human
Solve in Python:
Given n orders, each order consist in pickup and delivery services. 
Count all valid pickup/delivery possible sequences such that delivery(i) is always after of pickup(i). 
Since the answer may be too large, return it modulo 10^9 + 7.
 
Example 1:
Input: n = 1
Output: 1
Explanation: Unique order (P1, D1), Delivery 1 always is after of Pickup 1.

Example 2:
Input: n = 2
Output: 6
Explanation: All possible orders: 
(P1,P2,D1,D2), (P1,P2,D2,D1), (P1,D1,P2,D2), (P2,P1,D1,D2), (P2,P1,D2,D1) and (P2,D2,P1,D1).
This is an invalid order (P1,D2,P2,D1) because Pickup 2 is after of Delivery 2.

Example 3:
Input: n = 3
Output: 90

 
Constraints:

1 <= n <= 500
assistant
#Here is the Python program you requested.

class Solution:
    def countOrders(self, n: int) -> int:
        DP = [0]*(n+1)
        DP[1] = 1
        for i in range(2, n+1):
            newpos = ((i*2-1)+1)*(i*2-1)//2
            DP[i] = newpos*DP[i-1]
        
        return DP[-1]%(10**9+7)