# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach 1:

1. Consider N = 4, we have total of 8 events.
2. There are 4 events for pickup {P1, P2, P3, P4} and 4 events for delivery {D1, D2, D3, D4}.
3. If we consider only pickup events, there are no restrictions in arrangements between pickups. So, total possible arrangements 4!
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 any N, total valid arrangements: Below is the implementation of the above approach.

## C++

 `// C++ implementation of the above approach ` ` `  `#include ` `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*i-1) 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*i-1) 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*i-1) 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*i-1) 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 `

Output:

```2520
```

Time complexity: O(N)
Auxiliary Space complexity: O(1)

Approach 2:

1. For N number of orders, we have 2. So the total number of arrangements possible is 3. 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 ` `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 `

Output:

```2520
```

Time complexity: O(N)
Auxiliary Space complexity: O(1)

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