Ed
Whilst listing the various possibilities I will try and be as systematic as I possibly can.
ED DE 2 POSSIBILITIES
When I was listing out the different possibilities I split the word into two parts, as there are only two different letters. This is indicated above and so is the outcome.
I will now investigate how many possibilities there are for a 3 lettered word, with no repeats. The example I will use this time is ‘Raj.’
Raj
Whilst listing the different possibilities I will once again try and be as systematic as I possibly can.
RAJ ARJ JAR 6 POSSIBILITIES
RJA AJR JRA
In the effort of trying to be systematic I decided to split the word into three parts, as shown above. I split it into three parts as there were 3 different letters in this word, and my outcome is indicated above.
Now that I have worked out the how many possibilities there are for a 2 lettered word, a 3 lettered word, and a 4 lettered word, I think I may be able to figure a pattern out. Therefore I will list all my discoveries in a table on the following page.
Table of Discoveries
1 2 3 4 5
1 2 6 24 12
The Numbers on the top row indicate the words with different numbers of letters. For example, the number 4 can indicate the word ‘Lucy.’ The numbers on the bottom row indicate the various outcomes I obtained by using the method of listing. For example, the number 24 indicates that there are 24 different possible outcomes for a 4 lettered word, as it is placed under the number 4. The red arrow indicates a multiplication sign, and the blue arrow indicates an equals’ sign. This is the pattern I managed to work out. I will show you how to use this pattern below:
How TO Use The Pattern
I will show you how to use this pattern using an example. The example I will be using is the three lettered word. We know the outcome is 6, as I have worked this out previously, by using the listing method.
2 * 3
= 6 NUMBER OF POSSIBILITIES
I have colour coordinated this calculation so that I can explain it. The 2 is the previous answer I obtained, which can be checked in the table plotted above. The 3 is the number of letters in the word, and the 6 indicates all the various possibilities. As you can see, this answer is correct and can be double-checked in the table above.
I will also do another example to double-check that my method works, and that the pattern is correct. Therefore I will be using the four lettered word, and the outcome should be 24, as I have worked this out previously using the listing method. My working is shown below:
6 * 4
= 24 NUMBER OF POSSIBILITIES
This proves that my pattern is correct, and that the method works. This means that I can add on to my table of discoveries, using my method of working. Using the method I will now do a five lettered word, and then a six lettered word. My working will be shown below:
Five Lettered Word
24 * 5
= 120 NUMBER OF POSSIBILITIES
Six Lettered Word
120 * 6
= 720 NUMBER OF POSSIBILITIES
I will now show you a development of my table.
Whilst doing my working I also managed to find another pattern. This is shown on the following page.
1 2 3 4 5 6
1 2 6 24 120 720
1 * 1 * 2
1 * 1
1 * 1 * 2 *3
1 * 1 * 2 * 3 * 4
1 * 1 * 2 * 3 * 4 * 5
1 * 1 * 2 * 3 * 4 * 5 * 6
I then showed all my discoveries to my teacher, who then pointed out that what I had worked out is known as a factorial. I was then introduced to the factorial button on the calculator. This would enable me to workout the number of arrangements possible for a word with N number of different letters. I have come up with a conclusion of:
N!
When I got to this stage of my investigation I remembered that ‘Emma’ is a four lettered word, but the number of outcomes was only 12. Whereas ‘Lucy’ which is also a four lettered word had a different number of outcomes which was 24. This means that when there are repeated letters the number of possible outcomes varies. This led me to take my investigation a step further, and hence I will now have a fourth part to my investigation.
Part 4
In this part of the investigation I will be investigating the various outcomes of words with repeat letters, for example a 3 lettered word with 2 repeats, a 4 lettered word with 2 repeats etc. I also have an advantage as I have learnt about the factorial. I will be using ‘Emma’ to begin this part of my investigation, as it has 2 repeat letters.
Emma 12 POSSIBILITIES
This can be worked out by the following equation:
4! / 2
= 12
The reason for why I divided 4! by 2, is because it has 2 repeat letters. This gives me the answer 12, which is correct according my table of discoveries which has been shown previously. I worked this answer out by listing all the various possibilities.
Hypothesis
I am going to base my hypothesis on the information that I have already obtained. As we know that 4! / 2 gives us the correct answer, I predict that a five lettered word with three repeat letters will be:
5! / 3
= 40 40 POSSIBILITIES
The only way that I can prove this is by using the method of listing, as I have not found any trends whatsoever as yet. I will list all the various possibilities below, and will try to be as systematic as possible. The example I will be using is ‘AAABC’
AAABC
AAABC AACAB ACBAA BAAAC CAAAB
AAACB AACBA ACABA BAACA CAABA
AABAC ACAAB BACAA CABAA
AABCA ABCAA BCAAA CBAAA
ABACA
ABAAC
20 POSSIBILITIES
In the effort of trying to be systematic I decided to split the word up into five parts, as there are five letters to the word. The outcome obtain is shown above, and proves my hypothesis wrong. I will try to find a pattern like I did previously in part 3 of my coursework.
To work out a pattern I am going to use a constant number of letters. I will use a five lettered word, and then change the number of repeats. For example I will use a five lettered word with one repeat (ABCDE all different letters), a five lettered word with two repeats (AABCD), then a five lettered word with three repeats (AAABC), a five lettered word with four repeats (AAAAB), a five lettered word with five repeats (AAAAA), and finally a five lettered word with five repeats.
I know that a five lettered word with one repeat (ABCDE all different letters) has 120 different possibilities. This is because I have worked it out previously.
I also know that a five lettered word with three repeats (AAABC) has 20 different possibilities. This is because I have worked this out, and is shown above.
I only need to work out the number of possibilities for a five lettered word two repeats (AABCD), five lettered word with four repeats (AAAAB), and a five lettered word with five repeats (AAAAA). First I will work out all the various possibilities for a five lettered word with one repeat four times (AAAAA), this is because it is easier to find the possibilities for a word with more repeats, as the number of different combinations decreases. The example I will be using is ‘AAAAA’
AAAAA
AAAAA 1 POSSIBILITY
As indicated above there is only 1 possible outcome, and therefore the answer was straight forward and I didn’t have to use a systematic method.
I will now investigate now many possible outcomes there are for a five lettered word with one repeat three times (AAAAB). I will show my working below:
AAAAB
AAAAB AAABA AABAA ABAAA BAAAA
5 POSSIBILITIES
In the effort of trying to be as systematic as possible I decided to split the word up into five parts, as the word is five letters long. The number of outcome possible is indicated above.
I believe that there will be a lot of possibilities for the five lettered word and two repeats, so I will plot all the results I have obtained so far in a table. With these results I will try to work out a pattern and fill in the gap. This table of discoveries is shown below:
1 2 3 4 5
120 ? 20 5 1
I am going to try and use the same pattern as I used before in part three, but I will reverse on of the signs. The numbers on the top row indicate the number of repeats in a particular word. For example, the 3 indicates the three repeats in the word ‘AAABC.’ The numbers on the bottom row indicate the various outcomes that I have obtained previously by using the listing method. For example, the 20 indicates that there are 20 possible outcomes when there are 3 repeats in a five lettered word (AAABC). The red arrow will represent division, and the blue arrow will represent the equals sign like before. I will do a calculation to try and prove this pattern. The example I will use is the five lettered word with 4 repeats (AAAAB).
20 / 4
= 5 NUMBER OF POSSIBILITIES
I will do one more example jus to double-check that this pattern and method of working really works. Therefore I will use it to determine the number of possibilities for a five lettered word with five repeats (AAAAA).
5 / 5
= 1 NUMBER OF POSSIBILITIES
This proves that my pattern and method of working are correct, and I can now use them to fill the gap in my table, which is the answer for a five lettered word with 2 repeats. I will show my working on the following page.
Five Lettered Word With 2 Repeats
120 / 2
= 60 NUMBER OF POSSIBILITIES
I can now plot my table of discoveries with all the necessary information, as I have been able to work out what the gap in my table was. This table is plotted below:
Whilst doing part 3 of this investigation I found a pattern which is known as a factorial, and whilst working this pattern out I found that the pattern of factorials also works this time round. My findings are shown below:
1 2 3 4 5
120 60 20 5 1
1 * 1 * 2
1 * 1
1 * 1 * 2 *3
1 * 1 * 2 * 3 * 4
1 * 1 * 2 * 3 * 4 * 5
After I managed to get this far I knew that I could use the factorial button on my calculator. This meant that a five lettered word with four repeats could be expressed as:
5! / 4!
= 5 NUMBER OF POSSIBILITIES
As this equation worked I came up with generalised equation for my new findings, which is:
N!
T!
Using this equation I can find out the different number of possibilities for a word with any length with how ever many repeats of the same letter, for example ‘AAAAAAABCDEF.’ The working is shown below:
12! / 7!
= 95040 NUMBER OF POSSIBILITIES
This brings me back to my hypothesis, which I had proved wrong by using the listing method. I have now come to find the correct answer which is shown below:
5! / 3!
= 20 NUMBER OF POSSIBILITIES
After I got this far I thought to myself what if there are repeats of different letters, for example ‘AABB.’ This brought me to the fifth part of my investigation. This is also the last and final part of the investigation. This is shown on the following page.
Part 5
As mentioned above in this part of my coursework I will be investigating the various outcomes for words with more than one letter repeating. For example, ‘AABB.’
From previous work in the other parts of the investigation, I know that the first part of the equation for working this word out is going to be:
4! / ?
This is because there are four letters altogether in this word. Also from previous work I predict that the next part of the equation is going to be:
4! / 2! ...
This is true because the word has two repeats of the same letter ‘AABB.’ As it also has another two repeats I cannot be to sure what the next part of the equation is, but due to previous work I also predict that the next part of the equation is going to be:
4! / (2! * 2!)
= 6 NUMBER OF POSSIBILITIES
I was able to predict this because in part 4 of my investigation because there is a repeat I divided by a factorial, depending on the number of repeats. For example a four lettered word with 2 repeats (AABC) will be:
4! / 2!
As there are two letters repeated in this case (AABB) I decided to add on another 2!. I didn’t add the 2! as it would not make sense because it would have then been:
4! / 4!
Therefore I decided to multiply them together, hence giving me the equation mentioned above:
4! / (2! * 2!)
= 6
The only way to prove whether or not this is correct is by listing all the possible outcomes, and hence I have done so on the following page.
AABB
AABB BBAA
ABAB BABA 6 POSSIBILITIES
ABBA BAAB
I tried to be as systematic as I possibly could, and the number of outcomes is indicated above.
According to this my prediction seems to be correct, and hence I have managed to come up with another generalised equation, which can be used to work out the various combinations possible with a word with any length and more than one letter repeated. This is shown below:
AAA…A BBB…B
S T
N!
(S! * T!)
An example is ‘AAAABBB’
7! / 4! * 3!
= 35 NUMBER OF POSSIBILITIES
The equation can also be extended if necessary. This is done if there are more than two letters that are repeated, for example ‘AABBCCDD.’ The equation in this case will be:
N!
OR 8! / (2! * 2! * 2! * “
(S! * T! * U! * V!) = 2520
Conclusion
In conclusion I managed to find three different formulas which are developments of each other. As I worked on each part of the investigation I added a little part to the equation, and by the end of part five I have ended up with the final equation. This can be used to find all the possible combinations of any particular word, no matter how many repeats it has, nor how many letters are repeated. For example I can work out all the possible outcomes for a random word such as ‘MISSISSIPPI’ a shown below. Before I work out the suggested word I will show you the evolution of my equation.
- N!
- N! / S!
- N! / (S! * T!)
The working used to find how the number of possible outcomes for the word ‘MISSISSIPPI.’
Equation = N! / (S! * T! * U!)
= 11! / (4! * 4! * 2!)
= 34650
