In a country, there are some cities and the city named [i]Ben Song[/i] is capital. Each cities are connected with others by some two-way roads. One day, the King want to choose $n$ cities to add up with [i]Ben Song[/i] city to establish an [i]expanded capital[/i] such that the two following condition are satisfied:

(i) With every two cities in [i]expanded capital[/i], we can always find a road connecting them and this road just belongs to the cities of [i]expanded capital[/i].

(ii) There are exactly $k$ cities which do not belong to [i]expanded capital[/i] have the direct road to at least one city of [i]expanded capital[/i].

Prove that there are at most $\binom{n+k}{k}$ options to expand the capital for the King.
1. **Numbering the Cities**: 
   We first number all the cities as \( a_1, a_2, \ldots, a_m \). This will be the first number of each city.

2. **Constructing the Sequence**:
   For each possible expanded capital, we will construct an increasing sequence \( 1 \leq c_1 < c_2 < \ldots < c_n \leq n+k \). 

3. **Initial Setup**:
   Let \( S \) be the set of cities in the expanded capital, initially containing only Ben Song. 

4. **Assigning Second Numbers**:
   - In the first step, we look at the cities that are directly connected to Ben Song and give them a second number. If there are \( t \) such cities, we number them \( b_1, \ldots, b_t \) so that their second numbers are in the same order as their first numbers.
   - We then choose the city in the expanded capital with the least second number, give that value to \( c_1 \), and add the city to \( S \).

5. **Iterative Process**:
   - In each step, if the last city we added was \( b_{c_r} \), and we gave second numbers up to \( b_s \), we look at the cities that are directly connected to \( b_{c_r} \) and not to any other element of \( S \), and give them consecutive second numbers, starting with \( b_{s+1} \) and in increasing order of first numbers.
   - We then choose the city in the expanded capital with the least second number that is not yet in \( S \), give that value to \( c_{r+1} \), and include the city in \( S \).

6. **Ensuring Order and Bound**:
   - We know that \( c_{r+1} > c_r \) because in the previous step we chose the city in the expanded capital with the least second number, and it was \( b_{c_r} \).
   - We also know that \( c_{r+1} \leq n+k \) because each city with a second number is connected to some city in \( S \), and is not Ben Song.

7. **Uniqueness of Sequences**:
   - If two possible expanded capitals have the same sequence, then we can use the previous algorithm to show that both have the same cities, since the numeration in each step depends only on the previous steps.
   - Therefore, any two possible expanded capitals have different sequences, and there are at most \( \binom{n+k}{n} = \binom{n+k}{k} \) of them.

\(\blacksquare\)

The final answer is \( \boxed{ \binom{n+k}{k} } \)