How many ways can you arrange the letters of the word ADDITION so that no vowels are next to each other? 
Keeping in mind the repeated letters, I'm unsure how to approach this question with the nCr method.
 A: Ordering the vowels and consonants separately, they coincidentally have $\frac {4!}{2!} = 12$ arrangements each (the divisor of $2!$ there accounting for the repeats). Inserting the vowels into the consonants means choosing which of the five spaces to fit them into - the three spaces between letters and the two on the end - which can be done $\binom 54= 5$ ways.
Since these three stages are independent of each other - the choices made in one case do not limit the other cases - each stage will multiply up the options from the previous stage, so that overall there are: $$\frac {4!}{2!}\frac {4!}{2!}\binom 54 = 12\cdot 12\cdot 5 = 720\text{ options}$$
A: Thanks everyone for their help! I realized that the number of combinations was fewer than originally thought due to the possible placement of the consonants/vowels.
There are ${4!\over 2!}$ arrangements for consonants and ${4!\over 2!}$ for vowels. But there are also 2 ways where vowels can go in the spaces and both need to be accounted for.
A_I_I_O_
_A_I_I_O
Meaning that your final answer should be (${4!\over 2!}$ x ${4!\over 2!}$) x 2, which equals 288. 
