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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)

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

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  • $\begingroup$ Can you explain your working. Just putting down your calculation doesn't tell us why you chose to do them. $\endgroup$ – Ian Miller Nov 20 '16 at 14:30
  • $\begingroup$ @sorry, edited the calculation and added the details. pl help. $\endgroup$ – Kiran Nov 20 '16 at 14:31
  • $\begingroup$ I will point out that the solution in the excerpt solves a different problem. Your problem asks for "exactly two consecutive vowels", the excerpt's solution allows 3 consecutive vowels as well. As it says at the end "with at least two adjacent vowel" $\endgroup$ – ReverseFlow Nov 20 '16 at 14:36
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    $\begingroup$ @Kiran You answer is right and their answer is wrong. I have added my explanation below. $\endgroup$ – user940 Nov 20 '16 at 15:12
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    $\begingroup$ Checked with Python, the answer is indeed $90720$, deleted mine. $\endgroup$ – barak manos Nov 20 '16 at 15:13
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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 $105840-15120=90720.$

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  • $\begingroup$ yes, it is. I am able to understand now. thanks for your help, i spent over 6 hours for this problem only. I have to improve a lot :( $\endgroup$ – Kiran Nov 20 '16 at 15:13
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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 $$181440 - 75600 - 15120 = 90720$$

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