Third Arrangement:
I will be investigating the name “RITA”
I have created 24 different arrangements using the name RITA. This is the same amount of arrangements found when rearranging LUCY’s name. This is because they are quite similar, as they both have four letters and each letter is different from each other. I will now draw a table showing my results:
Looking at the table above, I have realised a pattern between the letters and arrangements. If I multiply the number of letters by the number of arrangements made from the previous pattern, I will get the result for my current arrangements.
E.g. To find out how many arrangements made with a three lettered word, I will have to do this:
3 x 2 = 6
E.g2. To find out how many arrangements made with a four lettered word, I will have to do this:
4 x 6 = 24
I am not going to use this way, as it needs too much information to be completed and an algebraic rule for this is quite hard to figure out.
I have realised a second way of finding out the answer, which is more practical than the previous.
If I multiply the letters consecutively, I will be left with the number of arrangements.
E.g. To find out how many arrangements in a word with 4 letters, I had to do this:
1 x 2 x 3 x 4 = 24
E.g2. to find out how many arrangements in a word with 5 letters, I have to do this:
1 x 2 x 3 x 4 x 5 = 120
Instead of writing “1 x 2 x 3 x 4”, I can use the shorter term: “!”. This sign is called Factorial. This term is used when the numbers are being consecutively multiplied until the number before “!”. This is normally seen as n! “n” is the number of letters in the name.
E.g. 5! = 1 x 2 x 3 x 4 x 5 = 120
Here is a table showing what the next few arrangements would be.
I have succeeded in finding an algebraic rule for this investigation. It is quicker and more efficient to find the number of arrangements.
Investigation Three:
I am now going to be investigating words with two letters in it, the same. I will be seeing if these words will have the same amount of different arrangements as the previous investigation.
First Arrangement:
I will be investigating the name “ANN”
I was only able to create three different arrangements from the word ANN. I have noticed that there are fewer patterns in this word, than in a word with different letters.
Second Arrangement:
I will be investigating the name “LILY”
Like the name EMMA in my first investigation, I have created 12 words using LILY. From the results in this investigation (ANN and LILY), I have noticed that the total of arrangements are half of those from my second investigation (KAT and RITA).
E.g. KAT(6) and ANN(3). 6 halved is 3. This is could be because ANN and LILY have two of the same letters, causing their arrangements to be halved.
To show this, I will compare the two sets of results.
Compared to
Comparing the two tables, I have confirmed the fact that the arrangements made from words with two of the same letters are half or that made from words with all different letters. I can now make an algebraic equation to use, to find out the number of arrangements made with two of the same letters. Since we use n! To find out words with all different letters, all I need to do is put ½ in front of n! as it halves n! The final equation would be ½n!
E.g. To find out the arrangements made from a word with 5 letters (two of which are the same), I do:
5! = 120 ÷ 2 = 60
Here is a table showing what the next few arrangements would be.
Investigation Four:
I am now going to investigate words with 3 of the same letters, and seeing how many arrangements they make. I expect that these words will make fewer arrangements than the words with two of the same letter. I will choose a 3, 4 and 5 lettered word to get enough information to figure out a rule.
First Arrangement:
Words created from the 3 lettered word “LLL”
- LLL
I have only been able to make one arrangement, as even if I do mix the letters around, they will all be “LLL” anyway.
Second Arrangement:
Words created from the four lettered word “ELLL”
I have created four words from ELLL. This is less than LUCY and LILY, which also have the same amount of letters. This is because there are fewer different letters to arrange.
Third Arrangement:
Words created from the five lettered word “LILLY”
I have created 20 different arrangements with the word LILLY.
I will now create a table showing the results of this investigation and also the results from my second investigation, to compare the arrangements with all different letters to the arrangements with three of the same.
After analysing these results, I have recognised a pattern. The results I got from the arrangements made from words with three of the same letters are a sixth (1/6) of the arrangements made from words with all different letters. So to get this investigation’s results, I would just need to divide the results of the second investigation (which the formula is n!) by 6.
E.g. To find the different arrangements for LILLY, I will need to do this:
1 x 2 x 3 x 4 x 5 = 5!
5! = 120
120 ÷ 6 = 20
E.g2. To find the different arrangements for a word with 6 letters (three are the same), I will need to do this:
1 x 2 x 3 x 4 x 5 x 6 = 6!
6! = 720
720 ÷ 6 = 120
The formula should be:
To get this formula, I took the formula from my first results, which is n!, and put it over 6, seeing as to get these results, I would need to divide n! by 6.
Investigation Five:
Now, I will investigate the different arrangements made from words with repeated letters.
First arrangement:
Words created from the word XXYY
I have created 6 different arrangements with XXYY. I realise that there is not many arrangements made from this, as there is a fewer range of letters to rearrange.
Second arrangement:
Words created from XXYYY
I have created 10 different arrangements with XXYYY. This too, has not got many different arrangements.
From these two results, I already have collected sufficient information to create a formula.
This is a more complicated algebra equation, which had me thinking for quite long, but I finally came up with a solution.
To find the algebraic equation, I got the number of letters and factorized it. I will demonstrate with a word with 5 letters (XXYYY). I now have 5! I then counted how many of each different letters there are in this word.
In this example, I have the letters: XX and YYY. Since I have two of X, I will make that into 2! I have three of Y that I will make into 3! I will put them under 5! As that is how many letters there are. I now have:
I now need to figure out the sign in between 2! And 3! At first, I tried the additions sign.
This equation is wrong. This is because there are 10 different arrangements made from XXYYY, and not 15. So, Instead of using the addition sign, I will try using the multiplication sign.
This equation is correct. It makes sense, as 2! = 2 and 3! = 6. 5! = 120
2 x 6 = 12
120 ÷ 12 = 10
XXYYY makes 10 different arrangements, and the formula shown above proves it.
Here are other formulas for what might be the next arrangements.
XXXYYY = 6!
XXXYYYY = 7!
XXXXYYYY = 8!
Here is a table displaying the information.
Using this information, the formula should be:
The letters in the formula, X and Y represent the letters there actually are in the word. So if there was six X’s in the letter, it would be 6! I am sure this is the correct formula to work out the arrangements for this particular arrangement.
This concludes my maths coursework on word arrangement. Overall, I have learnt that there are easier ways of working out how many different arrangements a word can be made out of. I have found that doing this without a formula, or a rule, is much longer than doing it with a formula. I have also learnt how to find out formulas quickly and work out if they are right or not.