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find the number of different 8 letter arrangements that can be made from the letters of the word DAUGHTER (8 letters) so that --All vowels occur together. I thought of a solution which is as follows-- if all the vowels are occurring together then wherever these three are they must be together so let us say they start occurring right from start or even from the second or third letter ((no special case or reason is there )) then no of choices for first letter=3 ////@@3 vowels are there. second letter=2 //@@ 2 vowels are remaining . third letter =1 //@@1 vowel is remaining which we must use. fourth letter =5 choices(D,G,H,T,R) fifth =4 choices • • • • so on till the last letter. so number of different possibilities by fundamental principle of counting= 3×2×1×5×4×3×2×1=5!×3!. But the answer is 6!×3!,which case I am missing in my solution because of which I am getting wrong solution(5!×3!)

Freelancer
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    You still have to multiply by the 6 possible positions the 3 vowels can occur in the word: 1-3, 2-4,...,6-8. – user84413 Oct 06 '15 at 18:22

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I think your basic method sounds ok, but the details get a little hazy.

Suppose we take $AUE$ as a block. Then you are looking at $6$ characters...namely, $\{D,G,H,T,R,AUE\}$. There are $6!$ ways to permute this collection hence there are $6!$ ways to arrange the letters so that the group $AUE$ occurs in that precise order. Now, of course there are also $3!=6$ ways to permute the vowels and, given any permutation, there are $6!$ ways to arrange all the letters so that the vowels occur in that precise order. Hence $6!^*3!$ ways altogether.

lulu
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  • You meant there are $\color{red}{3}!$ ways to arrange the letters of the block of vowels. – N. F. Taussig Oct 07 '15 at 10:01
  • @N.F.Taussig Thanks! I meant "there are $6!$ ways to arrange the letters so that the vowels appear in order as $AUE$ or any other permutation of the vowels" but I agree that what I wrote is unclear. I will edit. – lulu Oct 07 '15 at 11:10