How many words can be formed using all the letters of "DAUGHTER" so that vowels always come together? 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?

Consider two boxes to be filled with coins and we are given 7 different coins. For the first place you can select any of those 7 coins. Let we have chosen first coin for 1st box now you can fill the second box in 6 different ways. So with 1st coin in first box we have 6 permutations. Instead of having filled the first box with fist coin we could have filled it with 2nd coin, again the second box can be filled in 6 different ways. Then 6 ways if we put 3rd coin in first box, then 6 ways if we put 4th coin in first box, then 6 ways if we put 5th coin in first box, the 6 ways if we put 6th coin in first box and finally 6 ways if we put 7th coin in first box. So you see we have (6+6+.. upto 7 times permutations). In general if we can fill first box in $m$ ways and second box in $n$ ways then the total number of ways in which we can make different permutations is $m*n$, this is because for every 1 way out of m ways in which first box can be filled there are $n$ ways to fill the second box. In other words for 1st object in box one there are $n$ possible ways to fill the second box, for 2nd object in first box there are again $n$ possible ways to fill the second box. At last for the mth object in the first box there are $n$ possible ways to fll the second box. This fact is called "Fundamental Theorem of Multiplication".
If we had three boxes where first box could be filled in m, second in n and third in p ways then total no of permutations would be $m\times n \times p$, why? Because From every one possible permutation out of $m*n$ of first two boxes the third can be filled in $p$ different ways.
We can extend this theorem to any number of boxes. Now in place of boxes we can conceive of jobs where 1st box is 1st job which can be done in $m$ ways, 2nd box is 2nd job which can be done in $n$ ways and so on.
