ABC BAC CBA
ACB BCA CAB
There are six permutations with three different letters.
d) With four letters. * 5
ABCD BACD CABD DABC
ABDC BADC CADB DACB
ACBD BCAD CBAC DBAC
ACDB BCDA CBCA DBCA
ADCB BDAC CDAB DCAB
ADBC BDCA CDBA DCBA
There are twenty-four permutations with four different letters.
1! = 1
2! = 1 x 2 = 2
3! = 1 x 2 x 3 = 6
4! =1 x 2 x 3 x 4 = 24
From this table I predict that with five letters there will be 120 permutations.
5! = 1 x 2 x 3 x 4 x 5 = 120
e) With five letters. * 7
ABCDE BACDE CABDE DABCE EABCD
ABCED BACED CABED DABEC EABDC
ABDCE BADCE CADBE DACBE EACBD
ABDEC BADEC CADEB DACEB EACDB
ABECD BAECD CAEBD DAEBC EADBC
ABEDC BAEDC CAEDB DAECB EADCB
ACBDE BCADE CBADE DBACE EBACD
ACBED BCAED CBAED DBAEC EBADC
ACDBE BCDAE CBDAE DBCAE EBCAD
ACDEB BCDEA CBDEA DBCEA EBCDA
ACEBD BCEAD CBEAD DBEAC EBDAC
ACEDB BCEDA CBEDA DBECA EBDCA
ADBCE BDACE CDABE DCABE ECABD
ADBEC BDAEC CDAEB DCAEB ECADB
ADCBE BDCAE CDBAE DCBAE ECBAD
ADCEB BDCEA CDBEA DCBEA ECBDA
ADEBC BDEAC CDEAB DCEAB ECDAB
ADECB BDECA CDEBA DCEBA ECDBA
AEBCD BEACD CEABD DEABC EDABC
AEBDC BEADC CEADB DEACB EDACB
AECBD BECAD CEBAD DEBAC EDBAC
AECDB BECDA CEBDA DEBCA EDBCA
AEDBC BEDAC CEDAB DECAB EDCAB
AEDCB BEDCA CEDBA DECBA EDCBA
There are 120 permutations with five different letters
My prediction was correct.
When all the letters are different, the general rule is as follows:
Assuming ‘n’ letters, the choice for the first letter is ‘n’, for the second letter the choice is ‘n-1’ and so on.
The permutations are achieved by multiplying ‘n’ by ‘n-1’ until you reach 1. This can be shown by: ‘n ( n – 1 ) x n ( n – 2 )…. x 1’.
This sequence is given as ‘n factorial’ for which the mathematical expression is ‘n!’
Therefore the general rule for this sequence is tn = n!
This can also be expressed iteratively as: ‘n (tn-1)’ where ‘t1 = 1’.
Now I am going to try different combinations of letters with two repeating letters in them.
a) With two repeating letters.
AA
There is one permutation with two repeating letters.
b) With two repeating letters and one different letter. * 4
AAB BAA
ABA
There are three permutations with two repeating letters and one different letter.
c) With two repeating letters and two different letters. * 6
AABC BCAA CAAB
AACB BACA CABA
ABAC BAAC CBAA
ABCA
ACAB
ACBA
There are twelve permutations with two repeating letters and two different letters.
(2!) / 2 = (1 x 2) / 2 = 1
(3!) / 2 = 1 x 2 x 3) / 2 = 3
(4!) / 2 = (1 x 2 x 3 x 4) / 2 = 12
From this table I predict that with two repeating letters and three different letters there will be 60 permutations.
(5!) / 2 = 1 x 2 x 3 x 4 x 5 which equals 60
2
d) With two repeating letters and three different letters. * 8
AABCD BAACD CAABD DAABC
AABDC BAADC CAADB DAACB
AACBD BACAD CABAD DABAC
AACDB BACDA CABDA DABCA
AADBC BADAC CADAB DACAB
AADCB BADCA CADBA DACBA
ABACD BCAAD CBAAD DBAAC
ABADC BCADA CBADA DBACA
ABCAD BCDAA CBDAA DBCAA
ABCDA
ABDAC
ABDCA
ACABD BDAAC CDAAB DCAAB
ACADB BDACA CDABA DCABA
ACBAD BDCAA CDBAA DCBAA
ACBDA
ACDAB
ACDBA
There are sixty permutations with two repeating letters and three different letters.
My prediction was correct.
As I have previously shown, In a sequence of ‘n’ different letters, the first term in the sequence has n possibilities. Where one of these possibilities is duplicated and the position of the duplicated letters, in relation to each other, is not fixed. Then the number of permutations is halved, as for a given position the likelihood of the duplicate letters appearing is doubled, the effect of this is to halves the number of positions.
To formulate the general rule for this problem I am going to tabulate all of my results.All of the results have either been down previously or are below.
With two pairs of repeating letters.
AABB BBAA
ABAB BABA
ABBA BAAB
There are six permutations with two pairs of repeating letters.
With one triplet of repeating letters and one different letter.
AAAB BAAA
AABA
ABAA
There are four permutations with one triplet of repeating letters and one different letter.
With one quadruplet of repeating letters and one different letter.
AAAAB BAAAA
AAABA
AABAA
ABAAA
There are five permutations with one quadruplet of repeating letters and one different letter.
I will now tabulate all of my results on the following page.
The general rule to give the number of permutations is:
P = T _
L1! x L2! x L3! x …. Ln!
Where P is the number of permutations, T is the total number of letters, L1 is the number of occurrences of the first letter, L2 is the number of occurrences of the second letter, L3 is the number of occurrences of the third letter and Ln is the number of occurrences of the last letter.
Therefore I predict that the number of permutations for AABBC will be:
P = _ 5! _ = 120 = 30
2! x 2! x 1! 4
Where T = 5
L1 = 2
L2 = 2
Ln = 1
With two pairs of repeating letters and one different letter.
AABBC BBAAC CAABB
AABCB BBACA CABAB
AACBB BBCAA CABBA
ABABC BAABC CBAAB
ABACB BAACB CBABA
ABBAC BABAC CBBAA
ABBCA BABCA
ABCAB BACAB
ABCBA BACBA
ACABB BCAAB
ACBAB BCABA
ACBBA BCBAA
There are thirty permutations with two pairs of repeating letters and one different letter.
My prediction was correct.
