Explanation:
6 consonants - 6! / 2!
Self explanatory
Vowels have to be separated by
at least 1 consonant. (BTW, seems as if you're mistaking "separated by 1 vowel" with "separated by AT LEAST 1 vowel". at least means separation by 2 [or more] is fine too. So it's possible to have: VCVCCVCVCCV)
Method I'm going to use:
I'm going to fit in the vowels "in between the consonants". I've arranged the consonants above, but I've left a gap between each consonant, as well as 1 gap before and 1 gap after. After attaining 6!/2!, I now have to fit in 5 vowels into 7 spaces that have been created.
However, there is an ugly restriction: the vowels also have to be in alphabetical order.
Honestly, I couldn't figure out a short cut way, I just brute forced it.
Umbrella case 1: C C a C blank C blank C blank C blank <-- have to fit in 2o's and 2 i's into the 4 blanks. only 1 way to do this, since i's have to be before o's, and they're interchangeable.
Umbrella case 2: C a C blank C
ah cbf, just check my working out. If you don't understand, let me know and I'll elaborate more, but it should be self-explanatory. I just brute-forced it really.
I'm getting:
6!/2! x (1 + 5 + 15)
= 6!/2! x (21)
Probably made a silly mistake somewhere heh, but hope you see how I did it.
PS: how can I insert an image such that it appears as an image rather than a link?
edit million'th because not satisfied with ugly, brute-force method:
Alternative solution-
6!/2! x (7P5/2!2! / [5!/2!2!]) = 7560
6!/2! for consonants = 360
7P5/2!2! = ways to arrange vowels in between the consonants
then divide further by the number of ways to arrange the vowels amongst themselves, as only 1 of those will have them all in alphabetical order.
now THAT is fkn neat. IF it's correct. and I'm pretty sure it is. i think. i hope.
