$$\fbox{X}A\fbox{X}A\fbox{X}A\fbox{X}I\fbox{X}I\fbox{X}O\fbox{X}$$
Now, there are $\frac{6!}{3!2!}$ ways to arrange the vowels as I have placed them in the above schematic. Also, consonants will be fixed iin the places marked $X$. Again, there are $\frac{7!}{4!2!}$ ways to arrange the consonants. So, there are
$$\frac{6!}{3!2!}\cdot\frac{7!}{4!2!}$$
ways to arrange the letters of the word ASSASSINATION so that no two vowels are together. Am I correct?
Best Answer
You can arrange the consonants $SSSSNTN$ in $\frac{7!}{4!2!}$ ways.
Independently of this, you can insert the vowels $AAIAIO$ in the eight slots before, between and after each vowel. However that means two slots are unfilled - let’s denote these as $XX$.
Now you can arrange the slot-fillers $AAIAIOXX$ in $\frac{8!}{3!2!2!}$ ways.
Multiply these together and you have your total number of arrangements where no two vowels are together.