A three-letter word will have 6 variations. This can be proved by the following word
TOM
- TOM
- TMO
- OMT
- OTM
- MOT
- MTO
A five-letter word will comprise of 120 variations. The following word will prove this:
TEAMS
- TEAMS
- TEMAS
- TESMA
- TESAM
- TEMSA
- TEASM
- TAEMS
- TAESM
- TAMSE
- TAMES
- TASME
- TASEM
- TSAEM
- TSAME
- TSEMA
- TSEAM
- TSMAE
- TSMEA
- TMSAE
- TMSEA
- TMEAS
- TMESA
- TMAES
- TMASE
- MTSAE
- MTSEA
- MTEAS
- MTESA
- MTASE
- MTAES
- MATES
- MATSE
- MASET
- MASTE
- MAEST
- MAETS
- MSATE
- MSAET
- MSEAT
- MSETA
- MSTAE
- MSTEA
- MESTA
- MESAT
- METAS
- METSA
- MEATS
- MEAST
- EMAST
- EMATS
- EMTAS
- EMTSA
- EMSTA
- EMSAT
- ESMAT
- ESMTA
- ESTMA
- ESTAM
- ESATM
- ESAMT
- ETASM
- ETAMS
- ETMSA
- ETMAS
- ETSMA
- ETSAM
- EATSM
- EATMS
- EAMTS
- EAMST
- EASMT
- EASTM
- SAETM
- SAEMT
- SATME
- SATEM
- SAMET
- SAMTE
- SMATE
- SMAET
- SMTEA
- SMTAE
- SMETA
- SMEAT
- STEAM
- STEMA
- STMAE
- STMEA
- STAME
- STAEM
- SEATM
- SEAMT
- SETMA
- SETAM
- SEMTA
- SEMAT
- SMEAT
- SMETA
- SMATE
- SMAET
- SMTEA
- SMTAE
- SAMTE
- SAMET
- SAETM
- SAEMT
- SATEM
- SATME
- SEATM
- SEAMT
- SEMTA
- SEMAT
- SETMA
- SETAM
- STAME
- STAEM
- STEMA
- STEAM
- STMAE
- STMEA
We have investigated the three, four and five letter words. From this we can deduce the different variations for different letter words.
Now we can move onto investigating four and five letter-words with similar letters.
EMMA- This is a four-letter word with two same letters.
- EMMA
-
EMAM
- EAMM
- MEMA
- MAME
- MMEA
- MMAE
- AMME
- AMEM
- AEMM
- MAEM
- MEAM
EM1M2A - This is the same spelling as the one above, however the two M’s have been given numbers.
-
EM1M2A
-
EM2M1A
-
EAM2M1
-
EAM1M2
-
EM1AM2
-
EM2AM1
-
AEM1M2
-
AEM2M1
-
A M2EM1
-
A M1EM2
-
A M2M1 E
-
AE M1 M2
-
M2AEM1
-
M2EAM1
-
M2 M1 AE
-
M2 M1 E A
-
M2 E M1A
-
M2 A E M1
-
M1M2 A E
-
M1M2 E A
-
M1 A M2 E
-
M1 E M2 A
-
M1 A E M2
-
M1 E A M2
Both EM1M2A and EMMA have the same spelling except for the subscripts. However the one without the subscripts has 12 variations while the one with the subscripts has 24. This shows that in the word with the subscript, we will count both the M’s for the variation, while the one without we will take all 4 letters to find the factorial for the word but we will then divide it by two factorial as the number of letters repeated are 2. For the one without the subscripts, there is no letter repeated so the bottom part of the fraction is 0! which is 1.
This means that a four - letter word with two similar letters has half the number of variations of a four-letter word without any similar letters.
For a five-letter word with two similar letters of one letter and two similar letters for another letter, there are 30 variations.
AABBD
- AABBD
- AABDB
- AADBB
- ABBDA
- ABBAD
- ABABD
- ABADB
- ADBBA
- ADBAB
- ADABB
- ABDBA
- ABDAB
- BBAAD
- BBADA
- BBDAA
- BDBAA
- BDAAB
- BDABA
- BABAD
- BABDA
- BADBA
- BADAB
- BAABD
- BAADB
- DBBAA
- DBAAB
- DBABA
- DAABB
- DABAB
- DABBA
A1A2B1B2D
The above word has the same spelling as AABBD. AABBD has 30 variations as it has two similar As and two similar Bs. However in this word there are subscripts, which makes every letter different. Hence this word will have 120 variations (5!- 5*4*3*2*1). For AABBD there are 30 as there are five letters but two letters, which are repeated two times each (5! \ 2! *2!).
A five-letter word with 3 similar letters of one letter and 2 similar letters of another letter, e.g. AAABB. This word has 10 variations.
AAABB
- AAABB
- ABABA
- AABAB
- BABAA
- BBAAA
- ABBAA
- ABAAB
- BAABA
- BAAAB
- AABBA
A1A2A3B1B2
The above word has the same spelling as AAABB except for the subscripts. In this word we are taking all the letters separately so we get 120 variations. In the other word, there are 5 letters but 3 similar for one letter and two similar for the other. So the formula for A1A2A3B1B2 will be 5! \ 0! Which will be 5* 4*3*2*1 \ 1. the formula for the AAABB would be 5! \ 3! * 2! , which is 5* 4* 3* 2*1 \ (3* 2* 1) * (2*1). The 3! In the denominator is for the number of repetitive A letters and 2! Is for number of repetitive B letters. It is factorial because if the letters were different, they would form those many variations.
For a 6-letter word with all the letters different, I found 720 variations. However with a 6 letter word with 4 letters same of one alphabet and 2 letters same of another, I found only fifteen variations, e.g. AAAABB
AAAABB
- AAAABB
- AAABAB
- AAABBA
- AABBAA
- AABABA
- AABAAB
- ABAAAB
- ABAABA
- ABABAA
- ABBAAA
- BBAAAA
- BABAAA
- BAABAA
- BAAABA
- BAAAAB
These are the variations I have found for different letter words with no repetitive letters:
Two-letter word has two variations
Three-letter word has six variations
Four-letter word has twenty-four variations
Five-letter word has one hundred and twenty variations
Six-letter word has seven hundred and twenty variations
The mathematical word ‘factorial’ helped me calculate that. The word factorial is denoted by the symbol (!). Hence 5! Would be 5*4*3*2*1, which equals 120. This proves that there are 120 variations of a five-letter word with no repetitive letters. Similarly for a four letter word, 4! Would be 4*3*2*1, which gives 24 variations.
A five-letter word with three repetitive letters would have twenty variations. This means 5! Divided by 3! Gives this answer. It basically means 5*4*3*2*1/3*2*1. This formula applies for any word with repetitive letters of the same letter:
N! / R!
N denotes the number of letters and R denotes the number of repetitive letters. However the formula is different if there is 2 letters that are repeated e.g. AABBBC. This means there are repetitive letters for both A and B. The formula for this would be 6! Divided by (2! * 3!). 6 are the number of letters in the word, 2 is the number of repetitive A’s and 3 are the repetitive B’s.
EVALUATION
While discovering the formula, I went through many formulas, which were wrong. I’m going to state the formulas and show how I proved them wrong. While discovering the formula for letters with two letters, which are repeated, I thought we add the factorials in the denominator instead of multiplying it. However this was proved wrong when I tried this word:
PPPQQ
- PPPQQ
- PPQQP
- PQQPP
- PQPPQ
- PQPQP
- PPQPP
- QQPPP
- QPPPQ
- QPPQP
- QPQPP
If the denominators were being added, I would have had 15 variations.
I also started on finding a few big words variations but decided not to carry on, e.g. Mississippi. I realized it was a tedious process and there was no point in finding it out.
These are a few other words for which I calculated variations:
