It might be simpler just to find out the number of possible arrangments when two vowels are next to each other and subtract it from your answer to 1
As for your educated guess, I dont know. That's as simple as it gets right? you explained the logic yourself.
Last edited by Azland; 12-05-2012 at 23:51.
Your approach to (d) is along the right lines. The only thing you've forgotten is that there are 2 identical O's, hence divide by....
(Original post by vinvinvin)
(d) How many arrangements are there where no two vowels are next to each other?
To answer this Q I attempted the following. I imagined there was a space between each non-vowel and one before the beginning and end. Therefore 10 spaces.
To find the number of combinations of consonants I did the following,
I then multiplied this by 10 x 9 x 8 x 7 x 6 as there is one less space each time a vowel fills a spot. I thought this would give me the final number of arrangements but this was not correct.
Last edited by ghostwalker; 13-05-2012 at 10:17.
Reason: Corrected - thanks TenOfThem