Emma's dilemma.
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Introduction
Joe Grix 10H of Maths Coursework
For this investigation, I am going to investigate the number of different arrangments of letters in a word. An example of arrangments of a word can be JAM or MAJ etc, the same word but how many times the word can be mixed up to make different words or arrangments of the letters.
Here is a key to show the meanings to different letters and symbols
Letter  Meaning 
A  Number of arrangments 
L  Number of letters 
S  Number of letters the same 
+  Plus 
x  Multiply 
/  Divide 
  Subtract 
=  Equals 
The number of arrangments for words with the number of letters from 1 – 5, and no pairs of letters so there is only 1 same.
 J Number of arrangments = 1
 JO
 OJ Number of arrangments = 2
 JOE
 JEO
 OJE
 OEJ
 EJO
 EOJ Number of arrangments = 6


Number of arrangments = 24


24 arrangments so far. x 5 because there are 24 arrangments starting with ‘J’, so there will be 24 ways for each other letter ‘A,M,E,S’, and there are 5 letters including ‘J’.
5 x 24 = 120
Number of arrangments = 120
Table to show all of the number of arrangments with different number of letters, with only 1 letter the same.
Number of letters  Number of arrangments 
1  1 
2  2 
3  6 
4  24 
5  120 
Middle
The number of arrangments for words with the number of letters from 1 – 5, and 1 pair of letters so there is 2 same.
 JJ Number of arrangments = 1
 MUM
 MMU
 UMM Number of arrangments = 3
 EMMA
 EMAM
 EAMM
 AMME
 AEMM
 AMEM
 MMEA
 MMAE
 MEMA
 MEAM
 MAEM
 MAME Number of arrangments = 12
Number of arrangments = 24 + 36
 Here there are 12 arrangments starting with the letter ‘H’, as there are 3 single letters, and 1 pair, then 12 x 3 = 36, the number of arrangments starting with a single letter. 
123 4  Here there are 6 arrangments starting with there letter ‘R’, and second with the other letter ‘R’ there are 4 letters altogether that can be the second letter after the first ‘R’. so if there’s 6 arrangments with 1 letter in second after the first letter, then there will be 4 x 6 = 24 arrangements, because there are 4 other letters that can go second. 
= 60
The number of arrangments for words with the number of letters from 1 – 5, and 1 3 letters the same, so 3 same.
 JJJ Number of arrangments = 1
 NUNN
 NNUN
 NNNU
 UNNN Number of arrangments = 4

Number of arrangments = 20 
Conclusion
As each letter has its own number of arrangements i.e. there were 6 beginning with x, and 4 beginning with y, I think that factorial has to be used again.
As before, the original formula:
L! = the number of letters in the word
S! = the number of letters the same
From this I have come up with a new formula The number of total letters factorial, divided by the number of x's, y's etc factorised and multiplied.
For the above example:
A four letter word like aabb; this has 2 a's and 2 b's (2 x's and 2 y's)
So : 1x2x3x4 _ 24
(1x2) x (1x2) 4 = 6 different arrangements
A five letter word like aaaab; this has 4 a's and 1 b (4 x's and 1 y)
So: 1x2x3x4x5
(1x2x3x4) x (1)
= 120
24 = 5 different arrangements
A five letter words like abcde; this has 1 of each letter (no letters the same)
So : 1x2x3x4
(1x1x1x1x1x1)
= 24 = 24 different arrangements
A five letter word like aaabb; this has 3 a's and 2 b's (3 x's and 2 y's).
So : 1x2x3x4x5
(1x2x3) x (1x2)
= 120
12 = 10 different arrangements
This shows that my formula works:
L! = the number of letters in the word
X!Y! = the number of repeated letters the same
This student written piece of work is one of many that can be found in our GCSE Emma's Dilemma section.
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