A hotel has 6 single rooms, 6 double rooms and 4 rooms in which 3 persons can stay. In how many ways can 30 persons be accommodated in this hotel? I wanted to know if my answer and the method I used is correct or not...
First, we pick $6$ out of $30$ people for the $6$ single rooms = $^{30}C_6$
They can be in any of rooms $\implies ^{30}C_6 \cdot 6!$
Then, we pick $12$ people out of remaining $24$ and pair them up for $6$ dual rooms $= \dfrac{^{24}C_{12} \cdot 12!}{(2^6 \cdot 6!)}$
Next, we take remaining people and form groups of $\implies \dfrac{12!}{(4^3 \cdot 3!)}$
Finally,we multiply all three terms...
Ans $ = \dfrac{^{30}C_6 \cdot 6! \cdot  12! \cdot {^{24}C_{12} \cdot 12!}}{(2^6 \cdot 6!) \cdot {(4^3 \cdot 3!)}} $
But there can be other ways like first choosing triplets then pairs then singles or pairs then triplets then singles and so on, so do we ignore them or add them?
 A: As stated in the comments, the flaw in your answer is that while you treated the rooms as distinguishable for the single rooms, you did not do so for the rooms that accommodate more than one person.

In how many ways can a hotel with six single rooms, six double rooms, and four rooms in which three people can stay assign $30$ guests to those rooms?

Let's assume that the rooms are numbered and that we assign people to rooms of the same size in order of increasing room number. 
We can select six lucky people to receive single rooms in $\binom{30}{6}$ ways and assign them to those rooms in $6!$ ways.  Thus, we can assign six people to the single rooms in 
$$\binom{30}{6}6!$$
ways.
That leaves us with $24$ people.  We can assign two of them to each of the six rooms that accommodate two people in 
$$\binom{24}{2}\binom{22}{2}\binom{20}{2}\binom{18}{2}\binom{16}{2}\binom{14}{2}$$ 
ways.
That leaves us with $12$ people.  We can assign three of them to each of the four rooms that accommodate three people in 
$$\binom{12}{3}\binom{9}{3}\binom{6}{3}\binom{3}{3}$$
ways.  
Thus, the number of ways to assign $30$ people to the rooms in the hotel is 
$$\binom{30}{6}6! \cdot \binom{24}{2}\binom{20}{2}\binom{18}{2}\binom{16}{2}\binom{14}{2} \cdot \binom{12}{3}\binom{9}{3}\binom{6}{3}\binom{3}{3}$$
Simplifying this expression yields
$$\binom{30}{1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3} = \frac{30!}{1!1!1!1!1!1!2!2!2!2!2!2!3!3!3!3!}$$
To see why, imagine lining up $30$ people in some order, assigning the first six people to the singles in increasing room number order, assigning the next six (disjoint) pairs of people to the doubles in increasing room number order, and assigning the next four (disjoint) trios of people in increasing room number order.  The numbers in the denominator represent the number of orders in which we assign the same set of people to the same room without creating a different set of room assignments.
Note that this number does not depend on the order in which the people are assigned to the rooms.
Where did you make a mistake?
You correctly counted the number of assignments of six people to single rooms.
The number of ways we can assign twelve of the remaining twenty-four people to the six doubles is 
$$\binom{24}{2}\binom{22}{2}\binom{20}{2}\binom{18}{2}\binom{16}{2}\binom{14}{2} = \frac{24!}{2!2!2!2!2!2!12!}$$
which can be expressed in the form
$$\binom{24}{12} \cdot \frac{12!}{2!2!2!2!2!2!}$$
By dividing this number by $6!$, you treated the rooms as indistinguishable.  
You made the analogous error in assigning the remaining twelve people to the four triples.
A: MathJax goes over the top of my head, so I ask you to bear with me.
First, you choose 6 out of the 30 for the single rooms, giving you 30C6, and these can be arranged in 6! ways, giving you 30C6 * 6!
Next, choose 12 out of the remaining 24, 24C12, now, arranging them in duos for room-sharing, gives you 24C12 * 12C2 options, and these people can be arranged in 6 rooms, so multiply by 6!.
Entire thing now is (30C6 * 6!) * (24C12 * 12C2 * 6!)
Now for the remaining 12 people, 12C12 (selecting 12 people out of the 12 people)
, then 12C3 (finding how many ways they can be arranged in groups of 3), and multiplying by 4!, due to the four rooms, giving (12C12 * 12C3 * 4!)
Final answer comes out to be (30C6 * 6!) * (24C12 * 12C2 * 6!) * (12C12 * 12C3 * 4!).
That's what I got, at-least.
A: If the rooms are distinguishable, you can imagine putting the people in a line, which you can do in $30!$ ways.  Then assign them to to the rooms in order.  Swapping the two people in a double room does not change anything, nor does swapping the three people in a triple, so the number of arrangements is $$\frac {30!}{2!^63!^4}$$
If the rooms are not distinguishable you can permute the rooms as well for 
$$\frac{30!}{6!2!^66!3!^44!}$$
arrangements
