The rule is times the number of letters by the number of combinations of the previous number using this rule I can predict the 5 letters will have 120 different combinations because the previous number of combinations was 24 and the number of letters is 5 so 5 x 24 =120
There is an easier and quicker way to find how many combinations there are for a certain number this is called N factorial written as N! (N stands for the number of letters you have). N factorial times all the numbers from the number or letters you have all the way to 1 for example if you had 5 letters it would be 5!=5x4x3x2x1=120.
Or 10 would be 10!=10x9x8x7x6x5x4x3x2x1=3628800
For me the easiest way to rite out all the combinations by hand is to say I had 4 letters ABCD I would change the last 3 letters and keep the first one the same using N factorial I can tell that there will b 6 different combinations for them last 3 numbers (3x2x1=6) the six different combinations for A at the front are ABCD ACDB ADBC ACDB ACBD ADCB now I would do the same but this time with B at the front making sure I got 6 different combinations and then C and D then after I have done all of this I count my total combinations and make sure it is divisible by 6.
I used a different method of finding all the combinations for 5 letters what I done was I started with E as the started letter then I rite the combinations of 4 letters after it for example
EABCD
EACDB
EADBC
EACDB
EACBD
EADCB
EDCAB
EDCBA
EDBAC
EDBCA
EDACB
EDABC
ECBAC
ECBDA
ECDBA
ECDAB
ECABD
ECADB
EBADC
EBACD
EBCDA
EBCAD
EBDCA
EBDAC.
I then swapped around D with E for example
EABCD→ DABCE
EACDB→ DACEB
EADBC→ DAEBC
I done this until I had 124 different combinations or 24 different combinations for each starting letter this method went as good as the over one because it was confusing and took longer.
2 repeated letters
1 letter
No combinations
2 letters
AA
3 letters
AAB ABA BAA
4 letters
AABC AACB ABCA ABAC ACAB ACBA BAAC BCAA CBAA CABA CAAB
The rule for this is the amount of combinations with out repeats divided by 2 you can use N factorial so 5 letters would be 5!=5x4x3x2x1=120 =60
2
3 repeated letters
1 letter
No combinations
2 letters
No combinations
3 letters
AAA
4 letters
BAAA AAAB ABAA AABA
Using N factorial and dividing by six I can predict that 5 will have 20 different combinations because the rule is N! divided by 6.
4 repeated letters
1 letter
No combinations
2 letters
No combinations
3 letters
No combinations
4 letters
AAAA
Using N factorial and dividing by six I can predict that 5 will have 5 different combinations because the rule is N! Divided by 24.
CONCLUSION
Therefore, the rule for finding out number combinations is the number of letters factorial. So 4 would be 4! (4x3x2x1) which = 24 number of combinations. The rule for finding the combinations with a double or treble ECT. Is the number of repeats factorial divide the number or letters factorial so a 4 letter combination with two repeated letters for example abbc would be 4! (4x3x2x1) which equals 24 divided by 2! (2x1) which equals 2 so that means there will be 12 combinations for a 4 letter combinations with 2 repeats. Anther example could be an 8 letter combination with 4 repeated letters for example abcdeeee would be 8! (8x7x6x5x4x3x2x1) which equals 40320 dived by 4! (4x3x2x1) which equals 24. 40320 divided by 24 = 1680 so for an eight letter combinations with 4 repeats there are 1680 combinations
