In how many ways can the letters in WONDERING be arranged with exactly two consecutive vowels 
In how many ways can the letters in WONDERING be arranged with exactly
  two consecutive vowels

I solved and got answer as $90720$. But other sites are giving different answers. Please help to understand which is the right answer and why I am going wrong.
My Solution
Arrange 6 consonants $\dfrac{6!}{2!}$
Chose 2 slots from 7 positions $\dbinom{7}{2}$
Chose 1 slot for placing the 2 vowel group $\dbinom{2}{1}$
Arrange the vowels $3!$
Required number of ways:
$\dfrac{6!}{2!}\times \dbinom{7}{2}\times \dbinom{2}{1}\times 3!=90720$
Solution taken from http://www.sosmath.com/CBB/viewtopic.php?t=6126)

Solution taken from http://myassignmentpartners.com/2015/06/20/supplementary-3/

 A: The number of arrangements with 3 consecutive vowels is correctly explained in the original post: the number is $15120$.
To find the number of arrangements with at least two consecutive vowels, we duct tape two of them together (as in the original post) and arrive at $120960$.
The problem with this calculation is that every arrangement with 3 consecutive vowels was double counted: once as $\overline{VV}V$ and again as $V\overline{VV}$. To compensate for this we must subtract $15120$.
The correct number of arrangements with at least two consecutive vowels is 
$120960-15120=105840.$
Therefore, correct number of arrangements with exactly two consecutive vowels is $105840-15120=90720.$
A: The total number of ways of arranging the letters is $\frac{9!}{2!} = 181440$. Of these, let us count the cases where no two vowels are together. This is $$\frac{6!}{2!} \times \binom{7}{3}\times 3! = 75600$$
Again, the number of ways in which all vowels are together is 15120. Thus the number of ways in which exactly two vowels are together is 
$$181440 - 75600 - 15120 = 90720$$
