AB BA
Letters: AB, 2 Letters.
Combinations = 2
Letters: ABC, 3 letters.
ABC ACB BCA BAC CAB CBA
Combinations = 6.
Letters: ABCDE, 5 letters.
ABCDE ABCED ABECD ABEDC ABDCE ABDEC
ACBDE ACBED ACDEB ACDBE ACEDB ACEBD
ADCBE ADCEB ADBCE ADBEC ADECB ADEBC
AECBD AECDB AEBDC AEBCD AEDBC AEDCB
As you can see there are 24 combinations with A as the first letter, therefore as there are 5 letters, 24*5 gives 120 combinations.
Here are the results, which I know so far:
As you can see there is a connection between the number of letters, and the number of combinations. This is that the number of combinations is equal to the number of letters factorial, or ! This means a number multiplied by all of its previous whole numbers down to 1. For Example 6! = 6 x 5 x 4 x 3 x 2 x 1. Therefore the formula for working out the number of combinations for a name, which has no repeating letters, is:
No. Of Combinations = No. Of Letters!
Therefore if the table above were to continue I predict it would look like this:
There is however the problem of repeating letters, like the two m’s in EMMA. This make the number of possible formulas a lot less, especially in longer names. However an example for 3 letters is shown below:
ABC
ACB
BAC
BCA
CAB
CBA
= 6 Combinations
ABB
BAB
BBA
= 3 Combinations
As you can see the number of combinations change, and we therefore need a different formula to work it out, or we need to make some modifications to our current one. I will begin trying to find this formula, by using just one repeating letter:
AB ABB ABBB ABBBB ABBBBB ABBBBBB
BA BAB BABB BABBB BABBBB BABBBBB
BBA BBAB BBABB BBABBB BBABBBB
BBBA BBBAB BBBABB BBBABBB
BBBBA BBBBAB BBBBABB
BBBBBA BBBBBAB
BBBBBBA
2 3 4 5 6 7
= No. Of Combinations
As you can see from above, it seems that the number of combinations is just the same as the number of letters; this however is only true for one repeating letter. When you get two repeating letters it changes quite considerably. If you take 4 letters, as in LUCY, there are 24 combinations ordinarily, with no repeating letters. If however there is one repeating letter as in EMMA there are only 12 combinations, if there are two letters which repeat, as in AABB, there would only be 6 combinations. As you can see from the table below the number of combinations halve each time, but this is not the case for all different names, or combinations of letters.
Again it would be a lot easier for me to calculate the combinations if I had a formula which included the possibility of repeated letters. For the finding of the formula I have decided to use X’s and Y’s, and I will begin by gathering results for the number of combinations for 2, 3, 4, 5 and 6 letters, this is shown below. I will then proceed to calculate the formula by trying different ones.
XY
YX = 2 Combinations for 2 letters.
XXY
XYX
YXX = 3 Combinations for 3 letters.
XXYY
XYXY
XYYX
YYXX
YXYX
YXXY = 6 Combinations for 4 letters.
XXXYY
XXYXY
XXYYX
XYXXY
XYXYX
XYYXX
YYXXX
YXYXX
YXXYX
YXXXY = 10 Combinations for 5 letters.
XXXYYY YXXXYY
XXYXYY YXXYXY
XXYYXY YXXYYX
XXYYYX YXYXXY
XYXXYY YXYXYX
XYXYXY YXYYXX
XYXYYX YYYXXX
XYYXXY YYXYXX
XYYXYX YYXXYX
XYYYXX YYXXXY
= 20 Combinations for 6 Letters.
As you can see from the table I need to make the “Total Letters!” row to equal the “Combinations” row, to do this would therefore adjust my formula to work for repeated letters. To narrow down the possible formulas, as my first formula used the factorial function, I predict that the section which, I add onto my formula will also contain something to do with the letters factorial. I also know that the total! Figure needs to be reduced, therefore it has to either have something subtracted from it, or it needs to be divided by something. I will now try different formulas and then try them on my result for 5 letters to see if they work, and equal 10.
- Combinations = Total! - (No. of X! + No. of Y!)
= (120) – (2 + 6)
= 112
(This obviously does not work.)
- Combinations = Total! - (No. of X! * No. of Y!)
= (120) – (2 * 6)
= 108
(This obviously does not work.)
- Combinations = Total! / (No. of X! + No. of Y!)
= (120) / (2 + 6)
= 15
(This is closer but obviously is still wrong.)
- Combinations = Total! / (No. of X! * No. of Y!)
= (120) / (2 * 6)
= 10
(This works for this one but I will now try it for my other results to make sure that it is right.)
For XXY:
Combinations = 3! / (2! * 1!)
= 6 / 2
= 3 (The correct answer)
For XXXYYY:
Combinations = 6! / (3! * 3!)
= 720 / 36
= 20 (Again the correct answer)
As you can see from the calculations above, the formula to work out the number of combinations for 2 repeating letters is:
Number of Combinations = Total! / (No. of the one letters! * No. of the second letters!)
Or
C = T! / ( R! * R’!)
Where C is the total number of combinations, T is the total number of letters, R is the first repeated letters and R’ is the second repeated letters.
I have also discovered that by just adding on more R’’, and R’’’ you can work out the combinations for words or names with many repeating letters or just for all names by using the one formula. For example I know that AABBCC has 90 combinations and I will now prove it using the formula.
Combinations = T! / (R! * R’! * R’’!)
= 6! / (2! * 2! * 2!)
= 720 / 8
= 90 combinations.
I will now show how the formula can be applied to any name by incorporating all the letters, for STEVEN:
Combinations = T! / (No. of S’s! * No. of T’s! * No. of E’s! * No. of V’s! * No. of N’s!)
= 6! / (1! * 1! * 2! * 1! * 1!)
= 720 / 2
= 360 combinations of my name.
