In how many discrete ways can you arrange the word "challengers" if all the vowels should be separated with each other?
The total number of permutations is $\frac{11!}{2!2!}$ and I subtracted it with the total number of permutations when the vowels are together $\frac{3!}{2!} × \frac{9!}{2!}$
Is this correct or am I missing something?
How about we first arrange the consonants, there are $\frac{8!}{2!}$ ways of doing this. Then we choose where we are going to put the vowels, they can be between the consonants, be the first letter, or be the last letter, there are $\binom{9}{3}$ ways of doing this. Then from the $3$ vowel slots we pick, we choose which one will we put $a$ in, there are $3$ ways of doing this.
$$ \frac{8!}{2!}\times\binom{9}{3}\times 3 $$