5)AKN

6)ANK

I will now be investigating the number of different arrangements of the letters in AT’s name.

1)AT

2)TA

I will now be investigating the number of different arrangements of the letters in DAVIS’s name

1)DAVIS

2)DAVSI

3)DAISV

4)DAIVS

5)DASIV

6)DASVI

7)DVAIS

8)DVASI

9)DVISA

10)DVIAS

11)DVSAI

12)DVSIA

13)DSVIA

14)DSVAI

15)DSAVI

16)DSAIV

17)DSIAV

18)DSIVA

19)DIAVS

20)DIASV

21)DIVAS

22)DIVSA

23)DISVA

24)DISAV

Looking at my results I have created a results on the information.

Results

From the table of results I have found out that a 2 letter word has 2 arrangements, and a 3 letter word has 6.

Taking for example a 3 letter word, I have worked out that if we do 3 (the length of the word) x 2 = 6, the number of different arrangements.

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 40320 different arrangements for an 8 letter word.

The formula for this part of the investigation is: n! = d

n = the number of letters in the word, i.e. BIRKAN has 6 letters.

d = the number of different arrangements, i.e. DAVID has 120 arrangements.

Now I am going to investigate the number of different arrangements in a word with letters repeated e.g. 3 letters repeated and 4 letters repeated.

1)LTTA

2)LTAT

3)LATT

4)TLAT

5)TLTA

6)TALT

7)TATL

8)TTLA

9)TTAL

10)ALTT

11)ATLT

12)ATTL

1)CKK

2)KKC

3)KCK

1)AQMMM

2)AMQMM

3)AMMQM

4)AMMMQ

5)QAMMM

6)QMAMM

7)QMMAM

8)QMMMA

9)MAQMM

10)MAMQM

11)MAMMQ

12)MQAMM

13)MQMAM

14)MQMMA

15)MMAQM

16)MMQMA

17)MMMAQ

18)MMMQA

19)MMAQM

20)MMMQA

1)SPPP

2)PSPP

3)PPSP

4)PPPS

1)FFFFK

2)FFFKF

3)FFKFF

4)FKFFF

5)KFFFF

I have created a result table for all ways of combinations e.g. same letter repeated.

Table of Results

By looking at my results I have figured out that if a name has 5 letters to work out the number of combinations you do the formula: 5! What I have figured out is that I can work out if there is same letter repeated 2 times, 3 times, 4 times…etc.

The following is going to show how it works:

3 letter word = 6 combinations = 3!

3 letter word with 2 letters repeated = 3 = 3! /2

3 letter word with 3 letters repeated = 1 = 3! /2/3

4 letter word = 24 combinations = 4!

4 letter word with 2 letters repeated = 12 = 4! /2

4 letter word with 3 letters repeated = 4 = 4! /2/3

4 letter word with 4 letters repeated = 1 = 4! /2/3/4

5 letter word = 120 combinations = 5!

5 letter word with 2 letters repeated = 60 = 5! /2

5 letter word with 3 letters repeated = 20 = 5! /2/3

5 letter word with 4 letters repeated = 5 = 5! /2/3/4

5 letter word with 5 letters repeated = 1 = 5! /2/3/4/5

6 letter word = 720 combinations = 6!

6 letter word with 2 letters repeated = 360 = 6! /2

6 letter word with 3 letters repeated = 120 = 6! /2/3

6 letter word with 4 letters repeated = 30 = 6! /2/3/4

6 letter word with 5 letters repeated = 6 = 6! /2/3/4/5

7 letter word = 5040 combinations = 7!

7 letter word with 2 letters repeated = 2540 = 6! /2

7 letter word with 3 letters repeated = 840 = 6! /2/3

7 letter word with 4 letters repeated = 210 = 6! /2/3/4

7 letter word with 5 letters repeated = 42 = 6! /2/3/4/5

I have shown above that I have tested the formula and showed how it is worked out.

(Task 4)

Now I am going to investigate the number of different arrangement for words with 2 letters, one letter with repeated twice and the other letter repeated twice too.

## Four letter word

1)CCDD

2)CDCD

3)DCCD

4)CDDC

5)DCDC

6)DDCC

5 letter word

1)CCCDD

2)CCDDC

3)CDDCC

4)CCDCD

5)CDCCD

6)CDCDC

7)DDCCC

8)DCDCC

9)DCCDC

10)DCCCD

In the example above there are 3 c’s and 2 d’s

As each letter has its own number of arrangements i.e. if there were 7 beginning with c, and 4 beginning with d, I think that factorial has to be used again.

From this I have come up with a new formula the number of total letters factorial, divided by the number of c’s, d’s etc factorised and multiplied.

A four letter word i.e. cdcd, this has 2 c’s and 2 d’s.

Working out: n!(4)/ c!(2)/ d!(2)

1x2x3x4=24

(1x2)x(1x2)=4

24/4=6

A five letter word i.e. cccdd, this has 3 c’s and 2 d’s.

Working out: n!(5)/ c!(3)/ d!(2)

1x2x3x4x5=120

(1x2x3)x(1x2)=12

120/12=10

A six letter word i.e. ccccdd, this has 4 c’s and 2 d’s.

Working out: n!(6)/ c!(4)/ d!(2)

1x2x3x4x5x6=720

(1x2x3x4)x(1x2)=48

720/48=15

A seven letter word i.e. cccccdd, this has 5 c’s and 2 d’s.

Working out: n!(7)/ c!(5)/ d!(2)

1x2x3x4x5x6x7=5040

(1x2x3x4x5)x(1x2)=240

5040/240=21

On my previous pages I have explained, tested and showed how my formula works.

I have concluded by using examples on previous pages that N is the number of letters in the word and that C and D is the number of repeated letters the same. The letters I have chosen i.e. C and D can be different I just used those letters for a example.