One pattern I found from this was that if you divide the number of arrangements by the number of letters of a name, you would get the number of arrangements of the previous name.
For example: Lucy, a four-lettered name, has 24 arrangements. 24/4=6, which is the number of arrangements for the name Tom. This works for all names except J because no letters gives no arrangements. Tom has 3 letters and 6 arrangements. 6/3=2, which is the number of arrangements of the name Jo.
This gives us a formula of:
p=a/l
p= number of arrangements for previous name
a= number of arrangements
l= number of letters
I found that this formula should be changed so that you can find the number of arrangements for the next name only knowing the number of letters for the name and the number of arrangements from the previous name (p). I changed the subject of the formula to ‘a’, which is number of arrangements, by multiplying both sides with ‘l’ to leave ‘a’ on its own.
P=a/l
*l *l
lp=a
From this formula, we are now able to predict the number of arrangements for a 5-lettered name.
l=5
p=24
5*24=120
To prove that this is right, I found the arrangements of a 5-lettered name, Ahmed.
- AHMED
- AHMDE
- AHDME
- AHDEM
- AHEDM
- AHEMD
- AEHMD
- AEHMD
- AEMHD
- AEMDH
- AEDMH
- AEDHM
- ADMEH
- ADMHE
- ADHME
- ADHEM
- ADEMH
- ADEHM
- AMDEH
- AMDHE
- AMHED
- AMHDE
- AMEHD
- AMEDH
- HAMED
- HAMDE
- HADME
- HADEM
- HAEMD
- HAEDM
- HEADM
- HEAMD
- HEMDA
- HEMAD
- HEDMA
- HEDAM
- HMEDA
- HMEAD
- HMAED
- HMADE
- HMDEA
- HMDAE
- HDMEA
- HDMAE
- HDAEM
- HDAME
- HDEMA
- HDEAM
- MAHED
- MAHDE
- MADHE
- MADEH
- MAEDH
- MAEHD
- MHADE
- MHAED
- MHDEA
- MHDAE
- MHEAD
- MHEDA
- MEHDA
- MEHAD
- MEDAH
- MEDHA
- MEADH
- MEAHD
- MDEAH
- MDEHA
- MDHEA
- MDHAE
- MDAHE
- MDAEH
- EAHMD
- EAHDM
- EADHM
- EADMH
- EAMDH
- EAMHD
- EMAHD
- EMADH
- EMDAH
- EMDHA
- EMHDA
- EMHAD
- EDMAH
- EDMHA
- EDHAM
- EDHMA
- EDAMH
- EDAHM
- EHMAD
- EHMDA
- EHDMA
- EHDAM
- EHAMD
- EHADM
- DAHME
- DAHEM
- DAMEH
100.DAMHE
101. DAEMH
102. DAEHM
103. DEAMH
104. DEAHM
105. DEHMA
106. DEHAM
107. DEMAH
108. DEMHA
109. DMEAH
110. DMEHA
111. DMHEA
112. DMHAE
113. DMAHE
114. DMAEH
115. DHEAM
116. DHEMA
117. DHAEM
118. DHAME
120. DHMEA
From this, we can see that the formula, lp=a works because the 5-lettered name Ahmed, gives 120 arrangements, as I predicted. However, the letter ‘p’ may not be very useful because you do not know the number of arrangements from the previous name in all cases. Now I must find a simpler formula for this. I wrote the arrangements out again.
J-1
JO-2
TOM-6
LUCY-24
I noticed that ‘l’ was fine, as it is in the formula but ‘p’ needed to be broken down a little. I then wrote down a formula for all the names substituting the values in for ‘l’, ‘p’ and ‘a'
JO 1*2=2
TOM 2*3=6
LUCY 6*4=24
From this, I could see that in the formula for the name Tom, ‘p’ which is 2, is the same as the calculation for Jo; therefore, the calculation for Tom’s name is 1*2*3. This works for Lucy’s name as well because ‘p’ in the calculation of the name Lucy is 6, which is the same calculation as the name Tom, which is now 1*2*3. This now gives us a new a new calculation for the name Lucy, 1*2*3*4. This is known as finding the factorial of a number (this function is represented as x! on a calculator). Therefore, the new formula for finding the number of arrangements for a name with all the letters being different will be
a=l!
To prove this, I again chose the name Ahmed.
Ahmed has 5 different letters. 1*2*3*4*5 OR 5!
=120 which is the number of arrangements the name Ahmed has.
This formula however, does not work in all cases. For example, the name Emma has four has four letters, the factorial of 4 is 24 but is not the number of arrangements for the name Emma. This is because Emma has 2 of the same letter, which is ‘m’. I will now try to investigate a formula for a name like Emma. I investigated the arrangements of 3 different names, which have 1 letter twice, and the rest of the letters different.
- EMM
- MEM
- MME
EMMA- (refer to first page) 12 arrangements
- SUSIE
- SUSEI
- SUESI
- SUEIS
- SUIES
- SUISE
- SSUIE
- SSUEI
- SSEUI
- SSEIU
- SSIEU
- SSIUE
- SIUSE
- SIUES
- SIEUS
- SIESU
- SISEU
- SISUE
- SESUI
- SESIU
- SEISU
- SEIUS
- SUIES
- SEUSI
- USSIE
- USSEI
- USESI
- USEIS
- USIES
- USISE
- UISSE
- UISES
- UIESS
- UEISS
- UESIS
- UESSI
- ISUSE
- ISUES
- ISEUS
- ISESU
- ISSEU
- ISSUE
- IESSU
- IESUS
- IEUSS
- IUESS
- IUSES
- IUSSE
- ESUSI
- ESUIS
- ESIUS
- ESISU
- ESSIU
- ESSUI
- EUSSI
- EUSIS
- EUISS
- EIUSS
- EISUS
- EISSU
I then produced a table of results for all the names I investigated.
From this table, I could see that there is a formula amongst the names that the names that are in bold. It is the factorial of ‘l’, divided by ‘x’. This therefore gives us the formula, a=l!/x
I could see that ‘x’ is not needed for the names that have all the letters different. I investigated a five-lettered name with three repeated letters. I added them to my table after I had found the number of arrangements.
- SUSIS
- SUSSI
- SUISS
- SIUSS
- SISUS
- SISSU
- SSSIU
- SSSUI
- SSUSI
- SSUIS
- SSIUS
- SSISU
- ISSSU
- ISSUS
- ISUSS
- IUSSS
- UISSS
- USISS
- USSIS
- USSSI
From the table, we can see that the number of arrangements for the name Susis does not agree with the formula a=l!/x. This is because 120/3=40 and not 20 which is the value of ‘a’. I then looked at other names in which the same letter occurs 3 times.
- EMMM 1. MMM
- MEMM
- MMEM
- MMME
I then arranged the names in size order and wrote ‘l’ factorial for each one.
MMM-1 3!=6
EMMM-4 4!=24
SUSIS-20 5!=120
I noticed that all these arrangements with 1 letter repeated 3 times were the quotient of l! and 6 because 6 is the factorial of 3 and 3 is the number of same letters. This gives us a new formula of a=l!/x!. I shall test this with a 5-lettered name with 4 same letters. Firstly, I will predict my answer, then I will write all the different arrangements.
l! =a l=5 5!= 120
x! x=4 4!= 24
- MMMME
- MMMEM
- MMEMM
- MEMMM
- EMMMM
From this we can see that the formula l!/x!=a works for names which contain one letter which is repeated.
This formula works because the number of arrangements for a name that repeats a letter will obviously be less.
Now I will investigate a general formula by seeing AABBBCD
b= another pair of same letters or more. C=number of different letters i.e. ‘c’
We know that the formula a=l!
x!b!c!
EMMA 4! / 2!*1!*1!=12
AHMED 5! / 1!*1!*1!=120
LUCY 4! / 1!*1!*1!=24
SUSIE 5! / 2!*1!*1!=60