Explanation-
This is because once you have picked one letter there are then only two more letters and then one letter. This means that you get 3x2x1 and this gives you 6 which is equal to n!
This formula will allow me to work out the number of combinations of any word without a repeated letter by using this basic idea I will be able to modify it in order to discover the more complicated combinations, for example words with multiple repeats
One Repeated Letter-
2 letters- AA
3 letters- AAB
ABA
BAA
4 letters- AABC
AACB
ABAC
ABCA
ACAB
ACBA
BAAC
BACA
BCAA
CAAB
CABA
CBAA
No. of letters 2 3 4
No. of combinations 1 3 12
This gives the formula = n!/2!
Explanation-
If you have a double letter it means that each possible combination from before will have one that is exactly the same this time e.g. ACB and CAB will both be AAB and this will be true for all the possible combinations so you will have to divide the number of possible combinations, n!, by the factorial of the number of repeated letters in this case that is 2!.
One Triple Letter-
3 letters- AAA
4 letters- AAAB
AABA
ABAA
BAAA
5 letters- AAABC
AAACB
AABAC
AABCA
AACBA
ABAAC
ABACA
ABCAA
ACAAB
ACABA
ACBAA
BAAAC
BAACA
BACAA
BCAAA
CAAAB
CAABA
CABAA
CBAAA
No. of letters 3 4 5
No. of combinations 1 4 20
This gives the formula = n!/3!
Explanation-
If you have a triple letter, then you must divide the n! by 3! because there are 3 repeated letters and so the same will apply as above but in this case there would be three repeats of each combination with triple letters than if there were different letters, e.g. AAAB, if it was ABCD there would be 3! or six times the amount of combinations as there are with ABCD.
This is almost the same formula as above and has only changed in the fact that you have to divide by 3! instead of 2! and this leads me to think that I can easily form a generalised formula, this formula is n!/r! which is true because you need to divide the number of letters by the number of repeated letters because each repeated letter removes 1! options and so three repeated letters removes 3! options .
The number of possible combinations is now very high and so I will not put my results into a table because it would make the investigation longer than I am allowed to make it.
Two Double Letters-
No. of letters 4 5 6
No. of combinations 1 5 30
This gives the formula = n!/2! x 2!
Explanation-
If you take more than one set of double letters then the formula will be n!/the number of letters of each repeat times each other
because for each set of double, if you treated it as different letters you would have n! but because you have the two doubles you have the effect of a double i.e. dividing by 2! twice.
This tree diagram shows why you must divide by 2! each time because each repeated letter can be swapped over twice.
Multiple Repeated Letters-
For this we need to use a generalised formula for the one we use in the above example, this formula is n!/r! x s! where r equal the first group of repeated letters and s equals the second group of letters as we need to divide the factorial of the number of letters by the factorial of the number of each repeated letters multiplied together.
For this you do not know how many possible combinations there may be until you have the letters, but it uses the general formula above, this being n!/ the number of letters of each repeat times each other. For example xxxyyyyzz, this has 9 letters so if the letters were different you would have n! ways. In this case though x is repeated 3 times, y is repeated 4 times and z is repeated 2 times so in this specific case the formula would be 9!/3!x4!x2!.
Random Selection of Letters-
The other possible variable is the results you get when you take the letters you use from a bigger group and see the number of combinations you achieve by using this method. For example if you have a group of five letters and each time you pick three of these letters what will the pattern be and more importantly the formula.
Example-
ABC
AB
AC
BA
BC
CA
CB
This gives six possibilities
This is what you get if you pick two letters out of three. This example can be extended to if you are picking two letters out of five.
ABCDE
AB
AC
AD
AE
BA
BC
BD
BE
CA
CB
CD
CE
DA
DB
DC
DE
EA
EB
EC
ED
This gives 20 possibilities
If you were to pick two letters out of five letters, then the first time you pick a letter there would be three choices and then two choices, this is also the same as saying 5x4x3x2x1/3x2x1 so this in turn can be written as 5!/3!. We now need to generalise this formula and in order to do so we must find out how we arrived at the 3!. If we subtract the number of letters picked from the number of letters in total we can get the bottom line of the formula by doing (n-r)!, this means we get the whole formula n!/(n-r)!
Random Selection of Letters without order
In this process I am going to assume that AB is the same BA, this will obviously reduce the number of possible selections, after writing out the possibilities for 5 different letters and 6 different letters I quickly saw that the number was just divided by two, thus giving the general formula (n(n-1)/2
Development of formula-
1st - n! – Where n = number of letters
2nd - n!/ r! – Where r =number of repeated letters
3rd - n!/ r!xs!xt! Where r= 1st group of repeats s= second group and so on
4th - n!/(n-1)! This an be extended to form the formula n!/(n-r)! – Where r = number of letters selected from n