Applying Combinatorics on Combinatorics

There are 8 consanants, in which there is one repeat (C). Therefore, there are $\frac{8!}{2}$ ways to arrange just the consanants.

There are 9 spaces between the consanants. We must choose 2 places for the O's, 2 more for the I's, and one more for the A.

Therefore, the answer is $\frac{8!}{2}{9\choose 2}{7\choose 2}\cdot5=\boxed{76204800}$ .

There are 8 consonants and 5 vowels.Now the problem asks us to permute the string without putting any two or more vowels together. There will be 9 gaps created by these 8 Consonants, 5 vowels are to be placed in those 9 gaps. This can be done by (8!)* (9P5) ways, but there are duplications of C,I and O. So this whole quantity stated above has to be divided by (2!x2!x2!)=6.