This question has already been answered (how is it combination prob, Trouble in finding number of permutations, Trouble in finding number of permutations), I first seen it in my discrete maths book, and I understood it.
But, since this is a really difficult topic (combinatorics) for me, I was watching this video and the same problem came up but they changed it a little bit.
Given the word TALLAHASSEE, how many arrangaments can be made with no consecutive vowels?
He kinda uses the same approach as the book, but the last thing he does is to order the case of $AAAEE$, which is obviously $\frac{5!}{3!2!}$. But this doesn't make sense to me at all, the problem says that you can't have any conscutive vowels, so he's counting them by doing that?
For me, the correct answer would be $\frac{6!}{2!2!}{7\choose 5}$, where $\frac{6!}{2!2!}$ is the number of arrangements with no $A's$ and no $E's$ and ${7\choose 5}$ are the positions where the vowels could be in between the rest of the letters.
Can someone explain this to me a little bit?
Best Answer
In the original post you cite, all arrangements of the A's are equivalent (that is, all the symbols are the same). Here, though, putting AAAEE into five distinct holes gives a different word that putting EEAAA into those same holes (in order).
To think about it a different way, suppose all the vowels were some symbol, say $*$. Then the number of arrangements with no $*$'s together would be, as you computed, $\frac{6!}{2!2!}\binom{7}{5}$. Now you have five $*$'s that you need to replace with three $A$s and two $E$s. In how many ways can you do this? Well, you just need to choose the locations for the $A$s, which can be done in $\binom{5}{3}$ ways.