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Keeping in mind the repeated letters, I'm unsure how to approach this question with the nCr method.

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    How many ways to arrange just the consonants? How many ways to choose 4 spaces between the consonants for the vowels? How many ways to arrange the vowels? – 2017-01-25
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    I know there are 4!/2! ways for the consonants (considering repeats) as well as 4!/2! for the vowels. As for the spaces between consonants, would that be 4C4 ways? I know there is a repeat for the vowels so I am unsure how to account for that. – 2017-01-25
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    You have all the elements to the correct answer (which is posted below), and your thinking is on the right track. You just need to practice asking yourself the right questions: I can permute the consonants in 4!/2! ways. 'For each of these,' there are 5C4 ways to choose 4 spaces between the consonants for the vowels, and 'for each of these' I can permute the vowels in 4!/2! ways. Then the answer is the product of these... – 2017-01-25

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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: $$\frac {4!}{2!}\frac {4!}{2!}\binom 54 = 12\cdot 12\cdot 5 = 720\text{ options}$$

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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.

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    That would be correct if the problem also states that no two consonants are allowed next to each other, as well. But if there is no ban on adjacent consonants, the $720$ answer stands. – 2017-01-25