Emma's Dilemma Coursework.

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

Number of A's

Number of B's

Number of arrangements

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