Total number Of valid Home delivery arrangements
Given number of orders, find the number of valid arrangements of orders where delivery of ith order is always after the pickup of ith order.Examples:
Input: N = 1
Output: 1
Here total event is 2. They are {P1, D1}.
Total possible arrangement is 2! = 2. [P1, D1] and [D1, P1].
So only valid arrangement possible: [P1, D1].
[D1, P1] is invalid arrangement as delivery of 1st order is done before pickup of 1st order.Input: N = 2
Output: 6
Here total event is 4. They are {P1, D1, P2, D2}.
Here total possible arrangements are 4! = 24.
Among them 6 are valid arrangements:
[P1, P2, D1, D2], [P1, D1, P2, D2], [P1, P2, D2, D1], [P2, P1, D2, D1], [P2, P1, D1, D2], and [P2, D2, P1, D1].
Rest all are invalid arrangements.
Some invalid arrangements:
[P1, D1, D2, P2] – Delivery of 2nd order is done before pickup
[P2, D1, P1, D2] – Delivery of 1st order is done before pickup
[D1, D2, P2, P1] – Delivery of both order is before pickup
Approach 1:
 Consider N = 4, we have total of 8 events.
 There are 4 events for pickup {P1, P2, P3, P4} and 4 events for delivery {D1, D2, D3, D4}.
 If we consider only pickup events, there are no restrictions in arrangements between pickups. So, total possible arrangements 4!
 Now we consider delivery. We start from the last pickup we made.
 For D4, We can place D4 only after P4.
That is P1, P2, P3, P4, __. So only 1 valid position.  For D3, We can place D3 in any one of this following position.
They are P1, P2, P3, __, P4, __, D4, __ . So 3 valid position.  For D2, We can place D2 in any one of this following position.
They are P1, P2, __, P3, __, P4, __, D4, __, D3 __ .So 5 valid position.  For D1, We can place D1 in any one of this following position.
They are P1, __, P2, __, P3, __, P4, __, D4, __, D3 __, D2, __ .So, 7 valid positions.
So total valid arrangements: 4! * (1 * 3 * 5 * 7)
 For D4, We can place D4 only after P4.
For any N, total valid arrangements:
Below is the implementation of the above approach.
C++
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std; // Function to find arrangements int Arrangements( int N) { int result = 1; for ( int i = 1; i <= N; i++) { // Here, i for factorial and // (2*i1) for series result = result * i * (2 * i  1); } return result; } // Driver code int main() { int N = 4; cout << Arrangements(N); return 0; } 
Java
// Java implementation of the above approach class GFG{ // Function to find arrangements public static int Arrangements( int N) { int result = 1 ; for ( int i = 1 ; i <= N; i++) { // Here, i for factorial and // (2*i1) for series result = result * i * ( 2 * i  1 ); } return result; } // Driver code public static void main(String[] args) { int N = 4 ; System.out.print(Arrangements(N)); } } // This code is contributed by divyeshrabadiya07 
Python3
# Python3 implementation of the above approach # Function to find arrangements def Arrangements(N): result = 1 for i in range ( 1 , N + 1 ): # Here, i for factorial and # (2*i1) for series result = result * i * ( 2 * i  1 ) return result # Driver code N = 4 ; print (Arrangements(N)); # This code is contributed by Akanksha_Rai 
C#
// C# implementation of the above approach using System; class GFG{ // Function to find arrangements public static int Arrangements( int N) { int result = 1; for ( int i = 1; i <= N; i++) { // Here, i for factorial and // (2*i1) for series result = result * i * (2 * i  1); } return result; } // Driver code public static void Main(String[] args) { int N = 4; Console.Write(Arrangements(N)); } } // This code is contributed by AnkitRai01 
2520
Time complexity: O(N)
Auxiliary Space complexity: O(1)
Approach 2:
 For N number of orders, we have
 So the total number of arrangements possible is

Now each order can only be valid if delivery is one after pickup.
For each [Pi, Di], we can’t change this arrangement ie we can’t do [Di, Pi].There is only one valid arrangement for each such order. So we need to divide by 2 for each order. So total valid arrangement is
Below is the implementation of the above approach.
C++
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std; // Function to find arrangements int Arrangements( int N) { int result = 1; for ( int i = 1; i <= 2 * N; i += 2) result = (result * i * (i + 1)) / 2; return result; } // Driver code int main() { int N = 4; cout << Arrangements(N); return 0; } 
Java
// Java implementation of the above approach import java.util.*; class GFG{ // Function to find arrangements public static int Arrangements( int N) { int result = 1 ; for ( int i = 1 ; i <= 2 * N; i += 2 ) result = (result * i * (i + 1 )) / 2 ; return result; } // Driver code public static void main(String args[]) { int N = 4 ; System.out.print(Arrangements(N)); } } // This code is contributed by Code_Mech 
Python3
# Python3 implementation of the above approach # Function to find arrangements def Arrangements(N): result = 1 ; for i in range ( 1 , ( 2 * N) + 1 , 2 ): result = (result * i * (i + 1 )) / 2 ; return int (result); # Driver code if __name__ = = '__main__' : N = 4 ; print (Arrangements(N)); # This code is contributed by gauravrajput1 
C#
// C# implementation of the above approach using System; class GFG{ // Function to find arrangements public static int Arrangements( int N) { int result = 1; for ( int i = 1; i <= 2 * N; i += 2) result = (result * i * (i + 1)) / 2; return result; } // Driver code public static void Main() { int N = 4; Console.Write(Arrangements(N)); } } // This code is contributed by Code_Mech 
2520
Time complexity: O(N)
Auxiliary Space complexity: O(1)
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