MEAM MMEA
MAEM
MAME
Total: 6 arrangements
Starting with letter “A”
AEMM
AMEM
AMME
Total: 3 arrangements
Total arrangements in EMMA’s name = 12
Arrangements in LUCY’s name
Total arrangements in LUCY’s name = 24
Finding the formula to calculate the maximum number of arrangements of a word with no identical letters.
From this investigation I found out that a word 4-letters long with no identical letters can be arranged 24 times.
To make this theory correct I’ll test the name JOHN.
Arrangements in JOHN’s name
Total arrangements in JOHN’s name = 24
Arrangements in JO’s name
Starting Letter “J”
JO
Starting Letter “O”
OJ
Total arrangements in the name JO = 2
I found out that the total arrangements of a name 2 or more letters long with no identical letters is equal to the number of arrangements there will be if a letter was taken out of that name times the number of letters in the actual name.
For example, the name JOHN has 4 letters but if a letter was taken out of this name it’ll have 3 letters, which has a total of 6 possible arrangements.
Therefore; 4(the number of letters in the name) X 6(the total number of arrangements in a 3-letter word) = 24
Using this theory I have also noticed a pattern as a letter gets added to a word.
Number of arrangements in the name:
JO= (2X1)= 2
JOE= 3X(2X1)= 6
JOHN= 4X(3X2X1) = 24
By looking at this pattern the arrangements of a 5-letter word will be = 5X(4X3X2X1) = 120
A 6 letter word with no identical letters would equal = 6X(5X4X3X2X1) = 720
These type of sequences are called number FACTORIALS, which are represented by the ! Button on the scientific calculator.
So the formula for calculating the number of different arrangements in a word with no identical letters is:
A=N!
A = total number of arrangements
N = the number of letters in the word.
Testing and predictions
To test if the formula N! is correct I will use it to figure out the number of total arrangements in the name JOE.
Prediction:
There are 3 letters in the name JOE.
So 3! = 3X2X1 = 6
Testing
My Prediction is correct!
What if a name had a pair of identical letters
From the name EMMA, I already know a 4 letter word with a pair (2 identical letters) of identical letters can be arranged 12 times.
To make this theory correct I will investigate the number of arrangements there will be in the name ANNE.
Arrangements in AMMIE’s name
Arrangements in ANN’s name
So far….
Using this graph I’ve found a pattern that a word with 2 or more identical letters is equal to the factorial of numbers of letters in the word then divided by the number of identical letters, which is 2.
For example, AMMIE is a 5-letter word. So 5! = 5X4X3X2X1 = 120
Now since there are 2 M’s in the name we divide 120 by 2 which equals 60.
So the formula for finding the total number of arrangements for a word with 2 or more identical letters is:
A=N!/X!
A = Total number of arrangements
N = Number of letters in word
X = The number of identical letters in the word.
Test and prediction
Prediction:
To test out my formulae; A=N!/X! I will test it out on the name OTOJO
In this name there are 5 maximum letters and 2 identical letters so:
A=5!/3! = (5X4X3X2X1)/(3X2X1) = 120/6 = 20
Test
Arrangements in OTOJO’s name
My prediction is right.
The maximum number of arrangements in XXXXXXXXXXYYYYYYYYYY
The formula, A=N!/X! wouldn’t work because the arrangement above has two pairs of identical letters 10 X’s and 10 y’s. So therefore, looking back at my previous patterns, I added a new value to the equation:
A=N!/(X!)(Y!)
A = Total number of arrangements in word
N = Total number of letters in the word
X = Number of identical letters in the word
Y = Another different number of identical letters in the word
So the row XX……XXYY…….Y has a total of 20 letters (the dots represents the letter X or Y depending the area in which it is surrounded by). There are 10 X’s and 10 Y’s so:
A= 20!/(10!)(10!)
N= (20X19X18X17X16X15X14X13X12X11X10X9X8X7X6X5X4X3X2X1) =
2432902008176640000
X= (10X9X8X7X6X5X4X3X2X1) = 3628800
Y = (10X9X8X7X6X5X4X3X2X1) = 3628800
A = 2432902008176640000/(3628800)(3628800)
A = 2432902008176640000/13168189440000
A = 184756
So the total number of arrangements in the row XX……XXYY…….Y is 184756
Prediction
To test the formula A=N!/(X!)(Y!) if its right. I will predict the number of arrangements of the words AAABBB.
N = 6 (total number of words in the word)
X = 3 (number of A’s in the letter)
Y = 3 (number if B’s in the letter)
So A=6!/(3!)(3!)
A = (6X5X4X3X2X1)/(3X2X1)(3X2X1) = 720/(6)(6) = 720/36 = 20
My prediction is 20
Test
I’m correct so the formula A=N!/(X!)(Y!) is legitimate.
Observation
Word with no identical letters
I observed that the total number arrangements of these types of words are found out if you multiply the number of arrangements there will be if a letter was crossed out by the number of letters in the name. For example, the total number of arrangements of the name John is 24 which is equal to the number of arrangements in Joe’s name times by the number of letters in John’s name. The formula I found for solving the maximum number of arrangements of these types of words is: N! (N Factorial)
Words with a pair or more identical letters
I observed that to find the maximum number of arrangements of a word with 2 or more identical letters you have to follow the same sequence as to find the arrangements as if there were no identical letters. Then you have to divide the answer you get by the number of identical letters in the name. For example, Emma has 4 letters with 2 identical. Treat it as a name with no identical letters, such as Lucy and you’ll get 24 arrangements. Divide it by the number of identical letters in Emma’s name and you’ll get 24/2=12. However, if you’re dealing with a bundle of letters such as XXXXXXXXXXYYYYYYYYYY, you have to time the number of letters factorial in each groups of the word. For example, this one has 10 X’s and 10 Y’s so you have to do 10!X10!. The formula I got for these types of names are A=N!/X! and A=N!/(X!)(Y!).
There are many ambiguous answers I got through my investigation.
Firstly, I tried the formula; A=N!/X! on the name Lucy. Since there wasn’t any identical letters in the name, I substituted X for 0 which gave me the whole answer of 0. This gave me the idea that the A=N!/X! only suitable for words with 2 identical letters.
Secondly, I tried the same formula for the rows; XX……XXYY…….Y, I got the equation of 20!/10!, which gave me the ambiguous answer of 670442572800. This made me realise I was wrong so I created a new formula to help me figure this row out.
