200 delegates attend a conference is held in a hall. The hall has 7 doors, marked $A,B,\ldots, G$ . At each door, an entry book is kept, and the delegates entering through that door sign it in the order in which they enter. If each delegate is free to enter any time and through any door he likes, how many different sets of seven lists could arise in all?

If the answer can be expressed an $^nP_k$ for some positive integers $n,k$ , find (the minimium value of) $n+k$ .

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Details and Assumptions
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- Assume that every person signs only at his first entry.
- $\large ^nP_k=\dfrac{n!}{(n-k)!}$ .

The answer is 406.

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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 $\boxed{406}$ .