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?
You have not accounted for those arrangements such as EAUDCTOIN in which there are disjoint pairs of adjacent vowels.
What we need to exclude are adjacent pairs of vowels. Notice that we can partition $5$ in the following ways:
\begin{align*}
5 & = 4 + 1\\
& = 3 + 2\\
& = 3 + 1 + 1\\
& = 2 + 2 + 1\\
& = 2 + 1 + 1 + 1\\
& = 1 + 1 + 1 + 1 + 1
\end{align*}
Think of these numbers as the number of consecutive vowels.
The case $1 + 1 + 1 + 1 + 1$ in which no two vowels are adjacent is what we want to count.
The case $2 + 1 + 1 + 1$ has one pair of adjacent vowels.
The case $2 + 2 + 1$ has two disjoint pairs of adjacent vowels.
The case $3 + 1 + 1 + 1$ has two overlapping pairs of adjacent vowels (such as the string AEI, which has the pairs AE and EI).
The case $3 + 2$ has three pairs of adjacent vowels, two of which are overlapping.
The case $4 + 1$ has three pairs of adjacent vowels.
The case $5$ has four pairs of adjacent vowels.
We use the Incluson-Exclusion Principle to count the number of admissible choices for the positions of the vowels and consonants first, then arrange the vowels and consonants in the those positions.
There are $\binom{9}{5}$ ways to choose the positions of the vowels. From these, we must exclude those arrangements in which there are one or more pairs of adjacent vowels.
A pair of adjacent vowels: We have eight positions, one for a block of two vowels, three for single vowels, and four for consonants. Choose one position for the block and three of the remaining seven positions for the single vowels. The remaining four positions must be reserved for the four consonants. There are
$$\binom{8}{1}\binom{7}{3}$$
such choices.
Two pairs of adjacent vowels: This can occur in two ways. Either there are two overlapping pairs (a block of three consecutive vowels) or two disjoint pairs (two blocks of two vowels each).
Two overlapping pairs: We have seven positions to fill with a block of three vowels, two single vowels, and four consonants. Choose one position for the block and two of the remaining six positions for the single vowels. The remaining four positions must be reserved for the four consonants. There are
$$\binom{7}{1}\binom{6}{2}$$
such choices.
Two disjoint pairs: We have seven positions to fill with two blocks of two vowels, one single vowel, and four consonants. Choose two positions for the blocks and one of the remaining five positions for the single vowels. The remaining four positions must be reserved for the four consonants. There are
$$\binom{7}{2}\binom{5}{1}$$
such choices.
Three pairs of adjacent vowels: There are again two cases. Either there are three overlapping pairs of adjacent vowels (a block of four consecutive vowels) or two overlapping pairs of vowels (a block of three consecutive vowels) and a disjoint pair of adjacent vowels (a block of two consecutive vowels).
Three overlapping pairs of adjacent vowels: There are six positions to fill with a block of four consecutive vowels, a single vowel, and four consonants. There are six ways to choose the position of the block and five ways to choose the position of the single vowel. The remaining four positions must be reserved for the four consonants. There are
$$\binom{6}{1}\binom{5}{1}$$
such choices.
Two overlapping pairs of adjacent vowels and a disjoint pair of adjacent vowels: There are six positions to fill with a block of three consecutive vowels, a block of two consecutive vowels, and four consonants. There are six ways to choose the position of the block of three vowels and five ways to choose the position of the block of two vowels. The remaining four positions must be reserved for the four consonants. There are
$$\binom{6}{1}\binom{5}{1}$$
such choices.
Four pairs of adjacent vowels: There are five positions to fill with a block of five consecutive vowels and four consonants. Choose the position of the block. The remaining four positions must be reserved for the four consonants. There are
$$\binom{5}{1}$$
such choices.
By the Inclusion-Exclusion Principle, the number of ways of positioning the vowels and consonants so that no two vowels are consecutive is
$$\binom{9}{5} - \binom{8}{1}\binom{7}{3} + \binom{7}{1}\binom{6}{2} + \binom{7}{2}\binom{5}{1} - \binom{6}{1}\binom{5}{1} - \binom{6}{1}\binom{5}{1} + \binom{5}{1} = 1$$
namely, VCVCVCVCV.
There are $5!$ ways to arrange the vowels in their five positions and $4!$ ways to arrange the consonants in their four positions. Hence, the number of admissible arrangements is
$$\left[\binom{9}{5} - \binom{8}{1}\binom{7}{3} + \binom{7}{1}\binom{6}{2} + \binom{7}{2}\binom{5}{1} - \binom{6}{1}\binom{5}{1} - \binom{6}{1}\binom{5}{1} + \binom{5}{1}\right]5!4! = 5!4!$$
which agrees with your answer.
Best Answer
We first arrange the consonants, then insert the vowels.
The five consonants are $K, N, R, R, T$. Choose two of the five positions for the $R$s. The remaining three distinct consonants can be arranged in the remaining three positions in $$\binom{5}{2}3! = \frac{5!}{2!3!} \cdot 3! = \frac{5!}{2!}$$ ways.
Arranging the consonants creates six spaces in which we can place the vowels. For example, $$\square K \square N \square R \square R \square T \square$$ To ensure that exactly two of the vowels are consecutive, we must place a block of two vowels in one of these six spaces and a single vowel in one of the other five spaces.
Since exactly two of the vowels are together, there are three possible ways to group the vowels. They are
Since there are $\binom{5}{2}3!$ ways to arrangements the consonants, $3$ ways to group the vowels, $6$ ways to place the block of two consecutive vowels, and $5$ ways to place the single vowel, the number of admissible arrangements is $$\binom{5}{2}3! \cdot 3 \cdot 6 \cdot 5 = 5400$$
Check: The number of ways to arrange the letters of the word $TINKERER$ is $$\binom{8}{2}\binom{6}{2}4! = 10080$$ as you found, since we must select two of the eight positions for the $E$s, two of the remaining six positions for the $R$s, and then arrange the remaining four distinct letters in the remaining four positions.
Notice that since there are three vowels in $TINKERER$, either no two vowels are consecutive, exactly two vowels are consecutive, or all three vowels are consecutive.
No two vowels are consecutive: We arrange the five consonants as above, which creates six spaces. To ensure that no two vowels are consecutive, we fill two of these six spaces with $E$s and one of the remaining four spaces with the $I$. Since there are $\binom{5}{2}3!$ ways of arranging the consonants, $\binom{6}{2}$ ways to place the $E$s, and $4$ ways to place the $I$, there are $$\binom{5}{2}3!\binom{6}{2}\binom{4}{1} = 3600$$ such arrangements.
All three vowels are consecutive: We arrange the five consonants as above, which creates six spaces in which we can place the block of three vowels. The three vowels can be arranged within the block in three ways, depending on whether $I$ is on the left, on the right, or in the middle. Since there are $\binom{5}{2}3!$ ways to arrange the consonants, $6$ ways to place the block of vowels, and $3$ ways to arrange the vowels within the block, there are $$\binom{5}{2}3!\binom{6}{1}\binom{3}{1} = 1080$$ such arrangements.
Total: Adding the number of arrangements in which no two vowels are consecutive, exactly two are consecutive, and all three are consecutive yields $$3600 + 5400 + 1080 = 10080$$ as we would expect.