I also observed that Emma’s different combinations were half of Lucy’s.
As I have found that there were 24 arrangements for a 4 letter word with all different letters and that there were 6 combinations beginning with each of the letters. I predict that for a five letter word there will be 120 arrangements. For instance for Irena there would be 24 for I, 24 r, 24 for e and so on. 120 divided by 5 (number of letters) equals 24. Previously in the 4-letter word, 24 divided by 4 equals 6, the number of possibilities there were for each letter.
The different combinations with different letters
I can see that the numbers of different arrangements are going to increase as more different letters are used. So as a general formula for names with n number of letters all different I have come up with this formula. With Lucy's name; 1x2x3x4 = 24. With Irena; 1x2x3x4x5 = 120. This is called a factorial. There is a button on most scientific calculators which basically does this formula for you according to the number of letters in the word. If I key in (assuming all the letters are different) factorial 6, I get it gives me 720, which makes sense because 720 divided by 6 equals 120 which was the number of arrangements for a 5 letter word, and it continues to follow that pattern.
I am now going to find out how many different combinations there are when 2 letters are the same.
The different combinations with 2 letters the same
I figured out that the factorial nth for different combinations with two letters the same is n!/2!.
I noticed that the number of combinations with 2 letters the same was half the number of combinations with different letters.
The reason why I think this is because when the total number of letters that have a double letter in it are being arranged the double letter only counts as if it’s one letter. So therefore half of the combinations are reduced compared to the number of combinations with different letters.
I am now going to investigate the different combinations with 3 letters the same.
The different combinations with 3 letters the same
The factorial nth term is n!/3!
I’ve also noticed that if there are 3 letters with 3 the same then the number of combinations would be: 3! /3! =1 combination.
The reason why you divide by 3! is because the total number of letters that are the same is three, in the three letters. The factorial nth term formula is that you divide the total number of letters by the number of letters that are the same.
I am now going to investigate the different combinations with 4 letters the same.
The different combinations with 4 letters the same
The factorial nth term is n! /4!
I can state that if there are 8 letters with 5 the same then the number of combinations would be:
8! /5! which is 8*7*6*5*4*3*2*1 / 5*4*3*2*1=336
The different combinations with double, triple and etc letters the same
The factorial nth term for the number of possibilities with double, triple and etc letters the same is n!/(y!z!).
I noticed that the bottom two fractorial add up to the top fractorial, e.g: 4!=2!+2!. I also obsereved that the donomenator is how many letters there are the same and the numerator is how many letters there are all together
Also for example if I wanted to find out how many different combinations I could get with 2 double and 3 triple different letters together I would then do:
5!/(2!3!) which is 5*4*3*2*1/ ( 2*1*3*2*1)= 10 combinations.
