In how many arrangements of the letters of the word INTERMEDIATE 2 vowels never come together? I went on to proceed with this problem by subtracting the number of arrangements in which all vowels were together, from the total number of arrangement which proved time consuming. An alternative solution, given by the textbook was what they called the gap method.

They find the number of ways of arranging $6$ consonants. 
"Then they say that there are $7$ gaps remaining (how!!!!!)" in which the remaining $6$ vowels are arranged in $\frac{7p6}{3!2!}$. 
The answer is $151200$
The vowels are: AEEEII.

The non-vowels are: DMNRTT.

The number of ways to arrange the vowels is $\frac{(1+3+2)!}{1!\times3!\times2!}=60$.

The number of ways to arrange the non-vowels is $\frac{(1+1+1+1+2)!}{1!\times1!\times1!\times1!\times2!}=360$.

The number of ways to choose slots for the $6$ vowels between the $6$ non-vowels is $\binom{6+1}{6}=7$.

So the number of ways to arrange this word under the given restriction is $60\cdot360\cdot7=151200$.
how may gaps are there between the consonants..$6$ consonants so for vowels there are $7$ places.first arrange the consonants in $(6!/2)$ ways.now choose any 6 places from the 7 and arrange the 6 vowels that can be done in    ($7\choose 6$$×6!)/(2!×3!)$ and finally multiply both the arrangements