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 = 72)
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:
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
x = total number of letters, y = number of repeated letters (1), and
z = number of repeated letters (2).
If there were an arrangement of 3 x's and 3 y's, i.e. xxxyyy, and we separate the x's from the y's, so that we have (x + y). (x + y) is the total number of letters. When re-arranging, the first letter that we would chose would be (x + y), the second (x + y - 1) because one letter has already been chosen, and so on.
To find the first AND second AND third AND fourth AND fifth AND sixth one, you would need to multiply them all together to get part of our formula:
(x + y) (x + y - 1) (x + y - 2) (x + y - 3) (x + y - 4) (x + y - 5) = (x + y)!
(This is basically the total number of letters)
To get the final formula we need to divide it by (x) (x - 1) (x - 2) because this is the number of x's in the word. The same will have to be done with the y's and so it will have to be divided by:
(x) (x - 1) (x - 2) (y) (y - 1) (y - 2)
Therefore we get the formula:
(x + y) (x + y - 1) (x + y - 2) (x + y - 3) (x + y - 4) (x + y - 5)
(x) (x - 1) (x - 2) (y) (y - 1) (y - 2)
This can be simplified to:
(x + y)!
(x! x y!)
This formula can be abbreviated to:
x!
(y! x z!)
Now as you can see, it multiplies (x + y)…(x + y - 5). This because there are six letters. If it were 5 letters it would go from (x + y) to (x + y - 4). This is because each time one letter is taken out, so it has to be subtracted from the number being multiplied each time. So, if (x + y) = 5, then in other words we would have to multiply it by (x + y - 1) which is 4 and so on… giving us 5 x 4 x 3 x 2 x 1 or 5!
The bottom part now is similar, but the two parts, x and y are separated. If there are 3 x's, then we will multiply (x) by (x - 1) by (x - 2) which is
3 x 2 x 1 or 3! You will also notice that there are three brackets in this case as there are six brackets in the example above. I there are 2 y's, then (y) will be multiplied by (y - 1). This is the same as 2 x 1 or 2! Now these two will need to be multiplied giving us 3! x 2! The numbers 3 and 2, which are factorial, are the numbers that are different.
Proving that the formula works
1. Using the formula (taking that the word consists of six letters):
(x + y) (x + y - 1) (x + y - 2) (x + y - 3) (x + y - 4) (x + y - 5)
(x) (x - 1) (x - 2) (y) (y - 1) (y - 2)
· xxyy:
(2 + 2) (2 + 2 - 1) ( 2 + 2 - 2) ( 2 + 2 - 3)
(2) (2 - 1) (2) (2 - 1)
= 4 x 3 x 2 x 1
2 x 1 x 2 x 1
= 24
4
= 6
· xxxyy:
(3 + 2) (3 + 2 - 1) (3 + 2 - 2) (3 + 2 - 3) (3 + 2 - 4)
(3) (3 - 1) (3 - 2) (2) (2 - 1)
= 5 x 4 x 3 x 2 x 1
3 x 2 x 1 x 2 x 1
= 120
12
= 10
· xxxxy:
(4 + 1) (4 + 1 - 1) (4 + 1 - 2) (4 + 1 - 3) (4 + 1 - 4)
(4) (4 - 1) (4 - 2) (4-3) (1)
= 5 x 4 x 3 x 2 x 1
4 x 3 x 2 x 1 x 1
= 120
24
5
2. It can also be done simply by using the formula:
x!
y! x z!
· xxyy:
4! 4 x 3 x 2 x 1
(2! x 2!) = 2 x 1 x 2 x 1
= 24
4
= 6
· xxxyy:
5! 120
(3! x 2!) = 3 x 2 x 1 x 2 x 1
= 120
12
= 10
· xxxxy:
5! 120
(4! x 1!) = 4 x 3 x 2 x 1 x 1 x 1
= 120
24
= 5
Now suppose we were to take the letter z and put it into this context so that there would be three different letters.
Here are some examples:
· xyyzz
· xyyyz
…And so on…
Here are some arrangements for:
· xyyzz
There are 30 arrangements.
The formula can be changed slightly to
(x + y + z) (x + y + z - 1) (x + y + z - 2) (x + y + z - 3) (x + y + z - 4)
(x) (y) (y - 1) (z) (z - 1)
because it has 5 letters, 3 of which are different.
(1 + 2 + 2) (1 + 2 + 2 - 1) (1 + 2 + 2 - 2) (1 + 2 + 2 - 3) (1 + 2 + 2 - 4)
(1) (2) (2 - 1) (2) (2 - 1)
= 5 x 4 x 3 x 2 x 1
1 x 2 x 1 x 2 x 1
= 5!
1! x 2! x 2!
= 120
4
= 30
This proves that my formula works, and in future the 1! Does not need to be included because in names like THANCANAMOOTOO in which there are 3 letters that are not repeated (H, C, M) we do not need to write
1! x 1! x 1! Because 1! Is equal to simply 1 and this does not have any effect on the formula if we exclude it.
The formula can be simplified to
x!
y! x z! x a! …and so on, depending on the number of letters which are different.
x is the number of letters
y is the number of letters repeated, as are z and a.
This formula can go on forever depending on the number of letters repeated twice or more.
So, for example if in a name like THANCANAMOOTOO we would use the formula to work it out like this:
14!
2! x 3! x 2! x 4!
= 14!
2 x 6 x 2 x 24
= 151,351,200
· It is 14! Because that is the number of letters in the name.
· It is divided by (2! x 3! x 2! x 4!) because the letters T and N are both repeated twice, the letter A is repeated 3 times, and the letter O is repeated 4 times, and hence we multiply the factorials of them by each other.