Open in App Suggest Corrections 1 Answer VerifiedHint: 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: 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. |