How many words can be formed using all the letters of "DAUGHTER" so that vowels always come together?
I understood that there are 6 letters if we consider "AUE" as a single letter and answer would be 6!. Again for AUE it is 3!, but I didn't get why to do 6! * 3!.
Can't we just add (6! + 3!) to get final result?
Best Answer
Imagine your three vowels as a block. Ignore the ordering of the block for the moment. Now you have 6 remaining "letters"- the original consonants plus this "vowel block". Now, there are $6$ of these altogether. So you get $6!$ ways of ordering them- which is the standard permutation formula. Now for $each$ of these orderings, you can internally order the three vowels in the block. So you multiply by $3!$, which is the number of ways to order the vowels.
EDIT: The intuition for multiplication can be strengthened with some visualisation. Take a sheet of paper. Imagine all the "external" orderings of consonants and vowel block as being listed vertically. Now for each of these, write horizontally the 6 corresponding "internal" orderings of the vowels. You get a rectangular grid, each cell containing exactly one ordering. The area of this grid is clearly the breadth by its height. Hence $6!3!$.