The original question:
(i) In how many ways can the alphabet be ordered such that m and n aren't next to each other? (ii) And how many ways such that the five vowels aren't next to each other. (iii) And, such that no four or five vowels are next to each other.
I was able to answer the first two. ($26!$ – $2(25!)$ and $26!$ – $5!22!$) However, I'm not sure about the third. I'd say that it would equal ($26!$ – #combinations with $5$ vowels next to each other – #combinations with really only $4$ vowels next to each other (and not $5$).
Now, #combinations of really only 4 vowels = #combinations of just $4$ vowels – #combinations with $5$ vowels. This would mean that the answer would be $26!$ – #combinations of just $4$ vowels next to each other = $26! -$ $5 $($4!$)($23!$) = $26! – 5!23!$. Somehow I'm not convinced of this answer, because if I'd make all these combinations, I'd be counting some double I feel like. (…. a i e o u ….. and …. a i e o u ….. are the same combinations, but they are both counted). What am I doing wrong? Could it be $26!$ – #combinations with $4$ vowels next to each other + #combinations with $5$ vowels next to each other ?
Thanks in advance for any help.
Best Answer
Here’s a way to count the permutations with a block of exactly four vowels. (You’ve done everything else just fine.)
There are $21!$ permutations of the consonants. Each of them defines $22$ slots for the vowels, $20$ between adjacent consonants and $2$ on the ends. There are $5$ ways to choose the lonesome vowel, and $22$ ways to decide where to put it. The block of four vowels has to go in one of the other $21$ slots, and within that slot it can be permuted in $4!$ ways. Thus, there are
$$21!\cdot5\cdot22\cdot21\cdot4!=22!\cdot5!\cdot21$$
such permutations.