KTIEA AETKI TEIKA IKTEA EAKTI
KIATE AEIKT TEAKI IKATE EAIKT
KIAET AEITK TKAIE IKAET EAITK
KIETA AIEKT TKAEI IAKET ETAKI
KIEAT AIETK TKEIA IAKTE ETAIK
KITAE AITKE TKEAI IATKE ETIKA
KITEA AITEK TKIAE IATEK ETIAK
KETIA AIKTE TKIEA IAEKT ETKAI
KETAI AIKET TAKIE IAETK ETKIA
KEATI AKETI TAKEI ITAEK EITKA
KEAIT AKEIT TAEIK ITAKE EITAK
KEITA AKTIE TAEKI ITKEA EIKTA
KEIAT AKTEI TAIEK ITKAE EIKAT
KAEIT AKIET TAIKE ITEKA EIATK
KTAEI AKITE TIAEK ITEAK EIAKT
I have not shown the arrangements for the names with more than 5 letters because I believe that there would be too many of these to write.
From this data I then attempted to calculate a formula. To do this I did the following:
I realised that the number of arrangements had something to do with the number of letters in the name.
I did trial and error, trying to divide the number of arrangements by the number of letters. This did not work.
I tried to find a common factor in the both the number of letters and the number of arrangements, this was also not very successful.
I realised that the more arrangements there were, the number of letters being changed got smaller. e.g.~
KATEI
KAEIT
KAITE
This showed me that the numbers being used in the equation must get smaller towards the end of the equation.
I tried to find the factorial of the number of letters, the mathematical symbol for this is !
I found the factorial of 3 which to be calculated normally would be :
3 x 2 x 1 or ~ 3!
This when calculated, is 6. This is the same as the number of different arrangements I had found.
I found the factorial of 4 letters, the equation for this is 4! This worked out to be 24, also the same as the number of arrangements.
I found the factorial of 5, the equation for this being 5! This also worked out to be the same as the number of different arrangements, 120.
I had found the formula to work out the number of different arrangements for a name with none of its letters the same. The formula being:
a = x!
a = the number of arrangements x = the number of letters in the name and
! = the factorial of x
I can now predict that a 6 letter name without any letters the same will have 720 arrangements. The formula that I used to work this out is 6!. I also have worked out the number of arrangements for names with 6, 7, 8 and 9 letters.
If my previous calculations have been correct, this formula should work for these names also. I will not show any of the arrangements for these to prove my theory because there are too many to write.
Investigation 2
I am now going to investigate names which have 2 letters in them the same. I will show the different arrangements for certain names, and from this data try to calculate a formula.
I am using these names because they have 2 letters the same.
However, I cannot find a name which has 2 letters, and both of them being the same letter. Because of this I shall use AA. Although this is not a name it should still work as well as a name.
2 letter name ~ AA
AA
3 letter name ~ LEE
LEE EEL ELE
4 letter name ~ EMMA
EMMA EMAM EAMM MAME MEMA MMAE MMEA
AMME AMEM AEMM MAEM MEAM
5 letter name ~ ANNIE
ANNIE NNIAE NENIA IENNA EANNI
ANNEI NNIEA NENAI IENAN EANIN
ANEIN NNEIA NEIAN IEANN EAINN
ANENI NNEAI NEINA INNEA ENNAI
ANIEN NNIEA NEANI INNAE ENNIA
ANINE NNIAE NEAIN INEAN ENAIN
AINNE NIEAN NANIE INENA ENANI
AINEN NIENA NANEI INAEN ENIAN
AIENN NINAE NAIEN INANE ENINA
AENNI NINEA NAINE IENAN EINNA
AENIN NIANE NAENI IEANN EINAN
AEINN NIAEN NAEIN IENNA EIANN
I have not been able to show the arrangemants for the names with 7, 8 and 9 letters because this would take far too much time, but on from the data on this table I should be able to calculate a formula. This formula will tell me how to work out the number of arrangements for names with more than one letter the same.
This is the table to show my results:
I tried to find a formula to help me do this investigation without writing out all the different arrangements. This is how I went about it:
I realised that there were twice as many arrangements for the name when the arrangement started with the letter that was doubled.
Because of this something in the formula must be either divided or multiplied. This is because if a number is doubled, it is multiplied by 2. Division is the opposite of multiplication and therefore division may be used to too.
I knew that a four letter name without any letters the same would have 24 arrangements and that a four letter name with two letters the same would have 12 arrangements. So, for a four letter name with two letters the same, it has half the number of arrangements as a four letter name without any letters the same.
I also realised that a four letter name with two letters the same has half of its letters doubled. This shows that something in the formula must either be divided or multiplied by two.
I tried to see if the first formula I had calculated had something to do with this one. My first formula was: a = x!
I thought that x! would still be in the formula because the factoral of a number is needed to find the number of arrangements no matter how many letters in it are the same.
I now have a new formula, this is:
a = x!
2!
From this formula I can now predict the number of arrangements for names with 6, 7 and 8 letters with two of their letters being the same. However, I cannot think of an eight letter name with two letters the same so I shall have to use random letters. This will make no difference to the investigation and will not affect the outcome. These are:
I believe that from my formula and these results I can predict the results for names that have three letters the same. However as before, I cannot think of any names which have three letters the same. This will again, make no difference to the outcome of the experiment. These predictions are:
To back up my prediction I will show all the different arrangements of the letters in the table.
3 letter name ~ AAA
AAA
4 letter name ~ AAAB
AAAB
AABA
ABAA
BAAA
5 letter name ~ AAABC
AAABC BCAAA
AAACB CBAAA
ABCAA BACAA
ACBAA CABAA
AABCA CAABA
AACBA BAACA
ACABA BAAAC
ABACA CAAAB
ABAAC ACAAB
This proves that my prediction was right. The formula which I used to work these out is:
a = x!
3!
I worked out this formula by:
Linking it with the two other formulae which I have calculated.
I realised that there were three times as many arrangements of the name when the arrangement started with the letter which was tripled. This was the same in the other experiment, except that it was doubled.
So, again in this formula a three must be in it, and this must be the factorial of three, 3! So, I came up with a formula of:
a = x!
3!
I proved this correct as above. From my three formulae I can predict that to find the number of arrangements with 4 letters the same the formula is:
a = x!
4!
This is a table which shows all the different formulae for these experiments.
