x 2 x 1 / 1 x 2 = 720 / 2 = 360

If L = 6 and S = 3, then A = 6 x 5 x 4 x 3 x 2 x 1 / 1 x 2 x 3 = 720 / 6 = 120

If L = 6 and S = 4, then A = 6 x 5 x 4 x 3 x 2 x 1 / 1 x 2 x 3 x 4 = 720 / 24 = 30

Table to show results of the

Now I am going to investigate the number of different arrangement for words with 2 or more letters the same like, aabb, aaabb, or bbbaaa.

This is a 4 letter word with 2 letters the same, there are 6 different arrangements:

xxyy

 

I am going to use the letters x and y (any letter)

 

xxyy        

xyyx        

xyxy

yxxy

yxyx

yyxx  Number Of Arrangments = 5

This is a 5 letter word

 

xxxyy  

xxyxy  

xxyxx  

xyxyx  

xyxxy

xyyxx  

yyxxx 

yxxxy 

yxyxx  

yxxyx            Number Of Arrangments = 10

 

In the above example there are 3 x's and 2 y's

As each letter has its own number of arrangements i.e. there were 6 beginning with x, and 4 beginning with y, I think that factorial has to be used again.

 

As before, the original formula:

 

 L! = the number of letters in the word

 S! = the number of letters the same

 

 

From this I have come up with a new formula  The number of total letters factorial, divided by the number of x's, y's etc factorised and multiplied.

 

For the above example:

A four letter word like aabb; this has 2 a's and 2 b's (2 x's and 2 y's)

So : 1x2x3x4 _        24

       (1x2) x (1x2)         4     = 6 different arrangements

 

A five letter word like aaaab; this has 4 a's and 1 b (4 x's and 1 y)

So: 1x2x3x4x5

      (1x2x3x4) x (1)

           = 120

               24     = 5 different arrangements

 

A five letter words like abcde; this has 1 of each letter (no letters the same)

So : 1x2x3x4

   (1x1x1x1x1x1)

          = 24                      = 24 different arrangements

 

 

A five letter word like aaabb; this has 3 a's and 2 b's (3 x's and 2 y's).

So : 1x2x3x4x5 

       (1x2x3) x (1x2)

= 120

    12    = 10 different arrangements

 

 

This shows that my formula works:

                       

          L!   = the number of letters in the word

       X!Y! = the number of repeated letters the same