The way I did that probability question, it's kinda different:
You put the consonants as barriers, but instead of using 8C4, I put each vowel into a gap:
The first vowel can go in 8 places, the second vowel in 7, the third in 6, and the last vowel in 5 places. So we have 8x7x6x5.
But of the vowels, there are three O's, so I divided by 3! to get rid of the arrangements repeated:
Then we arrange the consonants, which can be done in 7! ways, but divide by 2! since there are 2 H's:
Which gives you the same answer.