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/
Best Answer
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 $105840-15120=90720.$