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Permutations question (1 Viewer)

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Hi this is the question:
Find the number of ways in which the letters of the word EPSILON can be arranged in a row so that the three vowels are all next to each other.
I let the three vowels be V, and the four constants be C
VCCCC
You can arrange the whole thing 5! ways, then the vowels 3! ways to get 5!3! which is 720. However the answer is (4x3!)4! where (4x3!) are the vowels and 4! are the consonants. I can't see how this is the case. My logic for trying to make this answer was have:
_C_C_C_C_ where underscores are where the Vowels can go and the c's are the consonants. Then the vowels can go in 5 positions, be arranged 3! ways and the consonants can be arranged 4! ways to get (5x3!)4! but that isn't the answer either. Can I have a detailed explanation please, thanks.
 

davidgoes4wce

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I draw a box of the 7 letters, one must also realise that the vowels must be treated as one whole group, so we have 5! (the group of words plus the remaining four consonants)
5!x 3!= 720
 
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I draw a box of the 7 letters, one must also realise that the vowels must be treated as one whole group, so we have 5! (the group of words plus the remaining four consonants)
5!x 3!= 720
That's what I got but as I said the answer was (4x3!)4! which is 576.
 

braintic

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That's what I got but as I said the answer was (4x3!)4! which is 576.
Whoever provided the other solution had a valid method, but got their numbers wrong.

The possible arrangements are:
Vcccc
cVccc
ccVcc
cccVc
ccccV

So the 4 times 3! for the vowels should have been 5 times 3!. They must have assumed 4 consonants equates to 4 places for the vowels - the old fence/fencepost problem.
 

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