How many words can be formed from the letters of I so the vowels come together II the vowels never come together?

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How many words can be formed from the letters of the word "d a u g h t e r" so that the vowels never come together?

There are $3$ vowels and $5$ consonants. I first arranged $5$ consonants in five places in $5!$ ways. $6$ gaps are created. Out of these $6$ gaps, I selected $3$ gaps in ${}_6C_3$ ways and then made the vowels permute in those $3$ selected places in $3!$ ways. This leads me to my answer $5!\cdot {}_6C_3 \cdot 3! = 14400$.

The answer given in my textbook is $36000$. Which cases did I miss? What is wrong in my method?

CiaPan

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asked Apr 18, 2017 at 13:23

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I believe you have misinterpreted what the question is asking you. It asks how many ways to arrange the letters so all 3 vowels aren't together, i.e. D$\color{red}{\text{AUE}}$GHTR has all three together. You want to avoid this. The number of ways in which they are all together is $6!\times3!=4320.$ The total number of ways to arrange DAUGHTER is $8!=40320$, so the number of ways to avoid the 3 vowels being together is $40320-4320=36000$.

What is wrong with your method is that you didn't even allow words like T$\color{blue}{\text{AU}}$HRDEG because of the AU, whereas the question allows this to count. Hope that made sense.

answered Apr 18, 2017 at 13:38

John DoeJohn Doe

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Solution : The letters of the word daughter are “d,a,u,g,h,t,e,r”.
So, the vowels are ‘a, u, e’ and the consonants are “d,g,h,t,r”.
(i)Now, all the vowels should come together, so consider the bundle of vowels as one letter, then total letters will be `6`.
So, the number of words formed by these letters will be `6!`
but, the vowels can be arranged differently in the bundle, resulting in different words, so we have to consider the arrangements of the `3` vowels.
So, the arrangements of vowels will be `3!`
Thus, the total number of words formed will be equal to `(6!×3!)=4320`
(ii)First arrange `5` consonants in five places in `5!` ways.
`6` gaps are created. Out of these `6` gaps, select `3` gaps in `6_(C_3)`​ ways and then make the vowels permute in those `3` selected places in `3!` ways.
This leads `5!×6_(C_3)​xx3!`=14400.

How many words can be formed from the letters of the word ‘DAUGHTER’ so that(i) The vowels always come together?(ii) The vowels never come together?

Answer

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Hint: The word daughter has $8$ letters in which $3$ are vowels. For the vowels to always come together consider all the $3$ vowels to be one letter (suppose V) then total letters become $6$ which can be arranged in $6!$ ways and the vowels themselves in $3!$ ways.Complete step-by-step answer:
Given word ‘DAUGHTER’ has $8$ letters in which $3$ are vowels and 5 are consonants. A, U, E are vowels and D, G, H, T, R are consonants.
(i)We have to find the total number of words formed when the vowels always come together.
Consider the three vowels A, U, E to be one letter V then total letters are D, G, H, T, R and V. So the number of letters becomes $6$
So we can arrange these $6$ letters in $6!$ ways. Since the letter V consists of three vowels, the vowels themselves can interchange with themselves. So the number of ways the $3$vowels can be arranged is $3!$
Then,
$ \Rightarrow $ The total number of words formed will be=number of ways the $6$ letters can be arranged ×number of ways the $3$ vowels can be arranged
On putting the given values we get,
$ \Rightarrow $ The total number of words formed=$6! \times 3!$
We know $n! = n \times \left( {n - 1} \right)! \times ...3,2,1$
$ \Rightarrow $ The total number of words formed=$6 \times 4 \times 5 \times 3 \times 2 \times 1 \times 3 \times 2 \times 1$
On multiplying all the numbers we get,
 $ \Rightarrow $ The total number of words formed=$24 \times 5 \times 6 \times 6$
$ \Rightarrow $ The total number of words formed=$120 \times 36$
$ \Rightarrow $ The total number of words formed=$4320$
The number of words formed from ‘DAUGHTER’ such that all vowels are together is $4320$.

(ii)We have to find the number of words formed when no vowels are together.
Consider the following arrangement- _D_H_G_T_R
The spaces before the consonants are for the vowels so that no vowels come together. Since there are $5$ consonants so they can be arranged in $5!$ ways.
There are $6$ spaces given for $3$ vowels. We know to select r things out of n things we write use the following formula-${}^{\text{n}}{{\text{C}}_{\text{r}}}$=$\dfrac{{n!}}{{r!n - r!}}$
So to select $3$ spaces of out $6$ spaces =${}^6{{\text{C}}_3}$
And the three vowels can be arranged in these three spaces in $3!$ ways.
$ \Rightarrow $ The total number of words formed=${}^6{{\text{C}}_3} \times 3! \times 5!$
$ \Rightarrow $ The total number of words formed=$\dfrac{{6!}}{{3!6 - 3!}} \times 5! \times 3!$
$ \Rightarrow $ The total number of words formed=$\dfrac{{6!}}{{3!}} \times 5!$
On simplifying we get-
$ \Rightarrow $ The total number of words formed=$\dfrac{{6 \times 5 \times 4 \times 3!}}{{3!}} \times 5!$
$ \Rightarrow $ The total number of words formed=$120 \times 5 \times 4 \times 3 \times 2 \times 1$
On multiplying we get,
$ \Rightarrow $ The total number of words formed=$14400$
The total number of words formed from ‘DAUGHTER’ such that no vowels are together is $14400$.

Note: Combination is used when things are to be arranged but not necessarily in order. Permutation is a little different. In permutation, order is important. Permutation is given by-
$ \Rightarrow {}^n{P_r} = \dfrac{{n!}}{{n - r!}}$ Where n=total number of things and r=no. of things to be selected.

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