UCLY
UYCL AMME
UYLC AMEM
AEMM
CULY
CUYL
CLYU
CLUY
CYLU
CYUL
YCUL
YCLU
YUCL
YULC
YLUC
YLCU
LUCY has 24 different arrangements for her name; whereas EMMA, with one repeat, only has 12. The equation (number of letters ×6)÷(number of repeated letters) works for these two words.
Other letter group arrangements
NOON
NONO
NNOO
ONNO
ONON
OONN
(4×6)÷(2)= 8
The above formula cannot work as NOON only has 6 arrangements not 8.
Rather than trying to guess random formulas and try to make them work I will, get more data.
TIM BOB
TMI BBO
IMT OBB
ITM
MTI
MIT
(Number of letters×2)÷(Number of repeated letters) works for the 3 letter words but can’t work for the 4 letter words because 6× does.
The formulae can’t work for any other letter words, and the 4 letter word equation didn’t work for 2 repeats in a 4 letter word. So forget repeats first, I’ll see if I can find an equation for no repeats that works for all by focussing on no repeated lettered words.
LUCY has 4 letters if I multiply that by the number of different arrangements of the 3 letter word TIM I get the number of different arrangements for LUCY. So if I multiply the number of arrangements of a 4 letter word by multiplying 4 by 6 (arrangements of a 3 letter word), will it work for other words. To make an easier comparison I will simply use the alphabet to represent each letter; i.e. LUCY would become ABCD and EMMA would become ABBC.
A 2 letter word would have 2 arrangements:
AB
BA
3×2=6 so it also works for 3 letter words.
So 5 lettered words should have 120 arrangements (5×24). 24 is the number of arrangements of a 4 letter word, so 4×6 would’ve given this. 6 can be shown as 3×2, as 6 is the arrangements of a 3 lettered word. A 1 lettered word would have 1 arrangement.
A
So working down a 5 lettered word would have 5×4×3×2×1 arrangements, the same with 4 or 3 lettered words. 4×3×2×1 for a 4 lettered word and 3×2×1 for a 3 lettered word. I have not yet shown that a 5 lettered words has 120 arrangements, so if it does then this should work.
Rather than write them all out I will show it like this.
A(BCDE) If BCDE was a 4 lettered then it would have 24 arrangements, so if I use each letter as a fixed then the 4 letter word that follows would have 24 arrangements.
So:
A(BCDE) 24
B(ACDE) 24
C(ABDE) 24
D(ABCE) 24
E(ABCD) 24
120
So it works! A 5 letter word does have 120 arrangements; this is 5×4×3×2×1. This is a factorial, which is expressed as 5!.
But that is with no repeats, it didn’t work when there was a repeat in EMMA. It came out as double the actual answer. So if I divide by number of repeated letters like earlier it would work once more. But would it work for other words? I’ll try 5 lettered words with one repeat.
A (BBCD) 12
B (ABCD) 24
B (ABCD) 24
C (ABBD) 12
D (ABBC) 12
84
The ones starting with B would be repeated so I need to ignore 1 of them. 84-24=60. So 5! (120)÷number of repeated letters (2) =60. This does work. What about other amounts of repeats in 4 letter words?
ABBB
BABB
BBAB
BBBA
So it would have 4 arrangements. 4! (24)÷number of repeated letters (3) =8, this is to much.
EMMA has 12 arrangements, which is 24÷2; ABBB has 4 arrangements, which is 24÷6. I had to use factorials before, and it seems like they are needed again. As said earlier 6 can be expressed as 3!, so number of letters!÷ number of repeats!.
A five lettered word with one repeated letter has 60 arrangements. So 5!(120)÷2!(2)=60. So this works, what about 5 lettered arrangements with 2 repeats.
A(BBBC)4
B(ABBC)12
B(ABBC)12
B(ABBC)12
C(ABBB)4_
44
Two of them are the same so ignore them, so a 5 lettered with 2 repeats would have 20 arrangements. 5!÷3!=20. So this also works. I will do one more test to make 100% sure it works. 6 lettered words have 720 arrangements (6×120). So:
A(BBCDE)60
B(ABCDE)120
B(ABCDE)120
C(ABBDE)60
D(ABBCE)60
E(ABBCD)60_
480
Ignore one of the 120, so a 6 lettered word with one repeat has 360 arrangements. 6!(720)÷2!(2)=360. This is also correct!
So to find the number of arrangements of any word you need to use the equation. n!÷r! Where n is the number of letters in the word and r is the amount of times a number is repeated. I.e. In EMMA n=4 and r=2!.
