200 Delegates and a Conference Hall

Suppose that $D$ delegates go through $d$ doors. Consider the numbers $1,2,3,\ldots,D$ and $d-1$ copies of $0$ . Since the $d-1$ copies of $0$ are indistinguishable, there are $\frac{(D+d-1)!}{(d-1)!} \; =\; {}^{D+d-1}P_D$ ways in which these symbols can be ordered.

The list of numbers before the first $0$ is the list of delegates through the first door (in order). The list of numbers between the first $0$ and the second $0$ is the list of delegates through the second door (in order), and so on. Thus ${}^{D+d-1}P_D$ is the number of possible sets of signing-in lists.

In the case that $D=200$ and $d=7$ , the number of signing-in lists is ${}^{206}P_{200}$ , making the answer $\boxed{406}$ .

Let a,b,c,d,e,f,g be the number of persons entering through the A,B,C,D,E,F,G respectively.
Then,
a+b+c+d+e+f+g=200
Therefore number of ways of dividing 200 into 7 parts = $^{7+200-1} C _{200}$ = $^{206}C_{200}$ = $^{206}C_6$
The total number of ways of dividing 200 persons into 7 sets containing a,b,c,d,e,f,g persons repectively = $\frac{200!}{a!b!c!d!e!f!g!}$
But the persons in the set of door A can be arranged in $a!$ ways. Same holds for the other sets.
So, total number of ways of dividing 200 persons into 7 sets = $\frac{200!}{a!b!c!d!e!f!g!} \times a!b!c!d!e!f!g!$ = $200!$
$\therefore$ Required number = $^{206}C_{200} \times 200!$ = $\boxed{^{206}P_{200}}$