With the four letter word above, I didn't need to retype this combination of letters ( four letters, with one repeated twice ), but to make sure that it didn't matter which letters were used, I decided to try them again.
I found out that my analysis was correct, and that the changing of letters didn't affect the number of different arrangements for four letters, with one repeated twice.
Five letters:
AABCD ACABD BAACD CAABD DAABC
AABDC ACADB BAADC CAADB DAACB
AACBD ACBAD BACAD CABAD DABAC
AACDB ACBDA BACDA CABDA DABCA
AADBC ACDAB BADAC CADAB DACAB Total:
AADCB ACDBA BADCA CADBA DACBA 60
ABACD ADABC BCAAD CBAAD DBAAC
ABADC ADACB BCADA CBADA DBACA
ABCAD ADBAC BCDAA CBDAA DBCAA
ABCDA ADBCA BDAAC CDAAB DCAAB
ABDAC ADCAB BDACA CDABA DCABA
ABDCA ADCBA BDCAA CDBAA DCBAA
Results:
Number of Letters: Number of different combinations:
2 1
3 3
4 12
5 60
Rule:
To find the number of all the different combinations possible, from a selected number of letters ( using every letter only once, ) with two of the chosen letters being the same, can be achieved by using the following formula:
To find the total number of combinations, take the number of letters ), and times that number by the number of letters minus 1, timed by the number of letters minus 2, and so on, until the number being times by is the number 1 itself. Then divide the number by the number of times the letter has been repeated.
No. of = No. of letters X ( No. of letters - 1 ) X ( No. of letters - 2 ) X etc…
combinations No. of times the letter has been repeated
For example:
To find the number of combinations which can be made from using 10 letters, with two of the ten being the same, can be found by using the following formula:
No. of = 10 X 9 X 8 X 7 X 6 X 5 X 4 X 3 X 2 X 1
Combinations 2
= 1814400
Justification
The reason why this rule occurs, is because there may-be the same number of letters, but one is used twice. This means that the total number of arrangements possible is reduced by a factor of 2, or "2 !" ( See later ).
For example
For 6 letters, all different, there are "6 X 5 X 4 X 3 X 2 X 1"arrangements.
For 6 letters, 1 repeated twice, there are "6 X 5 X 4 X 3 X 2 X 1"arrangements.
2
This is due to the fact that the six different letters can be arranged in any order, and because all of them are different, it dose matter which order they go in. However, with one letter repeated twice, the letter which is repeated can be swapped with itself, with-out making a new arrangement.
Answer, One letter repeated Three Times:
Two letters:
AAA Total:
1
Three letters:
AAAB Total:
AABA 4
ABAA
BAAA
Four letters:
AAABC BAAAC
AAACB BAACA
AABAC BACAA
AABCA BCAAA
AACAB
AACBA Total:
ABAAC 20
ABACA
ABCAA CAAAB
ACAAB CAABA
ACABA CABAA
ACBAA CBAAA
Results:
Number of Letters: Number of different combinations:
3 1
4 4
5 20
Rule:
To find the number of all the different combinations possible, from a selected number of letters ( using every letter only once, ) with one of the chosen letters being repeated three times, can be achieved by using the following formula:
To find the total number of combinations, take the number of letters ), and times that number by the number of letters minus 1, timed by the number of letters minus 2, and so on, until the number being times by is the number 1 itself. Then divide the number by the number of times the letter has been repeated, timed by the number of letters minus 2, and so on, until the number being times by is the number 1 itself.
No. of = No. of letters X ( No. of letters - 1 ) X ( No. of letters - 2 ) X etc…
Combinations Times repeated X ( Times repeated - 1 ) X etc…
For example:
To find the number of combinations which can be made from using 10 letters, with three of the ten being the same, can be found by using the following formula:
No. of = 10 X 9 X 8 X 7 X 6 X 5 X 4 X 3 X 2 X 1
Combinations 3 X 2 X 1
= 604800
Justification
The reason why this rule occurs, is because there may-be the same number of letters, but one is used three times. This means that the total number of arrangements possible is reduced by a factor of 6, or "3 !" ( See later ).
For example
For 6 letters, all different, there are "6 X 5 X 4 X 3 X 2 X 1"arrangements.
For 6 letters, 1 repeated three times, there are "6 X 5 X 4 X 3 X 2 X 1"arrangements.
3 X 2
This is due to the fact that the six different letters can be arranged in any order, and because all of them are different, it dose matter which order they go in. However, with one letter repeated three times, the letter which is repeated can be swapped with itself, with-out making a new arrangement.
Answer, Two letters repeated Twice:
Four letters:
AABB
ABAB Total:
ABBA 6
BAAB
BABA
BBAA
Five letters:
AABBC BAABC CAABB
AABCB BABAC CABAB
AACBB BAACB CABBA Total:
ABABC BABCA CBAAB 30
ABACB BACAB CBABA
ABCAB BACBA CBBAA
ABBAC BBAAC
ABBCA BBACA
ABCBA BBCAA
ACABB BCAAB
ACBAB BCABA
ACBBA BCBAA
Results:
Number of Letters: Number of different combinations:
4 6
5 30
Rule:
To find the number of all the different combinations possible, from a selected number of letters ( using every letter only once, ) with two letters being chosen twice, can be achieved by using the following formula:
To find the total number of combinations, take the number of letters, and times that number by the number of letters minus 1, timed by the number of letters minus 2, and so on, until the number being times by is the number 1 it-self, and then divide it by the 2 ( the number of times 'A' has been repeated ) times 2 ( the number of times B has been repeated ).
No. of = No. of letters X ( No. of letters - 1 ) X ( No. of letters - 2 ) X etc…
combinations ( times a letter X ( times a different letter
has been repeated ) has been repeated )
For example:
To find the number of combinations which can be made from using 10 letters, with two of the ten having two of the same letters, can be found by using the following formula:
No. of = 10 X 9 X 8 X 7 X 6 X 5 X 4 X 3 X 2 X 1
Combinations 2 X 2
= 907200
Justification
The reason why this rule occurs, is because there may-be the same number of letters, but two are used twice. This means that the total number of arrangements possible is reduced by a factor of 4, or "2 ! X 2!" ( See later ).
For example
For 6 letters, all different, there are "6 X 5 X 4 X 3 X 2 X 1"arrangements.
For 6 letters, 2 repeated twice, there are "6 X 5 X 4 X 3 X 2 X 1"arrangements.
2 X 2
This is due to the fact that the six different letters can be arranged in any order, and because all of them are different, it dose matter which order they go in. However, with two letters repeated twice, the letters which are repeated can be swapped with their opposites, with-out making a new arrangement.
Answer, Two letters repeated Three Times:
Six letters:
AAABBB BAAABB
AABABB BAABAB
AABBAB BAABBA
AABBBA BABAAB
ABAABB BABABA Total:
ABABAB BABBAA 20
ABABBA BBAAAB
ABBAAB BBAABA
ABBABA BBABAA
ABBBAA BBBAAA
Results:
Number of Letters: Number of different combinations:
6 20
Rule:
To find the number of all the different combinations possible, from a selected number of letters ( using every letter only once, ) with two letters being chosen three times, can be achieved by using the following formula:
To find the total number of combinations, take the number of letters, and times that number by the number of letters minus 1, timed by the number of letters minus 2, and so on, until the number being times by is the number 1 it-self, and then divide it by 3 X 2 ( 3 ! See later ), timesed by 3 X 2 ( 3 ! See later ).
No. of = No. of letters X ( No. of letters - 1 ) X ( No. of letters - 2 ) X etc…
combinations ( times a letter X ( times a different letter
has been repeated ) ! has been repeated ) !
For example:
To find the number of combinations which can be made from using 10 letters, with two of the ten having two of the same letters, can be found by using the following formula:
No. of = 10 X 9 X 8 X 7 X 6 X 5 X 4 X 3 X 2 X 1
Combinations ( 3 X 2 ) X ( 3 X 2 )
= 100800
Justification
The reason why this rule occurs, is because there may-be the same number of letters, but two are used three times. This means that the total number of arrangements possible is reduced by a factor of 6 X 6, or "3 ! X 3 !" ( See later ).
For example
For 6 letters, all different, there are "6 X 5 X 4 X 3 X 2 X 1"arrangements.
For 6 letters, 2 repeated three times, there are "6 X 5 X 4 X 3 X 2 X 1"arrangements.
( 3 X 2 ) X ( 3 X 2 )
This is due to the fact that the six different letters can be arranged in any order, and because all of them are different, it dose matter which order they go in. However, with two letters repeated three times, the letters which are repeated can be swapped with their opposites, with-out making a new arrangement.
Answer:
From the answers I have learned from the previous parts of this coursework, I should be able to produce a general rule to solve any question which follows the same basic patterns of the sequences of letters in the earlier parts of this coursework.
To make it easier to find, I have decided to show my findings again below:
Table 1, All different letters:
Number of Letters: Number of different combinations:
1 1 = 1!
2 2 = 2!
3 6 = 3!
4 24 = 4!
Table 2, One letter repeated twice:
Number of Letters: Number of different combinations:
2 1 = 2! / 2!
3 3 = 3! / 2!
4 12 = 4! / 2!
5 60 = 5! / 2!
Table 3, One letter repeated three times:
Number of Letters: Number of different combinations:
3 1 = 3! / 3!
4 4 = 4! / 3!
- 20 = 5! / 3!
Table 4, Two letters repeated twice:
Number of Letters: Number of different combinations:
4 6 = 4! / (2! X 2!)
5 30 = 5! / (2! X 2!)
Table 5, Two letters repeated three times:
Number of Letters: Number of different combinations:
6 20 = 6! / (3! X 3!)
Rule:
From these sets of results, I can see that if I expand the rule I found to solve "Two letters repeated twice", then this rule can be applied to the other two sets of results, still giving the correct answers.
The rule for "Two letters repeated twice" was:
No. of = No. of letters X ( No. of letters - 1 ) X ( No. of letters - 2 ) X etc…
combinations ( times a letter X ( times a different letter
has been repeated ) has been repeated )
From this rule, all that is needed to be done, is to introduce a small mathematical function, called "FACTORIALS ( ! )". All this term does is to remove the need for writing "No. of letters X ( No. of letters - 1 ) X ( No. of letters - 2 ) X etc…". This Whole line can be turned into a much shorter line, which means the same thing as the long line, but makes it easier to write down. I learned this formula form background research I did myself, and have decided to incorporate it in to my work. Factorials can be used, by instead of writing:
"No. of letters X ( No. of letters - 1 ) X ( No. of letters - 2 ) X etc…"
You can just write:
"No. of letters !"
Will have the same meaning as the original, longer line.
For example:
This equation:
No. of = No. of letters X ( No. of letters - 1 ) X ( No. of letters - 2 ) X etc…
combinations ( times a letter X ( times a different letter
has been repeated ) has been repeated )
Can now read as followed:
No. of = No. of letters !
combinations ( times a letter X ( times a different letter
has been repeated ) has been repeated )
By writing this, you still mean the previous line, but it is now a lot easier to write down, especially if the number of letters exceeds 10.
New Formula:
However for numbers larger than 2, the expression (times a letter has been repeated ) needs to be replaced with ( times a letter has been repeated )! as there are this many ways that the repeated letters could be used. My original equations worked because of the fact that 2! = 2.
My new, adapted formula is different from the previous formula, manly because it incorporates the "Factorial" function. By using this function, I will be able to work out the number of different arrangements for any group of letters, with as many letters repeated as is needed. This formula is shown below:
No. of = No. of letters !
combinations ( no. of times a letter X ( no. of times a letter
has been repeated ) ! different has been repeated ) !
As you can see, the factorial symbol has been added to the ends of each part of this equation. This allows the formula to be used to solve every question which needs to know the number of arrangements.
The reason why this wasn't added to the previous equation of "Two letters repeated twice":
No. of = No. of letters X ( No. of letters - 1 ) X ( No. of letters - 2 ) X etc…
combinations ( times a letter X ( times a different letter
has been repeated ) has been repeated )
Was that this equation, would have made the above equation more complicated. It could have been added, but I felt that because "2 !", which the two bottom answers were, would be easier to understand that way, mainly because "2 !", is "2 X 1", which is just 2.
This is the reason why I left it out until now, where a more in-depth understanding of the mathematics is involved.
Algebraic Rule
Due to this new mathematical symbol I have decided to use, I can now right the previous equation even more simply. If we take:
"n" to be the number of letters,
"A" to be any repeated letter,
"B" to be any repeated letter, and
"C" to be any other repeated letter,
the rule should look like the following:
( nA + nB + nC ) !
nA ! X nB ! X nC !
by using this general formula, you can find the number of combinations of any set of letters.
For Example
I will now show that this rule can find all of the different combinations for the letters of SASHA's name.
S = 2
A = 2
H = 1
Therefore,
( 2 + 2 + 1 ) ! = 5 X 4 X 3 X 2 X 1 = 120 30
2 ! X 2 ! X 1! 2 X 2 X 1 4
For this equation, there seems to be 30 different combinations of the name SASHA. To prove that this is correct, I will now write out all of the arrangements:
SASHA ASASH HSASA
SASAH ASAHS HSAAS
SAHSA ASSAH HSSAA
SAHAS ASSHA HAASS
SASHA ASHAS HASAS Total:
SASAH ASHSA HASSA 30
SSAHA AHSSA
SSAAH AHSAS
SSHAA AHASS
SHASA AASSH
SHAAS AASHS
SHSAA AAHSS
From these results, it proves that my rule works, and should work, not just for this combination of letters, but every one, and still produce the correct answers.
Justification
The reason why this rule occurs, is because there may-be the same number of letters, but any number can be used more than once. This means that the total number of arrangements possible is reduced by different amounts, according to the total number of letters, the number of repeated letters, and the number of times each one is repeated.
For example
For 6 letters, all different, there are "6 X 5 X 4 X 3 X 2 X 1"arrangements.
For 6 letters, 3 repeated twice, there are "6 X 5 X 4 X 3 X 2 X 1"arrangements.
2 ! X 2 ! X 2 !
OR
6 ! = 720 = 90
2 ! X 2 ! X 2 ! 8
This is due to the fact that the six different letters can be arranged in any order, and because all of them are different, it dose matter which order they go in. However, with more than one letter repeated more than once, the total number of arrangements is reduced by the number of times the first repeated letter is repeated FACTORIAL, timesed by the number of times the second repeated letter is repeated FACTORIAL timesed by "etc.". This is because when a letter is repeated, it can be swapped with its self, with-out making a new arrangement.
Question Four:
A number of X's and a number of Y's are written in a row such as
XX………XXYY………Y
Investigate the number of different arrangements of the letters.
Answer:
In this final part of this coursework, I will be investigating how the relation ship between X's and Y's affect my rule, and to see if I can find easier ones.
If in the row of letters there are "n" X's, and "p" Y's, the total number of letters "t" is given by the equation:
t = n + p
And the number of combinations of all the letters is:
T ! = ( n + p ) !
As all the X's are the same, the number of combinations of X's giving the same answer is "n !". similarly, the number of identical arrangements of Y's is "p !".
The number of different arrangements ( D ) possible is therefore given by:
D = t ! ( n + p ) !
n ! X p ! or n ! X p !
For Example
If I use 4 X's, and 3 Y's, giving a total of 7 letters, then from the formula above, the number of different arrangements should be:
D = t ! = 7 ! 35
n ! X p ! 4 ! X 3 !
The number of arrangements are for 4 X's + 3 Y's are:
XXXXYYY YXXXXYY
XXXYXYY YXXXYXY
XXXYYXY YXXXYYX
XXXYYYX YXXYXXY
XXYXXYY YXXYXYX
XXYXYXY YXXYYXX
XXYXYYX YXYXXXY Total:
XXYYXXY YXYXXYX 35
XXYYXYX YXYXYXX
XXYYYXX YXYYXXX
XYXXXYY YYXXXXY
XYXXYXY YYXXXYX
XYXXYYX YYXXYXX
XYXYXXY YYXYXXX
XYXYXYX YYYXXXX
XYXYYXX
XYYXXXY
XYYXXYX
XYYXYXX
XYYYXXX
This proves that my rule is correct.
Justification
The reason why this rule occurs, is because there may-be the same number of letters, but any number can be used more than once. This means that the total number of arrangements possible is reduced by different amounts, according to the total number of letters, the number of repeated letters, and the number of times each one is repeated.
For example
For 6 letters, all different, there are "6 X 5 X 4 X 3 X 2 X 1"arrangements.
For 6 letters, 3 repeated twice, there are "6 X 5 X 4 X 3 X 2 X 1"arrangements.
2 ! X 2 ! X 2 !
OR
6 ! = 720 = 90
2 ! X 2 ! X 2 ! 8
This is due to the fact that the six different letters can be arranged in any order, and because all of them are different, it dose matter which order they go in. However, with more than one letter repeated more than once, the total number of arrangements is reduced by the number of times the first repeated letter is repeated FACTORIAL, timesed by the number of times the second repeated letter is repeated FACTORIAL timesed by "etc.". This is because when a letter is repeated, it can be swapped with its self, with-out making a new arrangement.
To find Rules for increasing numbers of X's and Y's
In this very final part of my coursework, I will be investigating the number of arrangements that are possible for larger groups of letters.
I will start with a simple level, to show how the number of arrangements increases as the number of component letters increases. Many of the example I will give will repeat arrangements I have worked out for previous sections, but I will include them to show a complete set of results.
One X and an increasing number of Y's
One X and No Y's
X
One X and One Y:
XY Total:
YX 2
One X and Two Y's:
XYY Total:
YXY 3
YYX
One X and Three Y's:
XYYY
YXYY Total:
YYXY 4
YYYX
One X and Four Y's:
XYYYY
YXYYY Total:
YYXYY 5
YYYXY
YYYYX
One X and Five Y's:
XYYYYY
YXYYYY
YYXYYY Total:
YYYXYY 6
YYYYXY
YYYYYX
Results for One X and No. of Y's:
Number of Y's Number of different combinations:
0 1
1 2
2 3
3 4
4 5
5 6
Rule:
To find the number of all the different combinations possible, from one X, and a certain number of Y's ( using every letter only once, ) can be achieved by using the following formula:
To find the total number of combinations, take the number of letter Y's, and to that add One ( the X ). ( The total number of letters ).
No. of = ( Number of Y's + 1 )
combinations
For example:
To find the number of combinations which can be made from using 10 letter Y's and One X, can be found by using the following formula:
No. of = 10 + 1
Combinations
= 11
Two X's and an increasing number of Y's
Two X's and No Y's
XX Total:
1
Two X's and One Y:
XXY Total:
XYX 3
YXX
Two X's and Two Y's:
XXYY
XYXY
XYYX Total:
YXXY 6
YXYX
YYXX
Two X's and Three Y's:
XXYYY
XYXYY
XYYXY Total:
XYYYX 10
YXXYY
YXYXY
YXYYX
YYXXY
YYXYX
YYYXX
Two X's and Four Y's:
XXYYYY
XYXYYY
XYYXYY
XYYYXY
XYYYYX
YXXYYY Total:
YXYXYY 15
YXYYXY
YXYYYX
YYXXYY
YYXYXY
YYXYYX
YYYXXY
YYYXYX
YYYYXX
Results for Two X's and No. of Y's:
Number of Y's Number of different combinations:
0 1
1 3
2 6
3 10
4 15
Rule:
To find the number of all the different combinations possible, from Two X's, and a certain number of Y's ( using every letter only once, ) can be achieved by using the following formula:
To find the total number of combinations, take the number of letter Y's, and to that add ( the number of letter Y's + One ) plus ( the number of letter Y's - One ) plus ( the number of letter Y's - Two ) plus etc… , until the number being added is Zero.
No. of = No. of Y's + ( No. of Y's + 1 ) + ( No. of Y's - 2 ) + etc…
combinations
For example:
To find the number of combinations which can be made from using 10 letter Y's and Two X's, can be found by using the following formula:
No. of = 10 + 11 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1
Combinations
= 66
Three X's and an increasing number of Y's
Three X's and No Y's
XXX Total:
1
Three X's and One Y:
XXXY Total:
XXYX 4
XYXX
YXXX
Three X's and Two Y's:
XXXYY
XXYXY
XXYYX Total:
XYXXY 10
XYXYX
XYYXX
YXXXY
YXXYX
YXYXX
YYXXX
Three X's and Three Y's:
XXXYYY
XXYXYY
XXYYXY
XXYYYX
XYXXYY
XYXYXY
XYXYYX Total:
XYYXXY 20
XYYXYX
XYYYXX
YXXXYY
YXXYXY
YXXYYX
YXYXXY
YXYXYX
YXYYXX
YYXXXY
YYXXYX
YYXYXX
YYYXXX
Three X's and Four Y's:
XXXYYYY
XXYXYYY
XXYYXYY
XXYYYXY
XXYYYYX
XYXXYYY
XYXYXYY
XYXYYXY
XYXYYYX
XYYXXYY
XYYXYXY
XYYXYYX
XYYYXXY
XYYYXYX
XYYYYXX
YXXXYYY Total:
YXXYXYY 35
YXXYYXY
YXXYYYX
YXYXXYY
YXYXYXY
YXYXYYX
YXYYXXY
YXYYXYX
YXYYYXX
YYXXXYY
YYXXYXY
YYXXYYX
YYXYXXY
YYXYXYX
YYXYYXX
YYYXXXY
YYYXXYX
YYYXYXX
YYYYXXX
Results for Three X's and No. of Y's:
Number of Y's Number of different combinations:
0 1
1 4
2 10
3 20
4 35
I can't find a "simple" rule for 3 X's and an increasing number of Y's, so I have decide to go on to finding a rule based on factorials.
Answer:
From the answers I have learned from the previous parts of this coursework, I should be able to produce a general rule to solve any question which follows the same basic patterns of the sequences of letters in the earlier parts of this coursework.
To make it easier to find, I have decided to show my findings again below:
Table 6, One X + increasing number of Y's:
Number of Y's: Number of different combinations:
0 1
1 2
2 3
3 4
4 5
5 6
Table 7, Two X's + increasing number of Y's:
Number of Y's: Number of different combinations:
0 1
1 3
2 6
3 10
4 15
Table 8, Three X's + increasing number of Y's:
Number of Y's: Number of different combinations:
0 1
1 4
2 10
3 20
4 35
Conclusion
From my in-depth study into factorials, X's and Y's, and many more things, I conclude that the easiest, and simplest rule to find the number of arrangements for any set of letters, is the one I showed, and explained at the beginning of this piece of coursework.
This rule is:
D = T ! ( n + p ) !
n ! X p ! or n ! X p !
where,
D = number of different arrangements,
T = total number of letters,
n = number of letter X's, and
p = number of letter Y's.
I am now confident that I can extend this rule to even larger combinations of letters, as follows:
D = T !
m ! X n ! X p ! … X z !
Where lower case letters m, n, p, … z are the different numbers of each different letter used, and
T = m + n + p + … + z
Table at the end of this piece of coursework
Finally, I have also attached a tables, to make finding lower number of X's and Y's easier, and quicker to find. To use it, just count the total number of letter , and then the number of times the letter Y, and letter X have been repeated.
For example, to find the number of arrangements for:
XYYXYXXXYY
You count up the total number of letters, which is 10, then the number of Y's, which is 5, and finally the number of X's, which is also 5.
Letters = 10
Y's = 5
X's = 5
With these numbers, you need to find the right axis's. Then, read along the "X" axis, until you get to the number 5 ( the number of X letters for this set of letters ). Finally, read up the "Y" axis, until you get to the number 5 ( the number of Y letters for this set of letters ). Where the two numbers cross, is the total number of arrangements for that set of letters.
