Emma's Dilemma.
Introduction
The aim of this coursework is to investigate into the problem of different arrangements of a set of letters. By using the given guidelines and by widening the breadth of the operation I aim to find a rule or formulae that can represent and explain the situation. The basic parameters of the project are as follows:
"Emma is playing with arrangements of the letters of her name.
One arrangement is EMMA,
another is MEAM,
and another is AEMM.
Investigate the number of different arrangements of the letters of Emma's name."
This is the first task, which is then followed up by:
"Emma has a friend named Lucy.
Investigate the number of different arrangements of Lucy's name."
After these two tasks I can then lead the investigation in my own direction:
"Choose some different names.
Investigate the number of different arrangements of the letters of the names you have chosen."
An extension to the investigation is as follows:
"A number of Xs and a number of Ys are written in a row such as 'XX . . . XXYY . . . Y . .'
Investigate the number of different arrangements of the letters."
The last section will allow me even more freedom to change variables to find more complex solutions covering a greater variety of situations.
Finding
The mathematical term for this type of function where a set of values is rearranged to give the maximum number of possible combinations are called permutations. I will use this term through the rest of the investigation too.
Part One:
I will begin by manually working out the possible permutations of the letters in the name 'Emma'. To do this I have to distinguish between each character in the name. There are four characters ??E, M, M and A. However, there are two Ms and I must be aware of this repeating letter. I have found that if I replace the letters in the name and assign each to its own number then I can arrange combinations of the name numerically, thus avoiding errors from overlooking any combinations or from duplicating any permutations. Therefore:
E = 1 M = 2 A = 3
I have numbered both the Ms as 2, which indicates that they are indistinguishable. I can start with the lowest value combination of 1223 and go through up to 3221 (that is, EMMA through to AMME respectively):
Combination #
Combination
Letter Representation
2
2
3
E
M
M
A
2
2
3
2
E
M
A
M
3
3
2
2
E
A
M
M
4
2
2
3
M
E
M
A
5
2
3
2
M
E
A
M
6
2
2
3
M
M
E
A
7
2
2
3
M
M
A
E
8
2
3
2
M
A
E
M
9
2
3
2
M
A
M
E
0
3
2
2
A
E
M
M
1
3
2
2
A
M
E
M
2
3
2
2
A
M
M
E
As you can see my method of finding all the permutations allows them to be listed in ascending numerical order without missing out any. Altogether there are 12 possible arrangements of the letters in the name 'Emma' with no permutations missing or duplicated.
Part Two:
Moving onto the second task on the list, I have decided to adopt the method used in the first part in order to find all the possible combinations of the letters in the name 'Lucy' - i.e. by numbering the letters to avoid errors. This time there will be four individual numbers because there are four letters in the name, all of which are different. Therefore:
L = 1 U = 2 C = 3 Y = 4
Below are all the possible combinations:
Combination #
Combination
Letter Representation
2
3
4
L
U
C
Y
2
2
4
3
L
U
Y
C
3
3
2
4
L
C
U
Y
4
3
4
2
L
C
Y
U
5
4
2
3
L
Y
U
C
6
4
3
2
L
Y
C
U
7
2
3
4
U
L
C
Y
8
2
4
3
U
L
Y
C
9
2
3
4
U
C
L
Y
0
2
3
4
U
C
Y
L
1
2
4
3
U
Y
L
C
2
2
4
3
U
Y
C
L
3
3
2
4
C
L
U
Y
4
3
4
2
C
L
Y
U
5
3
2
4
C
U
L
Y
6
3
2
4
C
U
Y
L
7
3
4
2
C
Y
L
U
8
3
4
2
C
Y
U
L
9
4
2
3
Y
L
U
C
20
4
3
2
Y
L
C
U
21
4
2
3
Y
U
L
C
22
4
2
3
Y
U
C
L
23
4
3
2
Y
C
L
U
24
4
3
2
Y
C
U
L
This time there are 24 different possible combinations of the letters L, U, C, and Y. This is double the number of combinations for the letters in Emma's name. Both names contain four letters, however, the only difference is that Emma's name has got the letter 'M' twice and Lucy's name has all four letters different. So if the name contains a repeating letter then it will have fewer combinations than a name with all different letters, but with the same number of letters.
Part Three:
I will continue by individually investigating each type of arrangement - ones with and ones without repeating letters in the name. I will start by using a name with one letter then work upwards by adding more letters to try and find how the number of arrangements varies with the number of letters in the name.
With no repeating letters...
One Letter: 'J'
There is only one letter and it can only go in one position so there is only one possible arrangement.
Two Letters: 'JO'
J = 1 O = 2
Combination #
Combination
Letter Representation
2
J
O
2
2
O
J
With two letters there are two possible permutations. Either the first letter followed by the second or the second letter followed by the first.
Three Letters: 'LEN'
L = 1 E = 2 N = 3
Combination #
Combination
Letter Representation
2
3
L
E
N
2
3
2
L
N
E
3
2
3
E
L
N
4
2
3
E
N
L
5
3
2
N
L
E
6
3
2
N
E
L
With three letters the pattern of the number of permutations starts to widen out. Altogether there are now 6 different possible combinations for ...
This is a preview of the whole essay
Combination
Letter Representation
2
3
L
E
N
2
3
2
L
N
E
3
2
3
E
L
N
4
2
3
E
N
L
5
3
2
N
L
E
6
3
2
N
E
L
With three letters the pattern of the number of permutations starts to widen out. Altogether there are now 6 different possible combinations for a three-letter name.
Four Letters: 'LUCY'
I have already found the number of possible permutations for a name with four letters (Lucy) in part two. There are a total of 24 permutations, which shows the pattern widening out even more.
Five Letters: 'KACIE'
K = 1 A =2 C = 3 I = 4 E = 5
Combo #
Combination
Letter Representation
2
3
4
5
K
A
C
I
E
2
2
3
5
4
K
A
C
E
I
3
2
4
3
5
K
A
I
C
E
4
2
4
5
3
K
A
I
E
C
5
2
5
3
4
K
A
E
C
I
6
2
5
4
3
K
A
E
I
C
7
3
2
4
5
K
C
A
I
E
8
3
2
5
4
K
C
A
E
I
9
3
4
2
5
K
C
I
A
E
0
3
4
5
2
K
C
I
E
A
1
3
5
2
4
K
C
E
A
I
2
3
5
4
2
K
C
E
I
A
3
4
2
3
5
K
I
A
C
E
4
4
2
5
3
K
I
A
E
C
5
4
3
2
5
K
I
C
A
E
6
4
3
5
2
K
I
C
E
A
7
4
5
2
3
K
I
E
A
C
8
4
5
3
2
K
I
E
C
A
9
5
2
3
4
K
E
A
C
I
20
5
2
4
3
K
E
A
I
C
21
5
3
2
4
K
E
C
A
I
22
5
3
4
2
K
E
C
I
A
23
5
4
2
3
K
E
I
A
C
24
5
4
3
2
K
E
I
C
A
25
2
3
4
5
A
K
C
I
E
26
2
3
5
4
A
K
C
E
I
27
2
4
3
5
A
K
I
C
E
28
2
4
5
3
A
K
I
E
C
29
2
5
3
4
A
K
E
C
I
30
2
5
4
3
A
K
E
I
C
Combo #
Combination
Letter Representation
31
2
3
4
5
A
C
K
I
E
32
2
3
5
4
A
C
K
E
I
33
2
3
4
5
A
C
I
K
E
34
2
3
4
5
A
C
I
E
K
35
2
3
5
4
A
C
E
K
I
36
2
3
5
4
A
C
E
I
K
37
2
4
3
5
A
I
K
C
E
38
2
4
5
3
A
I
K
E
C
39
2
4
3
5
A
I
C
K
E
40
2
4
3
5
A
I
C
E
K
41
2
4
5
3
A
I
E
K
C
42
2
4
5
3
A
I
E
C
K
43
2
5
3
4
A
E
K
C
I
44
2
5
4
3
A
E
K
I
C
45
2
5
3
4
A
E
C
K
I
46
2
5
3
4
A
E
C
I
K
47
2
5
4
3
A
E
I
K
C
48
2
5
4
3
A
E
I
C
K
49
3
2
4
5
C
K
A
I
E
50
3
2
5
4
C
K
A
E
I
51
3
4
2
5
C
K
I
A
E
52
3
4
5
2
C
K
I
E
A
53
3
5
2
4
C
K
E
A
I
54
3
5
4
2
C
K
E
I
A
55
3
2
4
5
C
A
K
I
E
56
3
2
5
4
C
A
K
E
I
57
3
2
4
5
C
A
I
K
E
58
3
2
4
5
C
A
I
E
K
59
3
2
5
4
C
A
E
K
I
60
3
2
5
4
C
A
E
I
K
Combo #
Combination
Letter Representation
61
3
4
2
5
C
I
K
A
E
62
3
4
5
2
C
I
K
E
A
63
3
4
2
5
C
I
A
K
E
64
3
4
2
5
C
I
A
E
K
65
3
4
5
2
C
I
E
K
A
66
3
4
5
2
C
I
E
A
K
67
3
5
2
4
C
E
K
A
I
68
3
5
4
2
C
E
K
I
A
69
3
5
2
4
C
E
A
K
I
70
3
5
2
4
C
E
A
I
K
71
3
5
4
2
C
E
I
K
A
72
3
5
4
2
C
E
I
A
K
73
4
2
3
5
I
K
A
C
E
74
4
2
5
3
I
K
A
E
C
75
4
3
2
5
I
K
C
A
E
76
4
3
5
2
I
K
C
E
A
77
4
5
2
3
I
K
E
A
C
78
4
5
3
2
I
K
E
C
A
79
4
2
3
5
I
A
K
C
E
80
4
2
5
3
I
A
K
E
C
81
4
2
3
5
I
A
C
K
E
82
4
2
3
5
I
A
C
E
K
83
4
2
5
3
I
A
E
K
C
84
4
2
5
3
I
A
E
C
K
85
4
3
2
5
I
C
K
A
E
86
4
3
5
2
I
C
K
E
A
87
4
3
2
5
I
C
A
K
E
88
4
3
2
5
I
C
A
E
K
89
4
3
5
2
I
C
E
K
A
90
4
3
5
2
I
C
E
A
K
Combo #
Combination
Letter Representation
91
4
5
2
3
I
E
K
A
C
92
4
5
3
2
I
E
K
C
A
93
4
5
2
3
I
E
A
K
C
94
4
5
2
3
I
E
A
C
K
95
4
5
3
2
I
E
C
K
A
96
4
5
3
2
I
E
C
A
K
97
5
2
3
4
E
K
A
C
I
98
5
2
4
3
E
K
A
I
C
99
5
3
2
4
E
K
C
A
I
00
5
3
4
2
E
K
C
I
A
01
5
4
2
3
E
K
I
A
C
02
5
4
3
2
E
K
I
C
A
03
5
2
3
4
E
A
K
C
I
04
5
2
4
3
E
A
K
I
C
05
5
2
3
4
E
A
C
K
I
06
5
2
3
4
E
A
C
I
K
07
5
2
4
3
E
A
I
K
C
08
5
2
4
3
E
A
I
C
K
09
5
3
2
4
E
C
K
A
I
10
5
3
4
2
E
C
K
I
A
11
5
3
2
4
E
C
A
K
I
12
5
3
2
4
E
C
A
I
K
13
5
3
4
2
E
C
I
K
A
14
5
3
4
2
E
C
I
A
K
15
5
4
2
3
E
I
K
A
C
16
5
4
3
2
E
I
K
C
A
17
5
4
2
3
E
I
A
K
C
18
5
4
2
3
E
I
A
C
K
19
5
4
3
2
E
I
C
K
A
20
5
4
3
2
E
I
C
A
K
With five letters in the name there are considerably more possible combinations. Altogether I have come up with 120 permutations for a name with five different letters. Judging from the results I have got so far, if I continue with even longer names then there are going to be too many permutations to document. So I will stop at five letters and analyse the results I have obtained.
Results
For names with all the letters different these are the results I got back:
# Of Letters
2
3
4
5
6
7
Permutations
2
6
24
20
?
?
As the number of letters increases so does the number of permutations. Now I have to find a pattern for the rise in combination numbers.
I noticed that for a four-letter name there were four times the numbers of permutations than for a three-letter name. Seeing this, it also became obvious of the links between each name. To find the number of permutations for a name, you take the number of letters in the name and multiply that by the number of permutations for the previous name. So the formula is as follows:
Un = n x (Un - 1)
To make this formula simpler I can use the factorial function (!), which basically means that the integer is multiplied by all its preceding integers down till 1.
# Of Permutations = # Of Letters !
Using this formula I can also now work out how many permutations there would be for names with more than five letters:
# Of Letters
2
3
4
5
6
7
8
9
0
Letters !
2
6
24
20
720
5040
40320
362880
3628800
Now that I have found the formula for working out the number of permutations for names with letters that are all different, I can move onto finding the formula for names with letters that are repeated, such as 'Emma'. I know that that formula is going to be different because for 'Emma' there are 12 combinations, whereas when the existing formula is applied then there should be 24 combinations.
With one repeating letter...
One Letter: 'M'
As before, with one letter there will only by one possible arrangement, regardless of how many letters are repeating in the name, because with a one letter name you cannot have repeating letters. Therefore one-letter names are not affected by repeating letters.
Two Letters: 'BB'
When there is one repeating letter in the name then a two-letter name acts in the same way as a one letter name; it only has one possible arrangement: 'BB', because the two Bs are indistinguishable and would create the same combination if they were swapped around.
Three Letters: 'ANN'
A = 1 N = 2
Combination #
Combination
Letter Representation
2
2
A
N
N
2
2
2
N
A
N
3
2
2
N
N
A
A three-letter name has got three possible permutations.
Four Letters: 'EMMA'
As tested in part one, I already know that a four-letter name with one repeating letter has got twelve possible permutations.
Five Letters: 'LINNY'
L = 1 I = 2 N = 3 Y = 4
Combo #
Combination
Letter Representation
2
3
3
4
L
I
N
N
Y
2
2
3
4
3
L
I
N
Y
N
3
2
4
3
3
L
I
Y
N
N
4
3
2
3
4
L
N
I
N
Y
5
3
2
4
3
L
N
I
Y
N
6
3
3
2
4
L
N
N
I
Y
7
3
3
4
2
L
N
N
Y
I
8
3
4
2
3
L
N
Y
I
N
9
3
4
3
2
L
N
Y
N
I
0
4
2
3
3
L
Y
I
N
N
Combo #
Combination
Letter Representation
1
4
3
2
3
L
Y
N
I
N
2
4
3
3
2
L
Y
N
N
I
3
2
3
3
4
I
L
N
N
Y
4
2
3
4
3
I
L
N
Y
N
5
2
4
3
3
I
L
Y
N
N
6
2
3
3
4
I
N
L
N
Y
7
2
3
4
3
I
N
L
Y
N
8
2
3
3
4
I
N
N
L
Y
9
2
3
3
4
I
N
N
Y
L
20
2
3
4
3
I
N
Y
L
N
Combo #
Combination
Letter Representation
21
2
3
4
3
I
N
Y
N
L
22
2
4
3
3
I
Y
L
N
N
23
2
4
3
3
I
Y
N
L
N
24
2
4
3
3
I
Y
N
N
L
25
3
2
3
4
N
L
I
N
Y
26
3
2
4
3
N
L
I
Y
N
27
3
3
2
4
N
L
N
I
Y
28
3
3
4
2
N
L
N
Y
I
29
3
4
2
3
N
L
Y
I
N
30
3
4
3
2
N
L
Y
N
I
31
3
2
3
4
N
I
L
N
Y
32
3
2
4
3
N
I
L
Y
N
33
3
2
3
4
N
I
N
L
Y
34
3
2
3
4
N
I
N
Y
L
35
3
2
4
3
N
I
Y
L
N
36
3
2
4
3
N
I
Y
N
L
37
3
3
2
4
N
N
L
I
Y
38
3
3
4
2
N
N
L
Y
I
39
3
3
2
4
N
N
I
L
Y
40
3
3
2
4
N
N
I
Y
L
Combo #
Combination
Letter Representation
41
3
3
4
2
N
N
Y
L
I
42
3
3
4
2
N
N
Y
I
L
43
3
4
2
3
N
Y
L
I
N
44
3
4
3
2
N
Y
L
N
I
45
3
4
2
3
N
Y
I
L
N
46
3
4
2
3
N
Y
I
N
L
47
3
4
3
2
N
Y
N
L
I
48
3
4
3
2
N
Y
N
I
L
49
4
2
3
3
Y
L
I
N
N
50
4
3
2
3
Y
L
N
I
N
51
4
3
3
2
Y
L
N
N
I
52
4
2
3
3
Y
I
L
N
N
53
4
2
3
3
Y
I
N
L
N
54
4
2
3
3
Y
I
N
N
L
55
4
3
2
3
Y
N
L
I
N
56
4
3
3
2
Y
N
L
N
I
57
4
3
2
3
Y
N
I
L
N
58
4
3
2
3
Y
N
I
N
L
59
4
3
3
2
Y
N
N
L
I
60
4
3
3
2
Y
N
N
I
L
Five letters in a name with one repeating letter allows sixty possible permutations. Again, for the same reasons as before I will stop at five letters because any more would generate too many combinations to document.
Results
For names with one repeating letter in them, I got back these results:
# Of Letters
2
3
4
5
6
7
Permutations
3
2
60
?
?
When comparing these figures to those that I got for names with no repeating letters, I found that these names gave exactly half the number of combinations for the same number of letters. E.g. a four-letter name with no repeating letters has 24, whereas with a repeating letter it has 12 (half of 24) and similarly with a five-letter name with no repeating letters with 120 combinations, one with a repeating letter has only 60 (half of 120). Therefore the formula to find the number of permutations for names with one repeating letter is:
# Of Permutations = # Of Letters !
2
This formula can now also be used to work out values for the rest of the names:
# Of Letters
2
3
4
5
6
7
8
9
0
Letters !
3
2
60
360
2520
20160
81440
814400
I have now found two formulae, one for names without and one for names with one repeating letter. However, they do not cover names that have more than one repeating letter. There must be a general formula that can be used to find the number of permutations for any name with any number of letters repeating.
For two repeating letters...
One Letter: 'Q' ? 1 combination
Two Letters: 'OO' ? 1 combination
Three Letters: 'HHH' ? 1 combination
Four Letters: 'UMMM' ? 4 combinations
Five Letters: 'OKOLO' ? 20 combinations
Six Letters: 'AMANDA' ? 120 combinations
Seven Letters: 'AAABCDE' ? 840 combinations
# Of Letters
2
3
4
5
6
7
Permutations
4
20
20
840
Comparing these values to the values for names with no repeating letters, I found that this time, instead of only two times, they were six times smaller.
I continued in the same fashion for names with even more repeating letters and found that the values were all factors of values from names with no repeating values:
# Of Letters
2
3
4
5
6
7
Permutations
No repeats
2
6
24
20
720
5040
Repeat
3
2
60
360
2520
2 Repeats
4
20
20
840
3 Repeats
5
30
210
4 Repeats
6
42
The table of values has led me to the following conclusion.
The formula for names with no repeating letters is # of letters !
The formula for names with one repeating letter is (# of letters !) ? 2
The formula for names with two repeating letters is (# of letters !) ? 6
The formula for names with three repeating letters is (# of letters !) ? 24
and The formula for names with four repeating letters is (# of letters !) ? 120
I noticed that the denominators of the original formula in the formulae for names with repeating letters follow a familiar pattern ? 2, 6, 24, 120
These are ascending factorials. With this knowledge I can rewrite the formulae and create a general formula to cover any name:
# Of Letters ! ? (# Of Repeating Letters + 1) !
or
Permutations = n !__
(r+1) !
Explanation
I have noticed that the formula for some names is not exactly accurate. E.g. when there are two letters in the name and three repeating letters then to work out the permutations you would have to do 2 ! (r) 4 ! = 0.08333333333333.
So I need to add that if the value given is less than 1 it should be rounded up to 1 because you cannot have a fraction of a permutation and the lowest number of arrangements you can have is 1.
I have now worked out all the main formulas that I wanted to find. Now I need to try and explain why these formulae work.
Starting with the first formula I got for names with no repeating letters:
# Of Letters !
For this formula I need to try and find out why a factorial can calculate the number of permutations. Basically a factorial takes an integer and multiplies it by all the preceding integers down till 1. So the factorial of e.g. 5 would be
5 x 4 x 3 x 2 x 1 = 120. Or the factorial of 6 would just be the factorial of 5 x 6 (120 x 6 =720). So the permutations of one name will be the value for the name before multiplied by the number of letters in the name.
I have used the aid of a tree diagram to show why a factorial is used. As you can see on the next page, if you start the tree with a one-letter name then you will get one arrangement at the base. Then as you branch out from the one arrangement to the next level you are now showing the number of arrangements for a two-letter name. By use of colour coding, I have shown how in a two letter name, the arrangements from the name before are taken and the new letter is inserted into that arrangement in all the possible positions to create new permutations.
This kind of pattern continues on further branches. At three letters, all the two letter combinations are taken and the third letter is inserted into them all in every position. And so that is why the number of letters in the name multiplies the previous number of permutations.
DCAB
CDAB
CAB
CADB
CABD
DACB
ADCB
AB
ACB
ACDB
ACBD
DABC
ADBC
ABC
ABDC
ABCD
For five letters, E is then
placed in all the positions
in each of the four-letter
combinations
A
DCBA
CDBA
CBA
CBDA
CBAD
DBCA
BDCA
BA
BCA
BCDA
BCAD
DBAC
BDAC
BAC
BADC
BACD
Letter
2 Letters
3 Letters
4 Letters
5 Letters
Combo
2 Combos
6 Combos
24 Combos
20 Combos
The same kind of theory applies to names with repeating letters. From the base of the tree where there is one letter, as the tree branches out more letters are added on. If there is a repeating letter then at one of the branches there will be the case where that same letter is going to be introduced into the tree again. Since that letter is already in the combinations once, the number of places where the duplicate letter can be put into is now limited. If we go back to the first case, where I was working with the name 'Emma'. When it comes round to the point where the second M has to be inserted, if it is put into one combination in front of the first M then the same combination will be formed if it is put after the first M. E.g. if the second M is introduced into the tree, there will be existing combinations of EM and ME. It can then turn them into MEM, EMM, MME. If the M is put in like so: EMM, MME or MEM then these combinations already exist. One repeating letter halves the number of combinations. Two repeating letters divides it by 6 because there are even less places the duplicate letters can be put into, and three repeating letters divides it by 24, again because there are even less possible places.
AMEM
MAEM
MEM
MEAM
MEMA
AEMM
EAMM
EM
EMM
EMAM
EMMA
EMM
AMME
E
MAME
MME
MMAE
MMEA
ME
MME
MEM
Part Four:
This part of the coursework is where I can broaden the investigation of arrangements of letters a bit further. In this section I will be using arrangements consisting solely of the letters X and Y, which will give rise to combinations that may have two different repeating letters. I will start with the simplest cases and work my way upwards by adding more letters.
One Letter: 'X'
As before one letter will only give one combination.
Knowing that if all the letters are the same then only one combination is possible, I will continue by adding different letters, instead of pointlessly adding more Xs.
Two Letters: 'XY'
This is just like any two letter name with no repeating letters. So there are just two possible combinations.
Three Letters: 'XXY'
Three letters with one repeating letter as before gives 3 combinations.
Four Letters: 'XXYY'
I have added another Y, which now gives four letters - two Xs and two Ys.
X = 1 Y = 2
Combination #
Combination
Letter Representation
2
2
X
X
Y
Y
2
2
2
X
Y
X
Y
3
2
2
X
Y
Y
X
4
2
2
Y
X
X
Y
5
2
2
Y
X
Y
X
6
2
2
Y
Y
X
X
There are six combinations when two different letters are repeated. This is now 4 times less than if all the letters were different and 2 times less than if only one letter is repeated once.
Five Letters: 'XXXYY'
X = 1 Y = 2
Combo #
Combination
Letter Representation
2
2
X
X
X
Y
Y
2
2
2
X
X
Y
X
Y
3
2
2
X
X
Y
Y
X
4
2
2
X
Y
X
X
Y
5
2
2
X
Y
X
Y
X
6
2
2
X
Y
Y
X
X
7
2
2
Y
X
X
X
Y
8
2
2
Y
X
X
Y
X
9
2
2
Y
X
Y
X
X
0
2
2
Y
Y
X
X
X
3 Xs and 2 Ys give a total of 10 combinations. This is 12 times less than if there were no repeating letters, 6 times less than if there was one repeating letter, and 2 times less than if there were two repeating letters.
Six Letters: i) 'XXXXYY'
This set of six letters (4X and 2Y) gives 15 combinations. This is 48 times less than with no repeats, 24 times less than with one repeat, 8 times less than with two repeats and half of three repeats.
ii) 'XXXYYY'
This alternative set of letters, which has 3 of each letter, gives 20 combinations. This shows, compared to the previous set, that the less that any letter is repeated the more combinations are possible.
Results
Letters
Arrangement
Combinations
X
2
XY
2
3
XXY
3
4
XXYY
6
5
XXXYY
0
6
XXXXYY
5
6
XXXYYY
20
I think that all the formulae for permutations are going to be a derivative of n !. So I will work around that to try and find a master formula that would be able to work out the number of combinations for any type of name, with any set of duplicate letters.
I know the formula for if only one letter in the name is duplicated. However, with these both the X and Y may be duplicated a number of times. The formula for these then may also be a derivative of n ! ? r !. I am going to start testing using the 4-letter arrangement of XXYY.
I will break down the letters into two composites - XXYZ, which includes the two Xs, and WXYY, which includes the two Ys. Each of those separately would have 12 permutations. Together to make XXYY they have 6, which is half. This kind of occurrence, where two events together gives a smaller value, reminds me of probabilities. I.e. the probability of a coin being flipped twice and both flips landing on heads would be 0.5 x 0.5 = 0.25. In this case XXYZ gives half (0.5) the permutations of a 4-letter name with no repeats; WXYY also only gives half (0.5). Combining them together like a probability would give 0.5 x 0.5 = 0.25 the number of permutations of a 4-letter name with no repeats (24 x 0.25) = 6.
Constructing a formula to show this may go as follows:
P = n ! x ([n ! ? r+1 !] ? n !) x ([n ! ? d+1 !] ? n !)
(where d is the number of times the second letter is repeated)
This looks very complex but it should work.
E.g. 1: 'ABCD'
P = 4 ! x ([4 ! ? 1 !] ? 4 !) x ([4 ! ? 1 !] ? 4 !)
= 24 x 1 x 1
= 24
E.g. 2: 'AABCD'
P = 5 ! x ([5 ! ? 2 !] ? 5 !) x ([5 ! ? 1 !] ? 5 !)
= 120 x 0.5 x 1
= 60
E.g. 3: 'AABBBC'
P = 6 ! x ([6 ! ? 2 !] ? 6!) x ([6 ! ? 3 !] ? 6 !)
= 720 x 0.5 x 0.167
= 60
E.g. 4: 'XXXYY'
P = 5 ! x ([5! ? 3 !] ? 5 !) x ([5 ! ? 2 !] ? 5 !)
= 120 x 0.167 x 0.5
= 10
All of the above examples use the formula and all the values given are correct so this formula is viable for any name with up to two different letters duplicated. If you have a name with more than two different letters duplicated then you would probably have to add additional terms to the formula to incorporate those duplicated letters into the equation.
By Milan Shah - Form 11P - Set 01
Page 1
Mathematics Coursework:
Emma's Dilemma
Mathematics Coursework:
Emma's Dilemma
