Year 10 Maths Coursework
Emma's Dilemma
Emma is playing with different arrangements of her name.
One arrangement is
EMMA
A different arrangement is
MEAM
Another arrangement is
AEMM
I investigated the total number of different arrangements of the letters of Emma's name. I worked out all the arrangements and come up with the following
EMMA MAME MMAE AMME
EMAM MAEM MAEM AMEM
EAMM MEAM MAEM AEMM
From this I can see that there are 12 possible arrangements. It was also interesting to see that when starting with each different letter in Emma's name, there was the same amount of arrangements for each. We had to account double the M's with there being two M's in Emma's name and so their was two times as many arrangements for that letter.
Emma has a friend named Lucy. When investigating the number of different arrangements of the letters of Lucy's name, I came up with the following arrangements.
LUCY CULY ULCY YUCL
LCUY CLUY UCLY YCUL
LYUC CYUL UYLC YLUC
LYCU CYLU UYCL YLCU
LCYU CLYU UCYL YCLU
LUYC CUYL ULYC YULC
The total number of different arrangements for the letters in Lucy's name is 24. It is interesting to see that even though the two names both have the same amount of letters in them, the number of arrangements of the letters comes to different totals. We can see that this is due to the double M in Emma, meaning that some arrangements would not be counted as we would merle swap the two M's which does not count as an alternative arrangement.
We can also see how each letter at the start has the same number of arrangements. By seeing that this is the case in both Emma's and Lucy's names it is important to make sure that this rule applies to all other names. To show that this rule works, I checked it with a three-letter name. To begin with I took all the arrangements starting with the same letter for the three-letter name of Amy. These were
AMY
AYM
If my prediction was right then
No of letters in word (L) x arrangements for one letter = total number of arrangements (A)
3 ( A + M + Y) x 2 (AMY + AYM) = 6
I checked this by finding out the rest of the arrangements
AMY + MAY YMA
AYM MYA YAM
The total number of arrangements was 6 showing that the prediction and the rule also works with the three letter word Amy. This helps to make finding the number of arrangements a lot easier if you have the information about one letter. However, finding all the arrangements of one of the letters still takes time especially with a long name. To find out a quicker way I firstly looked into some other arrangements of letters in names of different length in order to collect more information. For the time being I decided to look just at names with all letters different rather than words like Emma with double letters, so as to keep it as simple as possible.
It was easy to show that the one-letter name J only had the arrangement of
J
I then looked at the arrangements of the two-letter name of Jo which were
JO OJ
Jo had two different arrangements.
I have already done three and four-letter names by using Amy and Lucy and so I then looked at the five-letter name of James. I decided just to look at the arrangements beginning with the first letter of J and then use my rule in order to find the total number more quickly. I came up with the following
JAMES JAMSE JASME JASEM JSMEA JSMAE
JSAME JSEMA JEASM JEAMS JESAM JESMA
JEMSA JEMAS JSAME JSEAM JAEMS JAESM
JMEAS JMESA JMASE JMAES JMSAE JMSEA
I can see that the arrangements for James starting with the first letter come to a total of 24. I applied this to my rule.
L x no of arrangements for one letter = A
5 x 24 = 120
This gives me all the information that I need about one, two, three, four and five letter names with all different letters. In order to see my information more clearly, I put the information of what I had into a table.
No of letters No of arrangements Increase
1 1
2 2 1
3 6 4
4 24 18
5 120 96
From the table I could also see by how much the arrangements increase. The increase between two numbers is shown next to the higher number. We can see the general trend that as the number of letters goes up so does the number of arrangements and so does the increase between the number of arrangements. This however does not help us with a way of finding the number of arrangements ...
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No of letters No of arrangements Increase
1 1
2 2 1
3 6 4
4 24 18
5 120 96
From the table I could also see by how much the arrangements increase. The increase between two numbers is shown next to the higher number. We can see the general trend that as the number of letters goes up so does the number of arrangements and so does the increase between the number of arrangements. This however does not help us with a way of finding the number of arrangements for every name.
For this we have to notice the relationship between the number of letters and the number of arrangements. By looking carefully I saw that the multiple of the
Number of letters (L) x previous number of arrangements = total number of arrangements (A)
Looking at any of the numbers of letters in names above can prove this prediction.
L x previous number of arrangements = A
e.g. 4 x 6 = 24
We see that this formula does work. From this we can continue the table without having to work out all the different arrangements. As long as we have the number of arrangements from the previous number of letters then we will be able to find the current number of arrangements. By this we can see that if we worked out the number of arrangements for six letters for example, then we would also be able to find it out for seven, eight, nine, ten and so on. By thinking of it like this I realised how this originally relates all the numbers back to the name with only one letter. By being able to see that 1 letter = 1 arrangement, we should then, in theory be able to simply find the infinitive number by working on through the other numbers.
For instance, No of letters Formula No of arrangements
= 1
2 2x1 = 2
3 3x2 = 6
4 4x6 = 24
5 5x18 = 120
We see that by using the original 1letter = 1 arrangement, we can obtain the same information as we obtained by collecting the information by writing out each arrangement and that the prediction was correct. By looking at this information, you can also see how you do not necessarily need to obtain the previous number of arrangements in order to know how many arrangements a name has. It can be simplified by continuing the information.
e.g. I want to find the total number of arrangements for a five-letter name. I could EITHER work it out using rule whereby I use the previous number of combinations like so:
2x1=2, 3x2=6, 4x6=24, 5x24=120 = total number of combinations = 120
OR, I could simplify this. This would be to remove the answers part seeing as, if I was to use a calculator, this is not needed and I could use the following, simpler method.
x2x3x4x5=120
We can see that the same sums are taking place as we have the same answer. To prove this prediction we can demonstrate with a different number of letters. This is the first method for a four-letter name.
2x1=2, 3x2=6, 4x6=24
The second method:
x2x3x4=24
Again the same results come up proving the prediction to be correct and showing how this is a simpler version of the same sums. We can see that for any number of letters we still go through exactly the same numbers until we reach the number of our letters. The numbers we are putting into the sum represent the number of letters in the word. If you have a ten-letter name for example, then your sum would continue until you multiplied your answer by 10.
The formula I had proved correct for any letter with all letters different was
L! (Factorial)
On a calculator, the sum is made even easier by using the factorial button. It is commonly marked as ' x! ' and is a simpler way of writing 1x2x3x4x5...etc to whatever number. It computes the factorial of this specific number.
With this formula, I decided to go back to the four-letter name Emma, with it's double 'M', to see if the same rule applies. From counting the number of different combinations earlier I had a figure of 12 different arrangements. I then tried the formula.
L! = total number of arrangements (A)
4!=24
We see how the rule which worked with Lucy, doesn't work for Emma. It is easy to see that this is due to the double letter. I decided to look into other names of different lengths with a double letter in order to gather more information. The smallest amount of letters I could start with was a two-letter name. I chose MM.
MM
We can see that there is only one arrangement for this.
I then looked at a three-letter word, al the time with a double letter and chose BOB. I came up with the following arrangements.
BOB OBB BBO
Bob has three arrangements. We can see how the first rules that the individual letters at the beginning have the same amount of arrangements. The double letter merle means that there is two of the double letter.
Emma shows the arrangements for a four-letter name and I also looked at the five-letter name of Kitty. I looked into the arrangements beginning with the letter K and then was able to use my rule.
KITTY KTITY KTYIT KYTTI
KITYT KTTIY KTTYI KYITI
KIYTT KTIYT KTYIT KYITT
If we put this into the rule
L x number of arrangements for one letter = A
5 x 12 = 60
We can then put the information for double lettered words into a table to display the information more clearly.
No of letters No of arrangements No of arrangements
(double letters) (all letters the same)
1 - 1
2 1 2
3 3 6
4 12 24
5 60 120
We can see how the same general patterns occur in the table as seen with all letters different. The increase between two numbers is shown next to the higher number. We can see the general trend that as the number of letters goes up so does the number of arrangements. When drawing out the table for names with all letters different, I included a column showing the increase between each number of arrangements. I decided not to include this in this column, as it did not play any important role. I have included the information about names with all letters the same so that I can easily compare the two and look for differences.
When looking at the information in the table, it is easy to see how we are able to use the same rule
Number of letters (L) x previous number of arrangements (PA) = number of arrangements (A)
e.g 4 x 3 = 12
It is also important to note how we can compare double lettered words with words with all the letters different. We can see that double lettered words have half the number of combinations as single lettered words.
The dash under number of arrangements for double lettered words for a one letter word is present to show how the minimum amount of letters in order for a letter to be doubled is two.
When looking into the formula, I firstly tried to apply the rule
L! = A
e.g 4! = 24
This is the incorrect answer. However, notice how the correct answer is half of what the factorial rule gives us. I predicted that this was common with all the other examples and tried it out.
I tried the first rule L x PA = A
5 x 12 = 60
and then the second L! = A
5! = 120
Again the answer is not correct but again we do see that the correct answer is half or the other. This is very important to note, as well as seeing that what we do get is the answers we had for single lettered words.
With this information I can predict that the rule for double lettered words is.
L! / ( divided by) 2 = A
To test that this works, I tested it with a three latter, double lettered name.
L! / 2 = A
3! / 2 = 3
This is the correct answer proving that the rule will work. I now needed to look into words with more than a double letter and looked into triple letter words. By choosing the next one on in step, I would have information that flowed and so would not have gaps. I began by choosing the minimum amount of letters to make a triple-letter word. I chose AAA and looked into its different combinations.
AAA
It is easy to see that it only has one combination.
I then looked into the four-letter word of AAAB and worked out its combinations.
AAAB AABA ABAA BAAA
It has four combinations, each letter at the beginning again having the same number of combinations.
I still needed more information and so I looked at the combinations for the five-letter word of AAABC.
AAABC ACAAB ACBAA BAAAC CAAAB
AAACB ABAAC ABCAA BAACA CAABA
AACAB ACABA AABCA BACAA CABAA
AABAC ABACA AACBA BCAAA CBAAA
In total there are 20 arrangements, again following the pattern of arrangements of
L x PA = A
5 x 4 = 20
With this information I created a table to clearly show all my information, including the information earlier collected from single and double lettered words.
No of letters No of arrangements No of arrangements No of arrangements
(triple letters) (double letters) (all letters the same)
- - 1
2 - 1 2
3 1 3 6
4 4 12 24
5 20 60 120
Again we can see how ' - ' represents where we are unable to include a figure, as where in double letters, in triple letters, we can not create words one and two letters long while still including a triple letter.
What I noticed is how the triple lettered words are a third of the double lettered words. We can also continue this to say that the double lettered are half of the single letters. We can write this in another way by saying that
x 2 x 3...etc (to number of same letters in word) = A
We can see that this is again the factorial. If I predicted this as the rule then I will check it with the five-letter word of ABBBB, with four letters the same.
My prediction would be
20/2 = 60, 60/3 = 20, 20/4 = 5
I therefore predict there to be 5 different arrangements. To check this is right, I then found out all the different arrangements
ABBBB BABBB BBABB BBBAB BBBBA
There are 5 different combinations in total, proving that this formula is correct.
Instead of working through each stage of number of letters, we can again simplify.
Step 1. We need to firstly apply the rule of
L! = A
Step 2. Once we have this answer, instead of dividing down between all the stages of double and triple letters, we can divide again by 1 x 2 x 3.... etc.
Step 3. We then have the formula of
Number of letters! (L!) / number of same letters! (SL!) = A
We can demonstrate this again with the five-letter word of ABBBB, with four letters the same
L! / SL! = A
5! / 4! = 5
This proves the rule to be correct.
I then can take this back to single lettered words. Here, the rule still applies. However, we can see that because all the letters are different, it makes no alteration to include the additional division section of the equation.
e.g L! / SL! = A
5! / 0! = 120
The question then arises of whether or not the rule will need to be altered if there were two or more lots of letters the same. To investigate this I looked at the example of AABBC. I firstly applied it to the rule
L! / SL! = A
5! / 2! + 2! + 1! = 1
I decided to add the separate numbers. By doing this, I produced the answer of 1. I then listed the arrangements.
AABBC AABCB BBAAC BBACA CAABB
AACBB ABABC BBCAA BABAC CABAB
ABACB ABCBA BABCA BACAB CABBA
ABCAB ACBBA BACBA BCAAB CBABA
ACBAB ACABB BCABA BCBAA CBAAB
ABBAC ABBCA BAACB BAABC CBBAA
Once again we see that this previous rule applies
L x no of arrangements for one letter = A
We then see that there are 30 combinations. This shows that my theory is not correct. I then decided to retry the formula by multiplying the separate numbers as follows
L! / SL! = A
5! / 2! X 2! X 1! = 30
This gives the correct answer of 30 arrangements. To make sure that this will work for all works, I predicted the number of arrangements for AAABB using the formula.
L! / SL! = A
5! / 3! X 2! = 10
To see if this is correct, I listed the arrangements.
AAABB AABAB AABBA BAAAB BAABA
ABABA ABBAA ABAAB BABAA BBAAA
This shows the 10 arrangements and shows that the formula works. Randomly choosing a name we have used previously will confirm it to be correct. I chose Kitty and put the name into the formula.
L! / SL! = A
5! / 1! X 1! X 2! X 1! = 60
This again is correct. T is interesting to see as well, that in all the words, the single letters, the 1! makes no difference to the equation. This explains why it was not needed to solve single letter words.
With any combination of letters, for instance Xs and Ys, I can now use my formula to work out all arrangements that can be made. To demonstrate, this, I will test three different arrangements.
. XYXXYYY
L! / SL! = A
7! / 3! X 4! = 35
I will now list all the arrangements.
XXXYYYY XXYYYYX XYYXYYX
XXYYYXY XYYYXYX XYXYYYX
XXYYXYY XYYYXXY XYXYYXY
XXYXYYY XYYXYXY XYXYXYY
XYYYYXX XYYXXYY XYXXYYY
These are all the combinations beginning with X. If I divide this by three because there are 3 Xs, I get the answer 5, which is how many combinations per letter. I can then add this into the formula to quickly find out haw many arrangements there are in total.
L x number of arrangements for one letter = A
7 x 5 = 35
This then gives us the correct total, proving the formula to work.
2. YXYYX
L! / SL! = A
5! / 3! X 2! = 10
Arrangements:
XXYYY XYYYX YYYXX YYXYX YYXXY
XYYXY XYXYY YXYYX YXYXY YXXYY
Again this shows that the formula works by showing ten arrangements.
Although on the previous test, I only wrote out a few arrangements and used the formula to complete the rest, I did not for this. This is because there are not as many for this example, and I feel safer by righting them out if it is possible. However, when there are a lot, it is not practical to write them all and so that is why I use the formula.
3. YYYXXYXY
L! / SL! = A
8! / 5! X 3! = 56
For this example it will be easier to work out the arrangements for one letter and work out the rest with the formula. This means that our total number needs to be divided by the number of letters. 56 / 8 = 7. I will use this to find out all the Xs, but first this number needs to be multiplied by however many Xs there are. 7 X 3 = 21. This means that I need to find 21 arrangements beginning with X.
Arrangements
XXXYYYYY XYYYYXYX XYYXYXYY
XXYYYYYX XYYYYXYX XYYXYXYY
XXYYYYXY XYYYXYYX XYXYYYYX
XXYYYXYY XYYYXYXY XYXYYYXY
XXYYXYYY XYYYXXYY XYXYYXYY
XXYXYYYY XYYXYYYX XYXYXYYY
XYYYYYXX XYYXYYXY XYXXYYYY
This shows 21 arrangements all beginning with X. It means that there is 7 arrangements for each letters. I will put this in my formula.
L x number of arrangements for one letter = A
8 x 7 = 56
This again proves that the formula works for any combination. By this I can say that the formula
Number of letters! (L!) / number of same letters! (SL!) = Arrangements (A)
Will give the correct number of arrangements for any given word or name.
Harriet Kemp
