Combinatorics with arrangements of the word UNIVERSALLY How many ways are there to arrange the letters in UNIVERSALLY so that the four vowels appear in two cluster of two consecutive letters with at least 2 consonants between the two clusters?
For this, we will look at the compliment of the statement. When doing this we have to look at when their are no consonants in between the cluster of vowels and when there is 1 consnant in between the vowel clusters.     
CASE 1: No consonants in between vowel cluster
Since there is no consonant in between the two pair of vowel clusters, we can line up all 4 vowels. We are going to treat UIEA as one letter for right now. We have UIEANVRSLLY. Since UIEA is one letter, we have a total of 8 letters. The ways to arrange this is $\cfrac{8!}{2!1!1!1!1!1!1!}$. Now lets revisit our vowel cluster. The ways to rearrange UIEA is 4!. Therefore the amount of ways to arrange UNIVERSALLY with no consonants in between vowel cluster is 4! $\cfrac{8!}{2!}$ 
CASE 2: 1 consonant in between vowel cluster
This is where I am stuck. The fact that there is 2 L's is confusing me and I do not understand how to proceed 
Any help would be appreciated 
Thank you! 
 A: Y is used as a vowel in the word UNIVERSALLY, but we will ignore that.
You handled the first case correctly.  
In how many ways can the letters of the word UNIVERSALLY be arranged so that the vowels appear in exactly two clusters consisting of two vowels apiece with exactly one consonant between the clusters?
Consider two cases, depending on whether or not that letter is L.  
The letter between the two clusters of vowels is an L:  The four vowels and the L form a block.  We have seven objects to arrange, the block and the six distinct consonants N, V, R, S, L, Y.  The objects can be arranged in $7!$ ways.  Since the L must be in the middle of the block, the letters in the block can be arranged in $4!$ ways.  Hence, there are $7!4!$ possible arrangements in this case.
The letter between the two clusters of vowels in not an L:  Choose which of the other five consonants lies between the two clusters of two vowels.  Then we have seven objects to arrange, the block, two Ls, and the four distinct consonants that do not appear in the block. The objects can be arranged in 
$$\binom{7}{2, 1, 1, 1, 1, 1} = \frac{7!}{2!1!1!1!1!1!}$$
distinguishable ways.  Since the consonant must be in the middle of the block, the letters in the block can be arranged in $4!$ ways.  Hence, there are 
$$\binom{5}{1}\binom{7}{2, 1, 1, 1, 1, 1}4!$$
possible arrangements in this case.
Hence, there are 
$$7!4! + \binom{5}{1}\binom{7}{2, 1, 1, 1, 1, 1}4!$$
possible arrangements in which exactly one consonant appears between two clusters of two vowels.
Can you take if from here?
A: The 7 letters not in the clusters can be arranged in $\frac{7!}{2!}$ ways.  The two clusters can straddle any of the 7 letters.  CASE2 can occur in $7(4!) \frac{7!}{2!}$ ways.  This answer agrees with the previous one. 
