e.g. Arrangements for EMMA = 24 = 12
2
Arrangements for EMMM = 12 = 4
3
Arrangements for MMMM = 4 = 1
4
Also, if 4! Is 24, then this can also be divided by the factorial of the number of letters which are repeated because 3! Is 6, double the number 3, so if we divide it by 24, double 12 we will get the same answer, proving that this formula works.
This gives us a formula:
x!/n!
x is the total number of letters in the name chosen, whereas n is the number of letters which are the same. I am going to test the formula on the name EMMA:
4! = 24
2! 2
= 12
Here are arrangements for some names that I have chosen:
· J:
·
JO:
·
JOE:
· JOEL:
I have found a pattern from the arrangements:
As you can see there is a pattern forming. The number of letters multiplies the previous number of arrangements each time.
e.g. -no. of letters = 4
no. of arrangements = 24 (previous number of arrangements)
-no. of letters = 5 (number of letters in this case)
Here are some different arrangements in terms of x's and y's:
Arrangements for xxyy:
As you can see there are 2 x's and 2 y's. There are 6 arrangements, 3 starting with each letter.
Arrangements for xxxyy:
There are 10 arrangements. There are 6 starting with x and 4 starting with y.
Arrangements for xxxxy:
There are only 5 arrangements. There are 4 starting with x and 1 starting with y.
Because there are two sets of same letters the previous formula cannot be used. However I have come up with another formula:
x!
(y! x z!)
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
Here is a table showing some more arrangements for x and y:
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.
