There was now definitely a correlation between the number of letters in a name and the number of ways in which they can be arranged:
In the next half of my investigation, I attempted to explain this. After looking at the result more closely, I could see the reason was evident in their layout. For four-letter names, the results were laid out in four different divisions, according to the four different options initially available:
As one of these options had to be taken, each of these groups were then divided into a further three groups according to the number of characters left over:
Each of these was halved as there were only two letters left:
Until finally, there was only one decision left to make. Therefore, there were no groups left; each of the arrangements was unique:
LUCY ULCY CLUY YLUC
LUYC ULYC CLYU YLCU
LCUY UCLY CULY YULC
LCYU UCYL CUYL YUCL
LYUC UYLC CYLU YCLU
LYCU UYCL CYUL YCUL
Hence, to find out the number of different arrangements of the letters of Lucy’s name, one could simply have multiplied 4 by 3 by 2 by 1 (or, as it is more commonly known, “4!”) instead of listing every single possibility of the letters L, U, C and Y as I did. This applies for fewer or more letter names, words or combinations. To find out the number of different arrangements of a three-letter word, one could multiply 3 by 2 by 1 and to figure that of a five-letter word, one could multiply 5 by 4 by 3 by 2 by 1. Again, these can be abbreviated to “3!” and “5!” respectively. To put it into a more mathematical context:
N.B. The formula above continues in sequence according to the number of letters in the combination. Hereafter, this will be known as “n”.
The number of different arrangements of the letters of a combination =
n × (n-1) × (n-2) etc.
or
n!
Part 2
My second assignment was to investigate the number of different arrangements for the letters in Emma’s name. I listed each arrangement as I did before. However, I did make some changes; after splitting the name up into the letters E, M, M and A, I decided to put a number next to the two M’s so as I could identify them later on (I would do this with any letters that occurred two times or more in a name hereafter). Having done that, I went on to continue as before. In all, I found a total of 24 different arrangements in which the M’s were unique:
From there, I went on to list the arrangements in which the M’s were interchangeable. Here, I found 12 different arrangements for Emma’s name:
I discovered 12 different arrangements in which the two M’s in Emma’s name could be exchanged and 24 in which they were individual. This tells me that the prior was half of the latter. Looking at my results now, this is quite understandable. I can see that for every two arrangements in which the M’s were separate, there was one of the opposite. For example, EM1M2A and EM2M1A could just as easily have been EMMA.
However, to verify the results, I decided to have a look at a similar name “Zeek”. Again, I found 24 different combinations where the M’s were identifiable and 12 where they weren’t so there shouldn’t be a problem:
I then decided to see if there was a trend between the number of repetitions there were in a word, and the number of different arrangements, in which those repetitions were interchangeable, that the letters in that word could form. In order to do this, I kept the length of the words constant (i.e. at four characters per word) so as it would not interfere with the results I got.
I decided to look at the word “baaa”. I found that, as with the names “Emma” and “Zeek” before it, the letters in it could appear in 24 different arrangements where the repeated letters were separable. However, there were only 4 different arrangements in which they were not:
Again, to credit my findings, I investigated a similar word “mooo”. And, again, I found that the letters it is spelt of could fit in 24 different arrangements where the repetitions of the letter O could be identified and 4 where they couldn’t:
It seemed that there was indeed a correlation between the number of times a letter is repeated in a word and the number of different arrangements in which the letters of that word could appear.
I continued to look at the trend by investigating a word with even more repetitions “oooo”. As with all those before it, the letters in the word “oooo” could fit in a number of 24 different arrangements where the repetitions (in this case of O) were unique. The interesting thing was that, this time, the four O’s in “oooo” would only form one arrangement in which they were anonymous:
Once more, simply to confirm the results, I looked at a similar word “mmmm”. And, once more, I found that the results, as far as figures go, were identical. There were 24 different arrangements in which the repeated letters M could appear if they were identified and 1 if they were not:
Although it may have seemed at first that the number of different arrangements, where the two M’s in “Emma” were interchangeable, possible for the letters in the name were completely dependant on the number of duplicate letters in that name – hence “the number of different arrangements for the letters in a name = n!/a” a here and indeed from now on being the number of repetitions in a name – this was not the case as can be seen in the later results. It may have applied for “Emma” where 24 divided by 2 is undoubtedly 12 but it doesn’t apply for the word “baaa” – 24 divided by 3 isn’t 4 – or “oooo” either – 24 divided by 4 isn’t 1. The number of different arrangements for the letters in a name is, in fact, equal to n!/a!:
The number of different arrangements for the letters in a name = n!/a!
Like those in any other word, the letters in words with duplicate letters can be arranged in n! different ways – that is, to begin with. The presence of a repeated letter narrows this down. If there were two duplicate letters in a word but they were somehow separated, there would be 2! different arrangements for them, just as there would be if they were a word in themselves. The same applies if there were three identical letters and each of them was distinguishable, there would be 3! different arrangements for them. Hence, the number of different arrangements for the letters in a name equals n! something a!. The “something” in the equation can be seen in my investigation into the number of different arrangements for the letters in the word “mooo”. On the whole, there were only really four different arrangements for the letters in the word but for each of these, there were six identical arrangements where the individual O’s were arranged in a different way. Hence, the “something” is “divided by” and the formula for the number of different arrangements for the letters in a name is equal to the n! divided by a!.
Part 3
My third and final task was to investigate the number of different arrangements for various groups of letters. I began by looking at the name “Abba”. I found that there were 24 different arrangements for the letters in that name where the two repetitions were separated. However, there were only 6 different arrangements for the letters in “Abba” where they were not:
I looked at a similar word “deed” to confirm the results. Again, I found the same thing. There were 24 different arrangements for the letters in the word when the each of the two sets of repeated letters were identified but just 6 when they were not:
The formula “the number of different arrangements for the letters in a name = n!/a!” obviously didn’t apply in this case as there were two completely unrelated repetitions. Therefore:
The number of different arrangements for the letters in a name = n!/a!×b!
N.B. b referring to the number of repetitions of the second variety.
The b! works on the same principle as the a! before it; that, just as if they were words themselves, the duplicate letters can be arranged in a number of different arrangements, taking into account that they are distinguishable, according to the number of them there were. Hence, the number of different arrangements for the letters in a name is equal to n! divided by a! b! or a! multiplied by b!. If there were more repeated letters, this could be expanded. c would be used to represent the number of times the third letter is repeated and d for the next and so on.
