Answer  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.
i) All are distinct letters so, and there are 7 letters, all permutations will work which is equal to $7!$. ii) L is first letter means, the rest letters are only allowed to be permuted, so 6 remaining letters so answer is all permutations of 6 letters i.e. $6!$. iii) All vowels are together, you have for vowels here U and O, so in a word they always come together (As a block) and in 2 ways either UO or OU, all the words with UO is permutation of 5 letters and this one block, i.e. $(5+1)!$ and similarly for the other block, hence you have $2\times 6!$ in total. iv) Note given a word, L either comes before U or after U. But suppose you have a word with U first and L coming afterwards, then if you switch the positions you have a corresponding word with L first and U coming afterwards, so they come in pairs, so they partition all words into two equal sets, so answer for this is $(7!)/2$. v) Using the same logic as before, in a word where ..L..U..W.., is needed, if the word contains in any other order then just interchange places to get teh required order. So here we have more possibilities i.e. $3! = 6$ ways, they come in a set of 6, hence the partion is into 6 equal size sets hence the answer will be $(7!)/6$. *In the above problem, I have assumed that I have to use all the letters to form the word, but if that is not the case, the values may vary but the logic still works. Open in App Standard IX Social Science
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How many words can be formed from letter of the word Lucknow when u c/k never come together?
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