total number of list of delegates in this question The question is  :
A conference is attended by 200 delegates and is held in a hall. The hall has $7$ doors labelled A,B,..G. At each door, an entry book is kept and the delegates entering through that door sign in the order in which they enter. If each delegate is free to enter any time and through any door he likes, then find the total number of different sets of seven lists would arise.
My approach:
Let $x_1,x_2,..x_7$ persons are entering into gate  A,B,C,....G gates respectively. now 
$$x_1 + x_2 +..+x_7 = 200$$
total number of ways to enter $={206\choose 6}$, now the arrangement of 200 people will also count so ans is ${206\choose 6}*200!$. However I am not sure if this is correct. If I am incorrect, kindly provide alternate solution.
 A: As I read the problem, you don't keep track of which list goes with each room because it talks about sets of lists.  It then does not matter if you swap the attendees in rooms A and B because you will get the same two lists.  You do have to keep track of the order that the people sign each list.  I don't think it is very clear.  In this reading, you can line up the people in $200!$ ways, then split them into rooms by a weak partition of $200$ into $7$ pieces.  The number of sets of lists will be the product of these two.
A: Each candidate can choose one of the seven doors. This doesn't restrict other candidates' choices. So there's a total of $7 \times 7 \ldots 7 = 7^{200}$ number of assigning candidates to doors. In this setup, the order of assignment doesn't matter, i.e. if all candidates choose door 1, the list will be $\{1:200, 2:0, \ldots  7:0\}$ (one assignment).
This comes from the fact that all doors and all candidates are unique. If the doors were unique and candidates identical, you should have used the stars-and-bars approach.
In your case, for example, the assignment $\{1:199, 2:1, \ldots  7:0\}$ can be done in 200 ways, because all candidates are different. If they were the same, that would be 1 choice.
