The formula for finding the number of arrangements for a word with any number of letters but without any repeated letters is n! and the n is the number of letters in a word. The n! means that there are n letters to begin with. For the second letter there are n-1 letters to choose. For the third letter there will be n-2 letters to choose. For the forth letter there will be n-3 letters to choose and so on.
So the equation would be n x(n-1) x(n-2) x(n-3) and so on. This is how n! works.
Part 2
These are all the different arrangements of letters of Emma’s name and it has 4 letters with 2 repeated letters.. I have written them out in an order so it is clear. Set1 begins with E, set2 begins with M and set3 begins with A. There are 12 arrangements of her name.
Table 1
Now I will write out the different arrangements again but treating the 2 repeated Ms separately naming them M1and M2. This will help me find out why there are half the number of combinations for a word of the same length but with two repeated letters.
Table 2
EMMA has 4 letters in her name so it would be expected that she has 24 arrangements because 4! is 24 (the 4! as explained in Lucy) but this is not the case because she has two Ms in her name. If the two Ms in Emma’s name were switched then this would make no difference to the number of arrangements as the name would be the same. So there are only 12 arrangements due to the fact that there are two Ms in the name. The 2 Ms have 2! arrangements. For each arrangement there is another the same with the Ms reversed.
In table 2 I have arranged them so that the corresponding arrangement can be clearly seen on the other side. Sets 1-3 have the M1 in front of the M2 and sets 4-6 have the M2 in front of the M1. There are 24 arrangements and 12 identical pairs. This means that there is half the number of arrangements for a word with 2 repeated letters.
Using the results I can say that the formula for finding out the number of arrangements of Emma is 4!/2 This means (4x3x2x1)/(2x1) = 24/2 = 12
Using the evidence gathered so far the formula might be n! divided by 2 if there are 2 repeated letters. The unique arrangements will only be half the total number of arrangements i.e. 24/2!
The problem with this is that the formula would not work for a 5 letter word with 5 repeated letters because 5!/5 = 24 as there is only one arrangement. This means that this formula is not the complete thing yet.
Examples
I am going to see if I can find the formula which lets me know the arrangements of a word with a letter repeated more than once based upon my results for Emma’s arrangements. I will use further examples carry this out.
AAAAB is a 5 letter word with 4 repeated As and this will be used as an example.
Again I have put the results in a table but it is impractical to make a table like the second one for Emma because there are will be 120 arrangements. Set 1 begins with A and set 2 begins with B. I have shown the arrangements of AAAAB when the As are all treated the same . With Emma’s name there are 2 Ms which meant that the number of arrangements was divided by 2. I have found that there are 5 different arrangements for AAAAB. This means that the formula for AAAAB is n! divided by 24. This can be put simpler by saying 5!/4! = 5.
It can be seen how 120 is divided by 24 because the As when written fully to give 120 arrangements would be sorted out with 4 identical arrangements.
If the As were taken out and labelled A1A2A3A4 and treated as a separate word it would have 24 different combinations because of 4! This is why 120 is divided by 24 in 5!/4!
If all letters are the same in a 5 letter word then obviously you cannot change it. If there are 4 letters repeated such as AAAAB as shown in the previous example then there are 5 arrangments. If there are 3 repeated letters in a 5 letter word then there are 20 examples. For example AAABC has 3 repeated letters and has 5 letters. There are 20 different arrangements.
I have done the same for ABBB as I did for the AAAAB previously. It has 3 repeated Bs and it is a 4 letter word. It has only 4 arrangements and you would expect a 4 letter word to have 24 arrangements because of 4! but this is not the case. You need to divide the number of arrangements by 6 which is 3!
For each of the unique combinations there will be 3! arrangements which is why there are 6 for BBBA.
I will now do more examples of 4 letter words. A 4 letter word with all the letters the same obviously cannot change. If it has 3 repeated letters as seen in ABBB then it has 4 arrangements. If it has 2 repeated letters such as EMMA then is can be seen previously that there are 12 arrangements. If it has not repeated letters like LUCY then it has 24 arrangements as seen previously.
Results
These are results tables that I have made to strengthen the claim that n!/m! is correct. Each of these tables has a letter that is repeated for the formula to work. The tables prove that the formula works and I have checked that they are right.
For any 4 letter word
For any 5 letter word
I can now calculate that the formula is n!/m! where m is number of times the letter is repeated. This is connected to Emma because the arrangements were divided by 2 which is also the same as 2! With AAAAB the As are repeated 4 times so it means 120/4!
n!/m! works for a word of any length with a letter that has been repeated.
I can conclude from these results that the number of arrangements for a word with a letter that is repeated can be found out correctly using the formula n!/m!
Part 3
I will now investigate the number of different arrangements of various groups of letters. I know that another formula must be found in order to predict the arrangements of a word with different letters that are repeated.
AABBC is a 5 letter word with 2 repeated As and 2 repeated Bs.
I have put the results in a table but it is impractical to make a table containing the alternate positions for the As, Bs and Cs in each arrangement because there will be 120 arrangements. Set 1-3 begins with A, Set 4-6 begins with B and Set 7-8 begins with C. I have shown the arrangements of AABBC where the As and Bs are treated the same. I have found that there are 30 different arrangements and this means that the formula for AABBC is 120 divided by 4. If the two As were switched around and the two Bs were switched it would make no difference to the actual arrangement of it.
The problem is that if the formula n!/m! was used then this wouldn’t work because it would mean 5!/2! which equals 60 as there are 5 letters and 2 repeated As.
In actual fact there are 30 different arrangements which means that n!/m! cant be used to find out the arrangements for AABBC. The reason this happens is because this formula only works when a single letter is repeated and not multiple letters.
By doing 5!/2! it will only show the combinations if the only repeated letters were 2 As. It is 2! at the bottom because 2 As have 2! arrangements. There is another repeated letter in the form of 2 Bs. This would mean that it would be divided by 2! as well because 2 Bs have 2! arrangements.
Using this evidence I can change the formula so it will be n!/ (a! x b!) where n is the number of letters in the word. a! is the number of As repeated and b! is the number of Bs repeated.
This formula can be used to find out the number of arrangements of AABBC by doing 5!/ (2! x 2!) = 120/4 = 30. This is how many arrangements there are which shows this is the correct formula.
I came to this formula because as mentioned earlier there are two different combinations for AA. It can be A1A2 or A2A1 and the 2 As have 2! arrangements. This is also the same with the two different combinations for BB. It can be B1B2 or B2B1 and the 2 Bs have 2! arrangements. This means that 120 is divided by 2 to equal 60 and 60 is divided by 2 to give 30 arrangements.
Examples
AABB has 4 letters with 2 repeated As and 2 repeated Bs.
This is the table showing me that there are 6 different arrangements of AABB.
I can prove this by doing 4!/ (2! x 2!)
AABBCC has 6 letters with 2 repeated As, 2 repeated Bs and 2 repeated Cs.
I will not do a results table for AABBCC because it would be impractical as I have found out that doing 6!/ (2! x 2! x 2!) = 90 arrangements
The reason for this is because the 2 As have 2! Arrangements, the 2 Bs have 2! arrangement and the 2 Cs have 2! arrangements.
I will now do this with other words and arrangements to prove that the formula works.
After doing this table the formula for finding the arrangements of a word with any length, any number of repetitions and any number of repeated letters is n!/ (a! x b! x c! and so on.)
A, B, and C represent each repeated the number of times a single letter is repeated.
