Given a 2-letter word, with 2 letters repeated, there is 1 arrangement.
1) AA
Given a 3-letter word, with 2 letters repeated, there are 3 arrangements.
1) ANN 2) NAN 3) NNA
Given a 4-letter word, with 2 letters repeated, there are 12 arrangements.
1) EMMA 2) AMME 3) AMEM 4) EMAM 5) AEMM 6) EAMM.7) MMEA 8) MMAE 9) MEMA 10) MAME 11) MEAM 12) MAEM
The connection between the two sets of data is that the words with two letters the same, have half the number of arrangements that the words with all different letters have.
With two letters the same (AA) there is only one arrangement that is seen. However if they were labelled A1A2 we would see that there are in fact two (or 2!) combinations in practise. Consequently the equation becomes
a= n! / 2! or n! / 2
Where n = the number of letters in the word and a = the number of different arrangements.
To confirm the prediction I will look at a five-letter word with one letter repeated twice. If the formula is correct there will be 5! / 2! different arrangements. (i.e. 60)
Assume the word is ABCDD.
If A B or C are the first letter there will be 3 x 12 arrangements as the last four letters will include one letter, D, twice. This is like EMMA where I have already shown there are 12 different arrangements.
If D is the first letter there will be 4! arrangements (i.e. 24) as the last four letters are all different.
This gives a total of 60 different arrangements as predicted.
I have already shown that LUCY has 4! arrangements.
I have shown that the formula for words with all letters different is a = n! where n = the number of letters in the word and a = the number of different arrangements.
The formula for words where two letters are the same is a= n! / 2 or a= n! /2!
b)To complete this part of the work I am going to investigate the number of different arrangements in words with three letters the same.
With a 3 letter word, with 3 letters repeated, there is 1 arrangement.
NNN
With a 4 letter word, with 3 letters repeated, there are 4 arrangements.
1) NNNA 2) NNAN 3) NANN 4) ANNN
With a 5 letter word, with 3 letters repeated, there are 20 arrangements.
1)AONNN 5)OANNN 9)NAONN 13)NONAN 17)NNNAO 2)ANONN 6)ONANN 10)NANON 14)NONNA 18)NNNOA 3)ANNON 7)ONNAN 11)NANNO 15)NNAON 19)NNAON 4)ANNNO 8)ONNNA 12)NOANN 16)NNONA 20)NNNOA
With three letters the same (AAA) there is only one arrangement that is seen. However if they were labelled A1A2A3 we would see that there are in fact six (or 3!) combinations in practise. Consequently the equation becomes
a = n! / 3!
Where n = the number of letters in the word and a = the number of different arrangements.
To confirm the prediction I will look at a six letter word with one letter repeated three times. If the formula is correct there will be 6! / 3! different arrangements (i.e.120)
Assume the word is ABCDDD
If A B or C are the first letter there will be 3 x 20 arrangements as the last four letters will include one letter, D, three times. This is like AONNN where I have already shown there are 20 arrangements.
If D is the first letter there will be 5! / 2! arrangements (i.e. 60) as the last five letters include one, D, that is there twice ( see table above)
This gives a total of 120 different arrangements as predicted.
I therefore conclude that the connecting formula is
a= N! / r!
Where n = the number of letters in the word, r is the number of letters repeated and a = the number of different arrangements.
The following pattern is developing and I have made my predictions for words with four letters the same based on the connective formula.
* = prediction
I will now investigate the number of different arrangements in words with four letters repeated. In order to see the accuracy of my prediction.
With a 4 letter word, with 4 letters repeated, there is 1 arrangement.
1)AAAA
With a 5 letter name, with 4 letters repeated, there are 5 arrangements.
1)AAAAN 2)AAANA 3)AANAA 4)ANAAA 5)NAAAA
With a 6 letter name, with 4 letters repeated, there are 30 arrangements.
1)NOAAAA 6)ONAAAA 11)AOAAAN 16)AAOANA 21)AAANOA 26)AANAAO 2)NAOAAA 7)OANAAA 12)AAOAAN 17)AAAONA 22)AAANAO 27)ANOAAA 3)NAAOAA 8)OAANAA 13)AAAOAN 18)AAAANO 23)AONAAA 28)ANAAOA 4)NAAAOA 9)OAAANA 14)AAAAON 19)AOANAA 24)AANOAA 29)ANAAAO 5)NAAAAO 10)OAAAAN 15)AOAANA 20)AAONAA 25)AANAOA 30)ANAOAA
This confirms my predictions in the table above and the validity of the connecting formula.
Part 3
Aim
Investigate the number of different arrangements of various groups of letters.
In order to do this I am first going to assess various 4 and 5 letter words to see if I can determine a universal formula covering various options. From the previous work summarised in the table on page 9. I already have some of the data.
- 4 letter words
The number of arrangements when all the letters are different is 24
When two letters are the same it is 12
When two letters are repeated twice there are 6 arrangements, namely
- AABB 2) ABAB 3) BABA 4) BBAA 5) ABBA 6) BAAB
- 5 letter words
The number of arrangements when all the letters are different is 120
When two letters are the same it is 60
When two letters are repeated twice there are 30 arrangements, namely
- AABBC 7)CAABB 13)BBAAC 19)ABCBA 25)BCABA
- AABCB 8)CBBAA 14)BBACA 20)ABCAB 26)BACBA
- AACBB 9)CABCA 15)BBCAA 21)ACBAB 27)BAABC
- ACBBA 10)CBABA 16)BCBAA 22)ABBAC 28)BAACB
- ABABC 11)CBAAB 17)BABAC 23)ABBCA 29)BACAB
- ABACB 12)CABBA 18)BABCA 24)ABCBA 30)BCAAB
From the pattern above I see that in each of the cases you are dividing the total number of arrangements of words with all the letters the same (n!) by the number of letters the same. In the case of two letters the same it is
a=n! / 2! or n! / 2
While in the case of two sets of 2 letters the same it is
a = n! / 2! X 2!
From this I can get a prediction for the universal formula of
a= n!
x! x y! x z! ….
where n is the number of letters in the word, x y z etc are the number of similar individual letters and a is the number of arrangements. It should be noted that
n = x + y + z …..
With the four letter words
When all the letters are different a = 4! / 1! x 1! x 1! x 1! = 24
When two letters are the same a = 4! / 2! x 1! x 1! = 12
When 2 sets of letters are the same a = 4! / 2! x 2! = 6
With the five letter words
When all the letters are different a = 5! / 1! x 1! x 1! x 1! x 1! = 120
When two letters are the same a = 5! / 2! x 1! x 1! x 1! = 60
When 2 sets of letters are the same a = 5! / 2! x 2! x 1! = 30
I will now look at two other combinations to see if the formula correctly predicts the number of arrangements there will be.
- A six letter word with two letters duplicated three times.
The formula would give a = 6! / 2! x 2! = 20 arrangements
- A seven letter word with one letter repeated three times and the other
four.
The formula would give a = 7! / 4! x 3! = 35 arrangements.
AAABBB
- AAABBB 6)ABABAB 11)BAAABB 16)BABAAB
- AABBBA 7)ABABBA 12)BAABAB 17)BBBAAA
- AABBAB 8)ABBABA 13)BAABBA 18)BBAAAB
- ABBAAB 9)ABBBAA 14)BABABA 19)BBAABA
- ABAABB 10)ABBAAB 15)BABBAA 20)BBABAA
AAABBBB
- AAABBBB 10)ABABABB 19)BAAABBB 28)BBAABBA
- AABABBB 11)ABABBAB 20)BAABABB 29)BBBBAAA
- AABBABB 12)ABABBBA 21)BAABBAB 30)BBBABAA
- AABBBAB 13)ABBAABB 22)BAABBBA 31)BBBAABA
- AABBBBA 14)ABBABAB 23)BABAABB 32)BBBAAAB
- ABBBBAA 15)ABBABBA 24)BABABAB 33)BBABBAA
- ABBBABA 16)BABBBAA 25)BABABBA 34)BBABABA
- ABBBAAB 17)BABBAAB 26)BBAAABB 35)BBABAAB
- ABAABBB 18)BABBAAB 27)BBAABAB
These two check words confirm my universal formula of
a = n!
x! x y! x z !…..
where n is the number of letters in the word, x y z etc are the number of different number combinations and a is the number of arrangements. It should be noted that n = x + y + z …….
Final Conclusion
From the work I have done I have developed a number of formula, some of which are for specific circumstances, leading to the universal formula which covers all possibilities.
- For words containing all different letters I have
a = n!
Where n = the number of letters in the word and a = the number of different arrangements.
This can also be written a = n! / 1! x 1! x 1!……….
- For words containing a letter repeated twice the equation is
a =n! / 2!
Where n = the number of letters in the word and a = the number of different arrangements.
This can also be written a = n! / 2! x 1! x 1! x 1!……
- For words containing a letter repeated three times the equation is
a = n! / 3!
Where n = the number of letters in the word, and a = the number of different arrangements.
This can be written as a = n! / 3! x 1! x 1! x 1!…..
The universal formula, covering all possibilities, is
a = n! / x! x y! x z!…….
Where n = the number of letters in the word, x y z…. are the number of similar individual letters and a is the number of arrangements.
n will be the sum of x + y+ z +….
It is seen that the formula 1), 2) and 3) fit the universal formula.
From this formula a master table can be achieved for any number of arrangements, for example :-
* shown in detail in text; the remainder direct from the universal formula.