Combinatorics and Probability. 3 vowels 4 consonants

So I am having some trouble with a few parts of my homework problem. The question gives us 3 vowels (A,E,O) and 4 consonants (B,C,D,F).

a) How many ways can you make a 7 letter word if each letter can only be used once. (Word doesn't have to be real). This was easy, as the answer is just 7!

b) If the vowels have to be together and the consonants have to be together?

My approach: 3 vowels can be rearranged in 3! ways and 4 consonants in 4! ways, and the vowels can be leading or ending, so the total is 3! * 4! * 2!.

c) If the vowels have to be together?

My approach: 3 vowels in 3! ways, and can be rearranged in 5! ways.

d) If B and C have to be together, but no other vowels or consonants can be together?

My approach: I first tried making a word that starts with a consonant, and then alternating between vowel and consonant, while keeping B and C last. My other word started with a vowel then a consonant, then B and C, then vowel, consonant and the vowel. This is all I could think of, and I'm not sure how the math checks out on this.

Thanks for any help!

Let's start by simplifying the problem a little bit. Let A, E, and O be represented by "V" and let (BC), D, and F be represented by "C". Then we see that we can arrange these letters as "VCVCVC" or "CVCVCV". This tells us that, assuming all vowels and all the consonants are the same (and here we are treating (BC) as one consonant), there are $2$ ways to arrange the letters.
But, the vowels and consonants are not all the same. There are $3!$ to arrange the distinct vowels and there are $3!$ ways to arrange the consonants (again treating (BC) as one consonant).
This gives $2 * 6 * 6 = 72$ ways to arrange the letters. Remember that we treated (BC) as a single letter. We can flip (BC) and write it as (CB) and that gives us twice as many combinations. Therefore, we have $72*2=144$ ways to arrange the letters so that BC are together and no other consonant pair or vowel pair are adjacent.