From these results I know that a 4 letter name, with no repeated letters gives 24 possible combinations, my initial thoughts where that you simply times the number of letters by 6 to get the number of possible combinations.
I will be investigating this is Part 3
Part 2
Emma
In this section I will be exploring the possible combinations of a name that includes a double letter.
Possible Combinations:
- emma
- emam
- eamm
- meam
- maem
- mame
- mema
- mmea
- mmae
- amme
- amem
- aemm
From this we can see that there are 12 possible combinations of the name Emma. This could mean that to find the number of combinations in a name with repeated letters, you take the number of cominations in that amount of letters without repeated letters and divide it by to.
For example
24/2 = 12
As Lucy had 24 combination and when adding a double letter the number of combinations were 12.
I will investigate into this further in Part 3
Part 3 – Further Research
Rohan
I know from investigation that Rohan has 120 combinations of his name, and has 5 letters in his name.
There for it is 24 x 5.
24 was the amount of combinations in a 4 letter name.
There fore a 6 letter name would have 720 combinations as it would be 720/6
This pattern works.
Also a 2 letter name like Jo has combinations of
Jo
Oj
To find this we can do 1x2 which = 2!
There fore to find a 5 and 6 letter name we can do
5= 5x4x3x2x1 = 5! = 120
6=6x5x4x3x2x1 = 6! = 720
This formula also works and we have proved it with the name Rohan. As Rohan has 5 letters in his name, and using 5! = 120 and we know that is the amount of combinations
We use! as to find the number of combinations in a persons name we times the number of letters in their name by the number of combinations of a name with 1 less amount of letters.
So to find the combinations in a 6 letter name we take the amount in a 5 letter name (120) and times it by 6.
To get 120 you would of times 24 x 5.
This is 1x2x3x4x5 and then again x6 to get the combinations for Rohan.
General Formula
From this we can use X! as the general formula, where X= the number of letters in the name.
Emma
As Emma has a repeated letter the combinations are halved.
There fore the formula could be 4!/2
But with a name with 3 repeated letters this would not work.
The formula for a name with repeated letters is
X!/R! where R= the amount a letter is repeated.
Using this, the word ABCBDB would be
6!/3!
If a name has 2 sets of repeated letters then you simply divide it by the amount of repeated letters. So a name like Hannah would be
6!/2!2! as there are 2 A’s and N’s.
This is true of any word. For example ABACABCBACABA would be
14!/6!4!3!
Also from using ! we can easily work out all the combinations of a name without having to write them all out:
To test this table we can take 9!
Which is 1x2x 3x4x5x6x7x8x9 which is 362880
Tom Edwards 10CHMY – Emmas Dilemma Page of