Emma's Dilemma

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 begging with one of the letter I predict that there will be 120 arrangements for lucyq, 24 for a, 24 for b, 24 for c; ect. 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 possibilities for 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. However this is expressed as factorial. There is a button on most scientific calculators with have embedded this factorial button feature generally sowing as an exclamation mark. All it does is save the time of having to put in to the calculator 1x2x3x4x5x6x7..... Ect. You just put in the number and press factorial and it will do 1x2x3... until it get to the number you put in. If I key in (lets say the number of letters all different) factorial 6 I get 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.

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 =

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 !!!

Total Letters Number Of X's Number Of Y's Number of Arrangements

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