So far (partial answer):
1. Since the RAP must always stand together, we can consider it as a single "letter" (a consonant).
2. So, now we have 16 letters, 10 consonants and 6 vowels. Let's substitute each consonant by 0 and each vowel by 1, and find all valid zero-one patterns.
3. We need to find number of permutations of 10 zeroes and 6 ones without 2 ones or 3 zeroes standing together. Each permutation is uniquely determined by the numbers of zeroes between adjacent ones (number of zeroes before 1st one, then between 1st and 2nd, etc.). Since there are 6 ones, there are 7 numbers which can be either 1 or 2 (since 0 would result into 2 ones standing together, and 3 or more will result into 3 zeroes standing together), however the 1st and last (7th) number can be zero. These numbers will sum up to 10 (since there are 10 zeroes in total), so there are either exactly 4 ones and 3 twos, or 1 zero, 2 ones and 4 twos, or 2 zeroes and 5 twos in these 7 numbers (since 1+1+1+1+2+2+2=0+1+1+2+2+2+2=0+0+2+2+2+2+2=10). The number of permutations with no zeroes will be exactly 7C3=7*6*5/(1*2*3)=35, with 1 zero will be 2*6C2=2*6*5/(1*2)=30 (since the zero can be either first or last), and with two zeroes 1 (only one: 0222220). Total number of ways is 35+30+1=66.
4. The vowels can be in any order, so there are 6!=720 ways to arrange them.
5. Now assign a digit 0 to 9 to each of the consonants according to their alphabetical order (here we can treat RAP as just P, since there are no letters between P and R (Qs are not present in the set)), and find number of valid permutations of 0123456789 for the all 66 cases.
6. Actually, the 66 cases can be reduced to 3 ones (with 3, 4 and 5 consonant pairs) since we can build a one-to-one correspondence between permutations for the cases having the same number of consonant pairs, so the number of valid sequences will remain the same.
7. So, the following 3 case classes are: find a permutation abcdefghij of 0123456789, where (1) a<b, c>d and e<f (35 cases); (2) the same as (1) and also g>h (30 cases); (3) the same as (2) and also i<j (1 case).
8. (Here I may be wrong) It's intuitively clear that the part of valid sequences will be 1/8, 1/16 and 1/32 of the total number of permutations (which is 10!=3628800), or 453600, 226800, and 113400 for the case classes (1), (2), and (3) respectively (since in each pair of digits, the first is either less or greater than the second, and this seems to be equally probable, but I cannot prove it) .
9. So, the final answer is 720*(35*453600+30*226800+1*113400)=16411248000
Edit: Tried to complete the answer, but step 8 may be incorrect.