Arrangements for names.

Total Letters (all different) Number of Arrangements

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 there are repeat letters? 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 more to it that just factorial in that way.

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

Arrangements for xxyy:

x=a, y=b

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 xxxyy?

x=a, y=b

aaabb aabab aabba ababa abaab

abbaa bbaaa baaab babaa baaba

There are 10 different arrangements for this instance.

What if I had an arrangement of xxxxy?

x=a, y=b

aaaab aaaba aabaa

abaaa baaaa

There are 5 different arrangements for this instance.

If I go back to xxxyy; there are 3 x's and 2 y's in a total of 5 unknowns. As each letter has its own number of arrangements i.e. there were 5 beginning with x, and 5 beginning with y, 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 I 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 x's, y's ect factorised and multiplied.

Formula =

n!/x!y!

For example:

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 !!!

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

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

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 !!!

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 !!!

The national lottery can be shown as 49x48x47x46x45x44=

In the national lottery you can get the numbers in any orde so therefore you divide by 6! This can be expressed as

N!/Nr(N-r)! or NCr.

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In a 4 letter word, to work out the amount of different arrangements you can do 4 x 3 x 2 = 24, or you can do 4! which is called 4 factorial which is the same as 4 x 3 x 2.

So, by using factorial (!) I can predict that there will be 5040 different arrangements for an 7-letter word.

The formula for this is: n! = A

Where n = the number of letters in the word and

A = the number of different arrangements.

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3 factors, 2 repetitions = 1 arrangements = (1 x 2) / 2 = 3!

6

4 factors, 2 repetitions = 4 arrangements = (1 x 2 x 3) / 2 = 4!

6

5 factors, 2 repetitions = 20 arrangements = (1 x 2 x 3 x 4) / 2 = 5!

6

There is a pattern in the denominatives, in that they are equal to the factorial of the number of repeated factors. In the sequence 1112, there are three repetitions, one number repeated twice. Therefore, the formula for this sequence would be:

4th = x! / r!

= 4! / 3!

= 24 / 6

= 4

and there are indeed 4 arrangements. In the sequence 11123, which also has three repetitions, the formula would be:

5th = x! / r!

= 5! / 3!

= 120 / 6

Emma's Dillemma - Rearranging Emma's Name in different permutations

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 to follow that pattern.

Total Letters (all different) Number of Arrangements

-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

0

2

-

-

2

3

-

-

6

4

-

-

24

5

-

-

20

3

2

3

4

2

2

6

5

2

3

0

6

2

4

5

3

3

0

4

3

4

5

3

2

0

6

3

3

20

* What about if a number has two same figures the formula is a=ni/2

* so the formula for three sames numbers of a number is:a= ni/6

* try 4 same number.a=n/24=(1*2*3*4*5)/24=5 the formula works

* What about if a number has 2 pairs of same numbera=ni/xixi

What about a number with different number of figures with the same numbers

For example: 11122,111122

The formula is a=ni/xixi

but we need to change the formula, because there are 2 pairs of same numbers with different number of figures. so we change the formula to a=ni/x1i*x2i

Let's try 5 figures with 3 same number, and 2 same number.

According to the formula, I expect the total arrangement for this is

a=(1*2*3*4*5)/(3*2*1*2*1)=10

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