SAM
SMA
MSA
MAS
ASM
AMS
There are 3 different letters in this name and 6 different arrangements. Again, as you can see there a pattern to how I have worked out the correct amount of arrangements, there are 2 different letters after the first letter, which means there are 2 different combinations for each first letter. However this is only the case with 3 letter words.
JO
OJ
There are 2 different letters in this name and there are 2 different arrangements.
From these investigations I worked out a method:
Step1: 1234---Do the last two numbers first then you get 1243.
1243---Do the last three numbers and try the possibility. 1423. 1432. 1342. 1324, because the number 2 has been the first number of last three numbers, so we don’t do it again.
Step2: we have list all arrangements of 1 go front, so we do 2 go front. 2134 and we do same thing to it, it will like this:
2134---2143, 2143---2431,2413,2314,2341
Step3: We have finished 2 go first, then let’s do 3 go ahead.
3124---3142, 3142---3241,3214,3412,3421
Step4: We have finished 3 go ahead, then try 4
4123---4132, 4132---4231,4213,4312,4321
We have list all arrangement of 1234, use this method we can arrange the number which has 5 figures or more.
We are trying to work out a formula which can calculate the number of arrangement when we look at a number.
Let’s list all the arrangement for 1234:
1234 4123
1243 4132
1324 4231
1342 4213
1432 4321
1423 4312
2134 3124
2143 3143
2341 3241
2314 3214
2413 3412
2431 3421
Table of Results
From the table of results you can see 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 is: n! = a
Where n = the number of letters in the word and
a = the number of different arrangements.
Now I am going to investigate the number of different arrangements in a word with 2 letters repeated, 3 letters repeated and 4 letters repeated.
EMMA
AMME
AMEM
EMAM
AEMM
EAMM
MMEA
MMAE
MEMA
MAME
MEAM
MAEM
4-letter word, 2 letters repeated, 12 different arrangements.
MUM
MMU
UMM
3-letter word, 2 letters repeated, 3 different arrangements.
PQMMM
PMQMM
PMMQM
PMMMQ
QPMMM
QMPMM
QMMPM
QMMMP
MPQMM
MPMQM
MPMMQ
MQPMM
MQMPM
MQMMP
MMPQM
MMQMP
MMMPQ
MMMQP
MMPQM
MMMQP
5-letter word, 3 letters repeated, 20 different arrangements.
SMMM
MSMM
MMSM
MMMS
4-letter word, 3 letters repeated, 4 different arrangements.
RRRRK
RRRKR
RRKRR
RKRRR
KRRRR
5-letter word, 4 letters repeated, 5 different arrangements.
Table of Results
I have worked out that if you do say 5! = 120, to find out how many different arrangement in a 3-letter word it would be 5! Divided by 6 = 20, so, a 6-letter word with 4 letters repeated would be 6! Divided by 24 = 30, as you can see in the “No letters repeated” column these are the numbers we are dividing by:
2 letters the same = n!
(2x1 = 2)
3 letters the same = n!
( 3x2x1 = 6)
4 letters the same = n!
(4x3x2x1 = 24)
5 letters the same = n!
(5x4x3x2x1 = 120)
6 letters the same = n!
(6x5x4x3x2x1 = 720)
From this I have worked out the formula to fine out the number of different arrangements:
n! = the number of letters in the word
p! = the number of letters the same
Now I am going to investigate the number of different arrangement for words with 2 or more letters the same like, aabb, aaabb, or bbbaaa.
This is a 4 letter word with 2 letters the same, there are 6 different arrangements:
xxyy
I am going to use the letters x and y (any letter)
xxyy xyxy yxxy
xyyx yxyx yyxx
This is a 5 letter word
xxxyy
xxxyy xxyxy xxyxx xyxyx xyxxy
xyyxx yyxxx yxxxy yxyxx yxxyx
There are 10 different arrangements
In the above example there are 3 x's and 2 y's
As each letter has its own number of arrangements i.e. there were 6 beginning with x, and 4 beginning with y, I think that factorial has to be used again.
As before, the original formula:
n! = the number of letters in the word
p! = the number of letters the same
From this I have come up with a new formula The number of total letters factorial, divided by the number of x's, y's etc factorised and multiplied.
For the above example:
A four letter word like aabb; this has 2 a's and 2 b's (2 x's and 2 y's)
So : 1x2x3x4
(1x2) x (1x2)
= 24
4 = 6 different arrangements
A five letter word like aaaab; this has 4 a's and 1 b (4 x's and 1 y)
So: 1x2x3x4x5
(1x2x3x4) x (1)
= 120
24 = 5 different arrangements
A five letter words like abcde; this has 1 of each letter (no letters the same)
So : 1x2x3x4
(1x1x1x1x1x1)
= 24
1 = 24 different arrangements
A five letter word like aaabb; this has 3 a's and 2 b's (3 x's and 2 y's).
So : 1x2x3x4x5
(1x2x3) x (1x2)
= 120
12 = 10 different arrangements
This shows that my formula works:
n! = the number of letters in the word
x!y! = the number of repeated letters the same
