2 letter – 0 same = 2
JO,
OJ
2 letters – 2 same = 1
DD
3 letters – 0 same = 6
JIM
JMI
IJM
IMJ
MJI
MIJ
3 letters – 2 same = 3
BOB
BBO
OBB
4 letters – 0 same = 24 (see pg1)
4 letters – 2 same = 12 (see pg1)
4 letters – 3 same = 4
BOBB
OBBB
BBOB
BBBO
5 letters – 0 same = 120
5 letters – 2 same = 60
JIMMY
JIMYM
JIYMM
JYMMI
JYIMM
JYMIM
JMMIY
JMMYI
JMYIM
JMYMI
JMIMY
JMIYM
IJMYM
IJMMY
IJYMM
IYMJM
IYJMM
IYMMJ
IMMYJ
IMMJY
IMJYM
IMYJM
IMYMJ
IMJMY
MJMIY
MJMYI
MJYIM
MJYMI
MJIMY
MJIYM
MIJMY
MIJYM
MIYJM
MIYMJ
MIMYJ
MIMJY
MMJIY
MMJYI
MMIYJ
MMIJY
MMJYI
MMYIJ
MYMJI
MYMIJ
MYJIM
MYJMI
MYIMJ
MYIJM
YJMMI
YJMIM
YJIMM
YMJMI
YMJIM
YMIMJ
YMIJM
YMMJI
YMMIJ
YIMMJ
YIMJM
YIJMM
Looking at my table of results I can see that a 3 letter word (with none the same) has 6 combinations, 3 x 2
- A 4 letter word (none same) has 24, 4 x 3 x 2
- A 5 letter word (none same) has 120, 5 x 4 x 3 x 2
I estimate that this pattern continues for all names with no letters the same, e.g. I predict that a name with 6 different letters will be 6 x 5 x 4 x 3 x 2 which equals 720. To work out larger numbers you can use the ‘!’ on a scientific calculator, it is called the factorial button and does the process automatically
-
If you key in 7! The calculator will perform the sum 7 x 6 x 5 x 4 x 3 x 2, which equals 5040.
The formula for this sum is n! = y
where n is the number of letters and y is the number of combinations
This table displays names with letters the same.
Names with 5 letters, none same = 120, with 2 letters the same it equals 60, which is half, if there are 4 letters the same (which gives 5 combinations) you must do 120/5= 24, then you divide 24 by 4 (because that is the amount of letters the same) which gives you 6, 6 divided by 3 = 2.
So 4 x 3 x 2 = 24, this is the factorial again.
This means that, in order to get the amount of combinations for a number (e.g 6) with more than 2 letters the same (e.g. 4) you must do the factorial of the number of letters, divided by the factorial of the number of letters the same (e.g. 6!/4!) 6 x 5 x 4 x 3 x 2 6!
4 x 3 x 2 = 4!
If the name is BARRRR (6 letters, 4 the same)
you must do 6!
4! = 720 / 24 = 30
So the formula is n!
a! = y
n being number of letters, a being the number of letters the same and y being the number of combinations found.
e.g. 9 letters in a name, 5 the same
9 x 8 x 7 x 6 x 5 x 4 x 3 x 2
5 x 4 x 3 x 2
this equals 9!
5! Which equals 3024
Names with two numbers appearing twice (e.g. xxyy) are different
xxyy xyxy yxxy
xyyx yxyx yyxx
there are 6 combinations
xxxyy has 5 letters
xxxyy xxyxy xxyxx xyxyx xyxxy
xyyxx yyxxx yxxxy yxyxx yxxyx
there are 10 combinations
I think factorials need to be used again because, with xxxyy, there are 6 combinations beginning with x and 4 with y, because of this I will need the old formula -
n!
a! = y
through investigation I have decided that I will need to factorise and multiply the factorials of the amounts of different letters
e.g. if there is 4 a’s and 2b’s (aaaabb)
I will do 6! (because there are 6 letters in total)
4! X 2!
This is 6 x 5 x 4 x 3 x 2 720
(4 x 3 x 2) X (2) which equals 48 = 15 combinations
This shows that the formula is n! *
a! x b!
If you include another letter (e,g, xxyyzz) the formaul continues
n! *
a! x b! x c!
for mmtookaanooba(2 x m, 3 x a, 4 x o) the formula would be
13! *
2! X 3! X 4! = 21621600
This formula is correct for all words and can be any length as long as you have the factorial of the amount of letters divided by the factorials of the letters which appear more than once multiplied together.
