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# 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?

Last updated date: 21st Jun 2024
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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.

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.