In how many ways the word hydrophobia can be arranged so that all vowels come together

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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)

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

asked Nov 20, 2016 at 14:25

KiranKiran

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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 $105840-15120=90720.$

answered Nov 20, 2016 at 15:11

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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 $$181440 - 75600 - 15120 = 90720$$

answered Nov 22, 2016 at 6:05

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In how many different ways, can the letters of the words EXTRA be arranged so that the vowels are never together?

1. 168
2. 48
3. 120
4. 72

Answer (Detailed Solution Below)

Option 4 : 72

Calculation:

EXTRA → Total number of words = 5 and total number of vowels = 2

The word EXTRA can be arranged in 5! ways = 120 ways

The word EXTRA can be arranged in such a way that the vowels will be together = 4! × 2!

⇒ (4 × 3 × 2 × 1) × (2 × 1)

⇒ 48 ways

The letters of the words EXTRA be arranged so that the vowels are never together = (120 - 48) = 72 ways.

∴ The letters of the words EXTRA be arranged so that the vowels are never together in 72 ways.

Answer

Verified

Hint: To solve this problem we have to know about the concept of permutations and combinations. But here a simple concept is used. In any given word, the number of ways we can arrange the word by jumbling the letters is the number of letters present in the word factorial. Here factorial of any number is the product of that number and all the numbers less than that number till 1.
$ \Rightarrow n! = n(n - 1)(n - 2).......1$

Complete step by step answer:
Given the word TRAINER, we have to arrange the letters of the word in such a way that all the vowels in the word TRAINER should be together.
The number of vowels in the word TRAINER are = 3 vowels.
The three vowels in the word TRAINER are A, I, and E.
Now these three vowels should always be together and these vowels can be in any order, but they should be together.
Here the three vowels AIE can be arranged in 3 factorial ways, as there are 3 vowels, as given below:
The number of ways the 3 vowels AIE can be arranged is = $3!$
Now arranging the consonants other than the vowels is given by:
As the left out letters in the word TRAINER are TRNR.
The total no. of consonants left out are = 4 consonants.
Now these 4 consonants can be arranged in the following way:
As in the 4 letters TRNR, the letter R is repeated for 2 times, hence the letters TRNR can be arranged in :
$ \Rightarrow \dfrac{{4!}}{{2!}}$
But the letters TRNR are arranged along with the vowels A,I,E, which should be together always but in any order.
Hence we consider the three vowels as a single letter, now TRNR along with AIE can be arranged in:
$ \Rightarrow \dfrac{{5!}}{{2!}}$
But here the vowels can be arranged in $3!$ as already discussed before.
Thus the word TRAINER can be arranged so that the vowels always come together are given below:
$ \Rightarrow \dfrac{{5!}}{{2!}} \times 3! = \dfrac{{120 \times 6}}{2}$
$ \Rightarrow 360$

The number of ways the word TRAINER can be arranged so that the vowels always come together are 360.

Note: Here while solving such kind of problems if there is any word of $n$ letters and a letter is repeating for $r$ times in it, then it can be arranged in $\dfrac{{n!}}{{r!}}$ number of ways. If there are many letters repeating for a distinct number of times, such as a word of $n$ letters and ${r_1}$ repeated items, ${r_2}$ repeated items,…….${r_k}$ repeated items, then it is arranged in $\dfrac{{n!}}{{{r_1}!{r_2}!......{r_k}!}}$ number of ways.

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