DAVID DAVDI DADIV DADVI DAIDV DAIVD DIAVD DIADV DIDAV DIDVA DIVDA DIVAD DVIAD DVIDA DVDIA DVDAI DVADI DVAID DDVAI DDVIA DDIVA DDIAV DDAIV DDAVI ADDVI ADDIV ADIVD ADIDV ADVDI ADVID AIDVD AIDDV AVDDI AVDID AVIDD AIVDD IAVDD IADVD IADDV IDADV IDAVD IDVAD IDVDA IDDVA IDDAV IVADD IVDAD IVDDA VIADD VIDAD VIDDA VDIAD VDIDA VDDIA VDDAI VDADI VDAID VADDI VADID VAIDD
ACIGR ACIRG ACRIG ACRGI ACGIR ACGRI AICGR AICRG AIGCR AIGRC AIRGC AIRCG AGIRC AGICR AGCIR AGCRI AGRCI AGRIC ARCIG ARCGI ARICG ARIGC ARGIC ARGCI CAIGR CAIRG CAGIR CAGRI CARGI CARIG CIARG CIAGR CIGAR CIGRA CIRGA CIRAG CGAIR CGARI CGIRA CGIAG CGRIA CGRAI CRAIG CRAGI CRIAG CRIGA CRGAI CRGIA IACGR IACRG IAGRC IAGCR IARGC IARCG ICAGR ICARG ICGAR ICGRA ICRGA ICRAG IGARC IGACR IGCAR IGCRA IGRAC IGRCA IRAGC IRACG IRCAG IRCGA IRGAC IRGCA GACRI GACIR GAIRC GAICR GARCI GARIC GCAIR GCARI GCIRA GCIAR GCRAI GCRIA GIACR GIARC GICAR GICRA GIRCA GIRAC GRAIC GRACI GRCAI GRCIA GRIAC GRICA RACGI RACIG RAIGC RAICG RAGIC RAGCI RCAIG RCAGI RCIAG RCIGA RCGIA RCGAI RIAGC RIACG RICAG RICGA RIGAC RIGCA RGAIC RGACI RGCAI RGCIA RGIAC RGICA
RESULTS TABLE FOR FIVE LETTER WORDS:
The results table for a five-letter word has proven my prediction. I have found a general rule to find the number of different combinations for the letters in words. The rule is: for a two letter word it would be 1x2=2, for a three letter word it would be 1x2x3=6, for a four letter word it would be 1x2x3x4=24, for a five letter word it would be 1x2x3x4x5=120 and so on. The rule is different for a word if it has two letters the same. The rule is: for a two letter word it would be 1x2÷2(number of same letters)=1, for a three letter word it would be 1x2x3÷2=3, for a four letter word it would be 1x2x3x4÷2=12, for a five letter word it would be 1x2x3x4x5÷2=60
This rule in algebraic terms would for a word with all letters different would be: n!
The rule is different for a word, which has two letters the same: n! ÷ 2
To check that n! works I will now try it on two, three, four and five letter words with different letters. Two letter word: 2!=2, three letter word: 3!=6, four letter word: 4!=24, five letter word: 5!=120
I will now check n! ÷ 2 works for a word with two letters the same. Two letter word: 2! ÷ 2=1, three letter word: 3! ÷ 2=3, four letter word: 4! ÷ 2=12, five letter word: 5! ÷ 2=60
I will now try to find a rule for a word with more than two letters the same.
FOUR LETTER WORD:
AABB ABAB ABBA BBAA BABA BAAB
FIVE LETTER WORDS:
AAABB AABBA ABBAA ABABA ABAAB AABAB BBAAA BABAA BAABA BAAAB
AAAAB AAABA AABAA ABAAA BAAAA
RESULTS TABLE:
PREDICTION:
I predict that the number of letters in the word divided by the number of A’s divided by the number of B’s will give the number of different combinations.
To test my prediction I will work it out using this rule: for a four letter word like AABB; it has 2 A’s and 2 B’s so: 1x2x3x4 ÷ 1x2 ÷ 1x2=24 ÷ 4=6, and for a five letter word like AAABB; this has 3 A’s and 2 B’s so: 1x2x3x4x5 ÷ 1x2x3 ÷ 1x2=120 ÷ 12=10, and for the five letter word AAAAB; this has 4 A’s and 1 B so: 1x2x3x4x5 ÷ 1x2x3x4 ÷ 1x1=120 ÷ 24=5
This rule explained in algebraic terms is: n! ÷ (A! x B!). I will now test this algebraic rule for AABB:
4! ÷ (2! x 2!)
24 ÷ (2 x 2)
24 ÷ 4=6 different combinations.