=120
In a 4-letter name such as Lucy, the way to work out the number of combinations is to do 4 x 3 x 2 x 1 or 4 factorial (4!). . Factorial is a number multiplied by the previous consecutive numbers. Factorial notation is symbolised using an exclamation mark! So I came to conclude that the formula to work out any the combinations is the numbers of letters factorial equals arrangements (l! =a).
With this formula in mind I can predict the number of arrangements for an 8-letter name. 8! (8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) =40320
When there are repeated letters like in EMMA there are 12 combinations.
EMMA
AMME
AMEM
EMAM
AEMM
EAMM
MMEA
MMAE
MEMA
MAME
MEAM
MAEM
Emma has 2 M’s in it which means that the formula that I used for LUCY would not work. Emma is a 4-letter word with 2 letters repeated with 12 different combinations. I investigated other word combinations with 1 or more letters the same.
DAD
DDA
ADD
Dad is a 3-letter word with 2 letters repeated and 3 different combinations.
IDMMM
IMDMM
IMMDM
IMMMD
DIMMM
DMIMM
DMMIM
DMMMI
MIDMM
MIMDM
MIMMD
MDIMM
MDMIM
MDMMI
MMIDM
MMDMI
MMMID
MMMDI
MMIDM
MMMDI
MMMDI is a 5-letter word with 3 letters repeated and has 20 different arrangements.
SIII
ISII
IISI
IIIS
IIIS is a 4-letter word with 3 letters repeated and 4 different arrangements.
BBBBD
BBBDB
BBDBB
BDBBB
DBBBB
DBBBB is a 5-letter word with 4 letters repeated and 5 different arrangements.
Using the evidence I gathered I made a table of results.
This table shows that if I take a 4 letter word for example and want to find the amount of combinations for 2 letters repeated, I do 4 factorial (24) divided by 2 factorial (2) equals 12. Another example would be 6 letter word with 4 letters the same (6!/4!) which is equal to (720 divided by 24)=30. A general formula for this would be number of letters factorial divided by letter repeated factorial (l! /r!)
I then investigated the number of arrangements with 2 or more letters the same. In my example I will use x and y.
xxyy
xyxy
yxxy
xyyx
yxyx
yyxx
This is a 4-letter word with 2 x’s and 2y’s and 6 different arrangements.
xxxyy
xxyxy
xxyxx
xyxyx
xyxxy
xyyxx
yyxxx
yxxxy
yxyxx
yxxyx
This is a 5-letter word with 3 x’s and 2 y’s and 10 different arrangements.
From this table I concluded that in order to predict the amount of combinations with 2 or more letters the same you do for example, for a 4 letter word with 2 x’s and 2 y’s you do number of letters factorial divided by the number of x’s factorial multiplied by the number of y’s factorial. So if I substitute figures into the figures form my table I get 4! (24)/2!(2) x 2!(2) =6
I am now going to try and predict the amount of combinations for Mississippi using the formula I have worked out above. It is an eleven letter word with 1 m 4 s’es, 4 i’s and 2 p’s. So 11! =39916800. I need to divide this by 1! =1 multiplied by 4! =24 multiplied by 4! =24 multiplied by 2! =2 which all equals 1152. So my final sum is 39916800/1152=34650.
