Dave's Dilemma

Calculation

2

2

x2

3

6

x2x3

4

24

x2x3x4

5

20

x2x3x4x5

6

720

x2x3x4x5x6

7

5040

x2x3x4x5x6x7

From this I have concluded that the formula n! is correct( n representing the number of letters).

Therefore, to find the number of arrangements for six letter word, you would multiply

the number of letters (6) by the number of arrangements of the previous number (120). This

gives seven hundred and twenty arrangements.

I then tried to simplify this. By doing so I found that multiplying all the numbers below

the number of letters in the given word together, the correct number of arrangements would be

calculated. For example, if I was to find the number of arrangements in a seven letter word I

would perform the following calculation :

x 2 x 3 x 4 x 5 x 6 x 7

This simplified again is 7!.

2. Investigate the number of arrangements of Emma's name.......

EMMA

EMAM

EAMM

AEMM

AMME

AMEM

MEMA

MAEM

MMEA

MMAE

MAME

MEAM

From Emma's name, a four letter word, we this time only get twelve arrangements. This exactly

half of twenty-four,(twenty-four being the total number of arrangements for a four letter word

with all the letters different, divided by two because two is the number of letters repeated). I

then tried a three letter (TOO) word with two letters repeated

TOO

OTO

OOT

From TOO we get three arrangements, which is the sum of six divided by two (again, six being

the number of arrangements for a three letter word with all letters different, and divide by two as

it is the number of letters repeated in the word.)

From this I worked out a formula for the number of arrangements in a word which has two

letters the same in it.

This was....

n!

2

This formula is demonstrated in the table below, which proves it is correct.

Number of letters

Number of letters

repeated

Arrangements

Calculation

2

2

2! divided by 2

3

2

3

3! divided by 2

4

2

2

4! divided by 2

5

2

60

5! divided by 2

6

2

360

6! divided by 2

7

2

2520

7! divided by 2

However, this formula was ineffective when trying to work out the number of arrangements for a

word which has three letters the same in it, and all the rest different.

AAAB

AABA

ABAA

BAAA

The word produced four arrangements, which is 24 divided 6(twenty-four being the number of

arrangements of a four letter word with all the letters different, and six being the total number of

arrangements of a three letter word and three is the number of letters repeated.) I worked out

that the correct formula for this was.......

n!

6

For example, if take the word ABACA, I found that there were twenty arrangements, these

were......

ABACA

ACABA

AAABC

AAACB

BCAAA

CBAAA

AABCA

AACBA

ACBAA

ABCAA

ABAAC

ACAAB

BAAAC

CAAAB

BACAA

CABAA

CAABA

BAACA

AACAB

AABAC

This proves my formula. Below is a table to show this.

Number of letters

Number of letters

repeated

Arrangements

Calculation

3

3

3! divided by 6

4

3

4

4! divided by 6

5

3

20

5! divided by 6

6

3

20

6! divided by 6

7

3

840

7! divided by 6

8

3

6720

8! divided by 6

I soon realised that a pattern was emerging. I found that the arrangements for a word with a

double letter was the same as a word with the number of letters of the letter repeated. For

example....

A three letter word, with all the letters different would have six arrangements.

A word with three letters the same in it would need to be divided by six to give the

correct number of arrangements.

? the formula 3! works.

6

I then tried a five letter word with four letters the same.

AAAAB

AAABA

AABAA

ABAAA

BAAAA

? The formula for four letters the same would be 4!

24

I the saw another pattern. The denominator in the formula was the same number of

arrangements as the number of repeated letters factorilised. For example........

To find the number of arrangements in a four letter word with three letters the same you would

calculate......

4!

3! = Four arrangements.

This can then be put into a general formula as........

n!

y!

( 'y' equalling the number of repeated letters in the word)

To test my formula I will use a six letter word with five letters the same(AAAAAB).

AAAAAB

BAAAAA

ABAAAA

AABAAA

AAABAA

AAAABA

There are six arrangements, which proves my formula is correct.

3. A number of X's and Y's are written in a row such as......

XX......XXXY......YXXY

Investigate the number of different arrangements of the letters.

I first investigated a number of different words to try and find if a common formula was present,

as in the tasks involving Dave's and Emma's names.

* A four letter word with three letters the same, the other different.

AAAB

AABA

ABAA

BAAA

There are four arrangements.

* A four letter word, with two different letters the same.

AABB

ABAB

BBAA

BABA

ABBA

BAAB

This gives us six arrangements, which

* A five letter word, with three letters and two letters the same.

AAABB

BBBAA

BABAA

BAABA

BAAAB

ABAAB

ABBAA

ABABA

AABAB

AABBA

There are ten arrangements.

* A six letter word, with two sets of three letters the same.

AAABBB

BBBAAA

ABBBAA

AABBBA

ABABAB

BABABA

BAAABB

BBAAAB

AABBAB

ABBAAB

BBAABA

BAABBA

ABAABB

BABBAA

AABABB

BBABAA

ABBABA

BAABAB

BABAAB

ABABBA

There are twenty arrangements of these letters.

From these words I have discovered a formula that can be used to work out the number of

arrangements for a word with more than one type of letter repeated, this formula is found from

expanding on the formula used to find the number of arrangements for Emma's

name (n!) . To find a word with more than one type of letter repeated you must multiply

(y!)

both 'y' values together. For example.........

To find a the number of arrangements of an eight letter word with two letters and two lots of

three letters the same (AAABBBCC) you would calculate

8!____

(3! x 3! x 2!)

I then set about trying to find a formula for this. I quickly came to the decision that all that is

needed is letters to represent the number of repeated letters. This gave me the formula:

____n!____

(x! X y! X z!)

('x' represents the number of repeated A's, 'y' represents the number of repeated B's

and 'z' represents the number of repeated C's)

To test my formula I used a five letter word (AABBC), with two lots of letters the same. This

gives me the following number of arrangements..........

CABBA

CBAAB

CAABB

CBBAA

CBABA

CABAB

AABBC

AACBB

AABCB

ABABC

ABBCA

ABCBA

ACBBA

ABBAC

ACABB

ACBAB

ABCAB

ABACB

BBAAC

BABAC

BBCAA

BAACB

BACAB

BCAAB

BAABC

BCBAA

BCABA

BACBA

BBACA

BABCA