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### Arrangements for Emma

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Arrangements for Emma

emma emam eamm mmae
mmea meam mema mame
maem amme amem aemm

There are 12 possibilities; note that there are 4 total letters and 3
different.

What if they were all different like Lucy?
Arrangements for Lucy:

lucy ucyl cylu ycul
luyc ucly cyul yclu
lcuy ulcy culy yulc
lcyu ulyc cuyl yucl
lyuc uycl clyu ylcu
lycu uylc cluy yluc

There are 24 different possibilities in this arrangement of 4 letters
that are all different. That's twice as many as EMMA, which has 4
letters and 3 different. I have noticed that with Lucy there are 6
possibilities beginning with each different letter. For instance there
are 6 arrangements with Lucy beginning with L, and 6 beginning with u
and so on. 6 X 4 (the amount of letters) gives 24.

What if there were 4 letters with 2 different?
Arrangements for anna:

anna anan aann
nana naan nnaa

There are 6 arrangements for aabb. From 2 different letters to all
different letters in a 4-letter word I have found a pattern of 6, 12
and 24. Maybe it would be easier to see what is happening if I used
larger words.

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 combinations
beginning with each of the letters in the word I predict that there
will be 120 arrangements for qlucy, 24 for q, 24 for l, 24 for u; and
so on. 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.

qlucy qucyl qcylu qycul
qluyc qucly qcyul qyclu
qlcuy qulcy qculy qyulc
qlcyu qulyc qcuyl qyucl
qlyuc quycl qclyu qylcu
qlycu quylc qcluy qyluc

I can see that the numbers of different arrangements are going to
dramatically increase as more different letters are used. So as a
general formula for names with x number of letters all different I
have come up with a formula. With Lucy's name; 1x2x3x4 = 24. With
qlucy; 1x2x3x4x5 = 120. This is called a factorial. There is a button
on most scientific calculators which basically does this formula for
you according to the number of letters in the word. If I key in
(assuming all the letters are different) factorial 6, I get it gives
me 720, which makes sense because 720 divided by 6 equals 120 which
was the number of arrangements for a 5 letter word, and it continues

Total Letters (all different) Number of Arrangements
1-1
2-2
3-6
4-24
5-120
6-720
So now that I've explained the pattern of general x lettered words,
what do I do if any letters are repeated? Like in Emma; it has 4
letters but 2 of which are the same. 4 factorial equals 24, but I
could only find 12, which means that there must be more to it that
just factorial in that way.

To make it a bit easier instead of using letters as such I will use
a's (any letter) and b's (any other letter). I will start with aabb:

Arrangements for aabb:

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

aabb abab baab
aaba baba bbaa

There are 6 arrangements. What if I had aaabb?

aaabb aabab aabba ababa abaab
abbaa bbaaa baaab babaa baaba

There are 10 different arrangements for this sequence.

What if I had an arrangement of aaaab?

aaaab aaaba aabaa
abaaa baaaa

There are 5 different arrangements for this sequence.

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 with 24 arrangements (120 divided by 5) for each letter.
As that seemed to work, I had a go at trying to work out a logical
universal formula. I came up with; The total number of letters
factorial, divided by the number of a's, b's ect factorised and
multiplied
Formula for emma=

4!/1!x1!x2!=48 /4=12

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.

Total letters

(all different!)

Number of A's

Number of B's

Number of arrangements

1

1

0

1

2

-

-

2

3

-

-

6

4

-

-

24

5

-

-

120

3

2

1

3

4

2

2

6

5

2

3

10

6

2

4

15

3

3

0

1

4

3

1

4

5

3

2

10

6

3

3

20

MLA Citation:
"Arrangements for Emma." 123HelpMe.com. 17 Apr 2014
<http://www.123HelpMe.com/view.asp?id=121735>.