If we use the name MAY to investigate the number of potential arrangements when there are only three factors, we get these arrangements:
M A Y
1 2 3
1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1
We only get six arrangements, a quarter of the amount when we use four factors.
If we use the word MOO to investigate the number of potential arrangements when there are only three factors, we get these arrangements:
M O O
1 2 2
1 2 2
2 1 2
2 2 1
We only get three arrangements, a half of the amount when there are no repetitions
If we only use two factors (why bother with letters when we can use numbers?) we only get two arrangements:
- 2
- 1
This is a third of the number of repetitions when we use three factors.
2 factors = 2 arrangements = 1 x 2 = 2!
3 factors = 6 arrangements = 1 x 2 x 3 = 3!
4 factors = 24 arrangements = 1 x 2 x 3 x 4 = 4!
Therefore, the number of possible arrangements is equal to the factorial of the number of factors.
nth = x! (x = no. of letters)
This only works for when there are no repetitions. The arrangements for when there is one repetition are:
1 1
1 1 2
1 2 1
2 1 1
1 1 2 3 2 1 1 3
1 1 3 2 2 1 3 1
1 2 1 3 2 3 1 1
1 2 3 1 3 1 1 2
1 3 1 2 3 1 2 1
1 3 2 1 3 2 1 1
2 factors, 1 repetition = 1 arrangements = (1 x 2) / 2 = 2!
2
3 factors, 1 repetition = 3 arrangements = (1 x 2 x 3) / 2 = 3!
2
4 factors, 1 repetition = 12 arrangements = (1 x 2 x 3 x 4) / 2 = 4!
2
If we investigate the arrangements when there are three repetitions, we get these results:
1 1 1
1 1 1 2
1 1 2 1
1 2 1 1
2 1 1 1
1 1 1 2 3 1 3 1 2 1
1 1 1 3 2 1 3 2 1 1
1 1 2 1 3 2 1 1 1 3
1 1 2 3 1 2 1 1 3 1
1 1 3 1 2 2 1 3 1 1
1 1 3 2 1 2 3 1 1 1
1 2 1 1 3 3 1 1 1 2
1 2 1 3 1 3 1 1 2 1
1 2 3 1 1 3 1 2 1 1
1 3 1 1 2 3 2 1 1 1
3 factors, 2 repetitions = 1 arrangements = (1 x 2) / 2 = 3!
6
4 factors, 2 repetitions = 4 arrangements = (1 x 2 x 3) / 2 = 4!
6
5 factors, 2 repetitions = 20 arrangements = (1 x 2 x 3 x 4) / 2 = 5!
6
There is a pattern in the denominatives, in that they are equal to the factorial of the number of repeated factors. In the sequence 1112, there are three repetitions, one number repeated twice. Therefore, the formula for this sequence would be:
4th = x! / r!
= 4! / 3!
= 24 / 6
= 4
and there are indeed 4 arrangements. In the sequence 11123, which also has three repetitions, the formula would be:
5th = x! / r!
= 5! / 3!
= 120 / 6
= 20
this is also the right amount. This also works for when there are two repetitions:
4th = x! / r!
= 4! / 2!
= 24 / 2
= 12
and for when there is only one repetition of each letter (i.e. 1234):
4th = x! / r!
= 4! / 1!
= 24 / 1
= 24
4) Multiple Repetitions
If there are a number of repetitions of different factors, then the arrangements look like this:
x x y y y x x y
x y x y y x y x
x y y x y y x x
With the standard formula, there should be:
4th = x! / r!
= 4! / 4!
= 24 / 24
= 1
this is clearly erroneous. To reach the right total using the formula, x! must be divided by 4.
If we take a more literal translation of the previous statement, then the two totals for the number of repetitions must be separate for each different factor, i.e. the factorial is divided by the two totals separately:
4th = (x! / r1!) / r2!
= (4! / 2!) / 2!
= 12 / 2
= 6
this provides us with the right answer. If there are five factors, and two separate repetitions;
x x y y z y x y z x
x x y z y y x z x y
x x z y y y x z y x
x y x y z y y x x z
x y x z y y y x z x
x y y x z y y z x x
x y y z x y z x x y
x y z x y y z x y x
x y z y x y z y x x
x z x y y z x x y y
x z y x y z x y x y
x z y y x z x y y x
y x x y z z y x x y
y x x z y z y x y x
y x y x z z y y x x
then the formula, if rearranged, should be
5th = x! / (r1! x r2!)
= (5! / (2! x 2!)
= 120 / 4
= 30
so this formula is right. If we take it one step further and account for each letter, we see that;
5th = x! / (r1! x r2! x r3!)
= (5! / (2! x 2! x 1!)
= 120 / 4
= 30
it doesn’t make much difference, as multiplying by 1 never has any effect anyway. If we test this formula on the first and second sequences (4 letters, no repetitions and 4 letters, 1 repetition);
4th = x! / (r1! x r2! x r3! x r4!)
= 4! / (1! x 1! x 1! x 1!)
= 24
4th = x! / (r1! x r2! x r3! x r4!)
= 4! / (2! x 1! x 1! x 1!)
= 24 / 2
= 12
The formula even works for these two, so it is safe to assume that it works for all of the combinations of every set of letters with any number of repetitions.
nth = n! / (r1! x r2! x r3! x r4!...) ; r1 + r2 + r3 + r4... = n
