How many different words can be formed from the word daughter ending and beginning are consonants?

Solution : No. of letters in the word 'DAUGHTER' =8
(i) If there is no restriction,
No. of permutations to fill 8 places with given 8 letters `= P_(8) = 8! = 40320`.
(ii) In the word 'DAUGHTER'. Vowels = A,U,E and Consonants = D,G,T,H,R
`rArr` No. of vowels and consonants are 3 and 5 respectively.
`:'` A,U,E are always together
`:.` Treating these letters as one letter, the letters are `5 +1 = 6`
Now the number of arrangement of 6 letters at 6 places `= .^(6)P_(6) = 6! = 720` and Number of arrangement of letters A,U,E at 3 places `= .^(3)P_(3) = 3! =6`.
`:.` Required arrangements `= 720 xx 6 = 4320`
(iii) For the words starting with A
A will be at first place and no of arrangements of remaining 7 letters at remaining 7 places
`=.^(7)P_(7) = 7! = 5040`.
(iv) For the words starting with A and ending with R,
No. of arrangements of remaining 6 letters at remaining 6 places
`=.^(6)P_(6) = 6! = 720`.

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Text Solution

Solution : No. of letters in the word 'DAUGHTER' =8
(i) If there is no restriction,
No. of permutations to fill 8 places with given 8 letters `= P_(8) = 8! = 40320`.
(ii) In the word 'DAUGHTER'. Vowels = A,U,E and Consonants = D,G,T,H,R
`rArr` No. of vowels and consonants are 3 and 5 respectively.
`:'` A,U,E are always together
`:.` Treating these letters as one letter, the letters are `5 +1 = 6`
Now the number of arrangement of 6 letters at 6 places `= .^(6)P_(6) = 6! = 720` and Number of arrangement of letters A,U,E at 3 places `= .^(3)P_(3) = 3! =6`.
`:.` Required arrangements `= 720 xx 6 = 4320`
(iii) For the words starting with A
A will be at first place and no of arrangements of remaining 7 letters at remaining 7 places
`=.^(7)P_(7) = 7! = 5040`.
(iv) For the words starting with A and ending with R,
No. of arrangements of remaining 6 letters at remaining 6 places
`=.^(6)P_(6) = 6! = 720`.

How many words with or without meaning can be formed from the letters of the word, 'DAUGHTER', so thati all vowels occur together?ii all vowels do not occur together?

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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 arrangedOn 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_RThe 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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