Emma's Dilemma.

2 letters

No different

Arrangements of jj

1) jj

There is only one combination of jj as they look the same either way you put as explained at the start of experiment.

3 letters

2 different

Arrangements of lee

1) lee

2) ele

3) eel

There are 3 combinations of lee and now I can put my results up to a four lettered word, into a table to see if there is any trends.

Number of Letters
All Different
1 Letter Repeated
1
1
-
2
2
1
3
6
3
4
24
12



5
120
60
6
720
360
7
5,040
2,520
8
40,320
20,160

I am predicting the 5, 6, 7 and 8 lettered names through my trends in my first 4 results. You can see that the number of all different is times by the next amount of letters to make the next amount of all different. E.g. 1 (all different for 1 lettered name)  x 2 (next name length) = 2 (all different for a 2 lettered name.)

I know know my first four lettered results so I will use this to quickhand any more names. E.g If I was looking at a 5 letter word, such as James, I could just bracket of the first letter to make a 4 lettered word, j(ames), then I will know the amount of combinations for a 4 lettered word with 5 different letters. Then I will do this with each letter of the name and add up the combinations to get the result.

What if there was a five-letter word? How many different arrangements would there be for that?

As I have found that there were 24 arrangements for a 4 letter word with all different letters and that there were 6 ways beginning with one of the letter I predict that there will be 120 arrangements for lucyq, 24 for a, 24 for b, 24 for c; etc. 120 divided by 6 (number of letters) equals 24. Previously in the 4-letter word, 24 divided by 4 equals 6, the number of possibilities there were for each letter. For the sake of convenience I have used lucy and put a q in front of it to show that there are 24 different possibilities with each letter of a 5 lettered name being all different.

5 letters

All Different

Arrangements for lucyq:

l(ucyq) = 24

u(lcyq) = 24

c(luyq) = 24

y(lucq) = 24

q(lucy) = 24

This word has 120 combinations.  I can see that the numbers of possibilities for different arrangements are going to dramatically increase as more different letters are used.

So as a general formula for names with n number of letters all different I have come up with a formula. With Lucy's name; 1x2x3x4 = 24. With qlucy; 1x2x3x4x5 = 120. However this is expressed as factorial. If I key in (lets say the number of letters all different) factorial 6 it gives me 720, with makes sense because 720 divided by 6 equals 120 which was the number of arrangements for a 5 letter word and it continues to fall in that pattern. This formula is expressed as n!. (n=number of letters). This now proves that my prediction was correct and lets me investigate further.

Now I have got results for words with all different letters and a four lettered word with a repeated letter (emma), I now look at my table again to see if I can figure out a pattern for repeted letters words.

Now I can predict from the table that a word like gemma might have 60 combinations. Gemma is a 5 lettered word with 1 repeated letter, making the amount of a 5 lettered word with all different letters be divided by 2 to get 60. This now also prooves my prediction was correct with 1 repeated letter. There could now be a formula to work this out and this could look like n!/2.

Another way of maybe finding the combinations for Gemma is to look at emma. Emma has 12 combinations and as we are looking for a 5 lettered word with 1 letter repeated, we can times n (the number of letters, in this case 5) by the result of emma to achieve the amount of combinations. 5 x 12 = 60. Again we get the same answer. This can be expressed as 4! (Four lettered word)/2 (one letter repeated) x 5 (number of letters = 60 (combinations for gemma)

Arrangements for gemma

G(emma) = 12

E(gmma) = 12

M(gema) = 24

A(gemm) = 12

Total combinations = 60.

I now know that my prediction of the formula for Gemma was true and I just need to proove that it works for all the 1 letter repeated words.

Number of Letters
All Different
1 Letter Repeated
1
1
-
2
2
1
3
6
3
4
24
12
5
120
60
6
720
360
7
5,040
2,520
8
40,320
20,160
n
n!
n!/2

After looking at my table I can see a link between all different and 1 letter repeated words. You can see that on every number of lettered word that the 1 letter repeated is oine less than the all different column. This then means that if you can find the combinations for a certain number of lettered word than you can find out how many combinations there is for the same amount of lettered word but with a repeated letter. The all different column is n! (which is the number of letters !). This then gives us the formula for 1 letter repeated; which is n!/2.

I now notice that the 2 relates to something. If I look at the word there is 2 letters the same, so this shows that I should divide n! by 2 to get the amount of combinations.

This shows you that factorial works and relates with the other combinations.

Total Letters (all different) Number of Arrangements

1-1

2-2

3-6

4-24

5-120

6-720

6 x 120 (5th answer)

       5 x 24 (4th answer)

                    4 x 6 (3rd answer)

                            3 x 2 (2nd answer)

                                    2 x 1 (1st answer)

I will now be finding out the combinations for words that have more repeated letters, hopefully finding a successful formula to find out these combinations without using a long process.

What if there was 3 letters with 3 the same

Arrangements of YYY

1) yyy

There is only one arrangement.

What if there was 4 letters with 3 the same

Arrangements of aaab

1) aaab

2) aaba

3) abaa

4) baaa

There are 4 combinatons of aaab.

What if there was 4 letters with 4 the same

Arrangements of aaaa

1) aaaa

There is one combination

What if there was 5 letters with 3 the same

Arrangements of aaabc

a(aabc)

1) Look at 4 lettered word with 2 same(emma) = 12

 

b(aaac)

2) Look at 4 lettered word with 3 same(aaab) = 4

c(aaab)

3) Look at aaab again = 4

12+4+4=20 combinations

What if there was 5 letters with 4 the same

Arrangements of aaaab

a(aaab)

1) Look at aaab = 4

b(aaaa)

2) Look at aaaa = 1

4 + 1 = 5 combinations

What if there was 5 letters with 5 the same

Arrangements of aaaaa

aaaaa

1) Look at aaaa = 1

 = 1 combination

I will now put all my results, for up to 5 lettered words, in a table. I will then try to draw conclusions from it with a true formula, to see if it backs the one I made earlier.

Number of letters
All Different
2 Same
3 Same
4 Same
1
1
-
-
-
2
2
1
-
-
3
6
3
1
-
4
24
12
4
1
5
120
60
20
5
Formula:
n!
n! / 2!
n! / 3!
n! / 4!

This shows that if you have some words with some of the same letters in you have to use the amount of all different factorial divided by the amount of the letter what are the same, factorial.

The All different column is divided by two to make the 2nd column and then the 2nd column is divided by 3rd to make the 3rd column and then this is divided by 4th to make the 4th column and so on.... But however I am trying to put this into a more mathematical formula.

I am going to try to find a way to find the amount of combinations for a word with 3 of the same letters without having to find out the combinations of the same amount of lettered word with 2 the same letters. If I use a five lettered word in this example I can see the the 3 same column is 6 times smaller than the all different column. So if I now look on my table I notice the number of letters column equalling 3 is linked up with the all different column which is 6. I now notice that the number of all different is divided by 3 to make a 3 the same lettered word. So now I think that if the number of all different is made by n! The amount of letters! Must go together with it. To make sure sure this point is true I calculate that 3! (3x2x1=6) equals 6 so now I have come to a prooven true formula.

Now to extend my experiment I will try to find out a formula to find out the combinations of a word with more than 1 letter repeated. My aim will be to work out the amount of combinations in mississippi.

I will now try to investigate how to find out the amount of combination with a word that has more repeated letters. E.g. aabb

Arrangements for aabb:

This is a 4-letter word with 2 repeated letters. I have done this with;

a(abb) = 3

b(aab) =3

There are 6 arrangements. This is half of emma which has one repeated letter.  

Now I will add another letter of the same to help me find a formula when there is more than 1 letter repeated.

Arrangements for aaabb

A(abbb) = 4

B(aabb) = 6

There are 10 different arrangements for this instance. Now if I look back at my last table I can see the the number of combinations for a 5 lettered word is 120. So now if I try to relate this to ten I might be able to find another formula. 120/12=10. This now allerts my attention to attempt to relate it to my table. If I look up in my table I see 2 and 6 above one and other and again this links to my formula. The use of numbers in the table have come up again.

What if I had an arrangement of aaaab?

1) aaaab

2) aaaba

3) aabaa

4) abaaa

5) baaaa

There are 5 different arrangements for this instance.

If I go back to aaabb; there are 3 a's and 2 b's in a total of 5 unknowns. As each letter has its own number of arrangements i.e. there were 5 beginning with a, and 5 beginning with b, I think that factorial has to be used again. Also in a 5 letter word there are 120 arrangements and 24 arrangements (120 divided by 5) for each letter. As there is a divide issue involved I had a go at trying to work out a logical universal formula. I came up with; The number of total letters factorial, divided by the number of a's, b's ect factorised and multiplied. I found this formula because if had letters the same I had to divide them by how many letters were the same factorial, so I believed this would still work if I just added another division if there was more than 1 repeated letter.

Formula =

n!/x!y!

I have come to this formula by looking at my result tables from previous investigations. I worked out earlier that n! Is how to work out how many combinations there are in that certain amount of lettered word and now have found out how to work out words with repeated letters and some more than others. E.g. If there is a 4 of one letter in a word it would be 4! And if there was 3 of another it would be 3!. You then times these two together and they are n! is divided by the sum of them.  

For example:

A five letter word like aaabb; this has 3 a's and 2 b's

So : 1x2x3x4x5 / 1x2x3 x 1x2 = 120 / 12 = 10

A four letter word like aabb; this has 2 a's and 2 b's

So : 1x2x3x4 / 1x2 x 1x2 = 24 / 4 = 6

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

So: 1x2x3x4x5 / 1x2x3x4 x 1 = 120 / 24 = 5

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

So : 1x2x3x4 / 1x1x1x1x1x1 = 24 / 1 = 24

All these have been proved in previous arrangements. This shows that my formula works!

Mississippi can be shown as:

11x10x9x8x7x6x5x4x3x2x1 / 4x3x2x1(s) x 4x3x2x1(i) x 2x1(p) x 1(m)

=

39916800 / 1152 = 34650

This can be expressed as:

n!/s! i! p! m!