In how many different ways can the letters of the word "CORPORATION"

There are several problems with this approach:
1. The number of arrangements of CORPORATION is 11!/(2!*3!) (we should divide by 2! and 3! because of 2 R's and 3 O's there);

2. When you consider the vowels as one units as {OOAIO}, you get 7 units {OOAIO}, {C}, {R}, {P}, {R}, {T}, {N}. The number of arrangements of these 7 units is 7!/2! (again we divide by 2! because of 2 R's). OOAIO within their unit can also be arranged in 5!/3! ways (5 letters with three O's). So, the number of arrangements when vowels are together is 7!/2!*5!/3!.

3. When you subtract the cases when the vowels are together from total ways you don't get the cases where no two vowels are together, you just get the cases when ALL 5 vowels are not together, which is not what we wanted. For example, one of the cases you get would be CORPORATION, where, as you can see, I and O are together.

So, the way you are doing is not a good way to approach this problem.

THEORY

Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:
\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\)

For example number of permutation of the letters of the word "gmatclub" is \(8!\) as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.

Hope this helps.