Emma's dilemma
Extracts from this document...
Introduction
Emma’s dilemma
THAMILINI KUGARAJAH
CATERHAM HIGH SCHOOL
CENTER NO: 13321
CANDIDATE NO: 9191
Introduction:
This investigation is about finding the different arrangements of letters in various words. I am going to look at the number of different arrangements for words with all letters different, words with two letters the same and the words with three letters the same and so on.
 Part 1
I am going to look at the number of different arrangements of the letters of Lucy’s name.
LUCY




As it shown above, there are 24 different arrangements in four letter word.
CY
U
YC
UY
L C
YU
CU
Y
U C
A tree diagram can also be used as I have used above. Here we have used “L” as a first letter; however the same process can be used with other 3 letter, so we can multiply the number of letter by the number of possibilities, 6*4=24
Each starting letter has six possibilities. Therefore four letters altogether 6+6+6+6=24
You will get 24 possibilities altogether.
First of all we have got 4 possibilities (4 letters) and then 3 possibilities(by leaving first letter “L”) and then 2 possibilities 9by leaving first two letters “L”U”) after that one possibility which is the last letter . Therefore 4*3*2*1=4! Which is equal to 24, where as we can see this from tree diagram (which is shown by using different colours)
 Part 2
I am going to investigate the number of different arrangements of the letters of EMMA’S name.
EMMA


Middle
 AEBCD
 AEBDC
 AEDBC
 AEDCB
 AECDB
 AECBD
There are 24 arrangements starting just with one letter “A”. therefore there are five letters altogether, so 5*24=120
Consequently there are 120 arrangements altogether.
51 = 5*4*3*2*1
= 120 arrangements.
Now I am going to draw a table of results and see whoat relationship between them.
Table 1
Number of letters in the word  Number of different arrangements  observation 
1  1  1=1! 
2  2  1*2=2! 
3  6  1*2*3=3! 
4  24  1*2*3*4=4! 
5  120  1*2*3*4*5=5! 
5*24 =120 (cross multiple)
I have noticed that for the word with all letters different, the factorial of the number of the letters in the word is equal to the number of different arrangements.
Let “N”=number of letters in the word
“A”= number of different arrangements
Therefore N! =A
For example: a three letter word with all letters different
Prediction: for the word with six letters different.
Using pattern noticed
1) N*(N1) =A
6*5! =6*120
=720
Using the formula
 A=N!
A=6!
A=720
The calculations using formula and the pattern noticed both agree.
I am going to investigate words with two letters the same.
 HH – this has one arrangement
 TOO
OTO this has three arrangements
OOT
 A word with four letters with two letters the same will give the same number of arrangements as EMMA i.e. 24 arrangements
 The arrangement for a word with five letters the two letters the same is shown on the next page. I have used the word “PAPER” and I have numbered each of the letter “p “to make them different from each other.
This is to enable me to work out the total number of arrangements when all the letters are different and to compare it when two of the letters are the same.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 

 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 


There are 60 different arrangements altogether, however
PAPER
PAPER 1 arrangement
We can see two arrangements, when we swap that “P”,”P” but as they look the same there are 120 arrangements when we use the P as P AND P. however as they look the same we divide 120 by two which is equal to sixty arrangements.
Now I am going to put the results on the table and see what relationships between them.
The word with two letters the same
Table 2
Number of letters in the word  Number of different arrangements  observation 
2  1  2!/2 
3  3  3!/2 
4  12  4!/2 
5  60  5!/2 
Conclusion
Therefore N! /R! =A
Prediction:
Testing the formula
N! R! =A
e.g. =MUMMY
=5!/3!=120/6=20 different arrangements
This proves my formula is right and it would work for the word that has any number of letters the same.
Multiple number of letters the same
Now I am going to look at the group of letters that has more than one letters the same, example: “ABBCC, AABBCC,” etc
First of all I am going to look at “AABB”
 AABB
 ABAB
 BABA
 BBAA
 BAAB
 ABBA
There are six different arrangements.
I know numerator should be N!. I know this from the previous section.
From this section I have noticed that for the word with multiple number of letters the same..
The number of different arrangements is equal to the factorial of the different no of multiple letters the same.
Example: there are four letters in the word and two different the same
So 4! /2!*2!
Let “N”=number of letters in the word
“A”= number of different arrangements
R
Q =number of letters the same.
p different number of letters appears this number of time e.g. P times.
In a “N” letter word same letters appears “P, Q, R” times. So the number of ways we could arrange is
N! /P!*Q!*R! =A
Conclusion
In this investigation, I have learned how to find the different number of arrangements for the word with any of different letters or the word with different number of letters the same.
I found three different formulae:
 Word with different letters N! /A
 Words with more than one letter the same N! /R! =A
 Words with multiple letters the same N! /R!*P! =A
This student written piece of work is one of many that can be found in our GCSE Emma's Dilemma section.
Found what you're looking for?
 Start learning 29% faster today
 150,000+ documents available
 Just £6.99 a month