Number of ways to distribute N Paper Set among M students

Given N students and a total of M sets of question paper where M ? N. All the M sets are different and every sets is available in sufficient quantity. All the students are sitting in a single row. The task is to find the number of ways to distribute the question paper so that if any M consecutive students are selected then each student has a unique question paper set. The answer could be large, so print the answer modulo 10⁹ + 7.
Example:

Input: N = 2, M = 2 
Output: 2 
(A, B) and (B, A) are the only possible ways.
Input: N = 15, M = 4 
Output: 24 

Approach: It can be observed that the number of ways are independent of N and only depend on M. First M students can be given M sets and then the same pattern can be repeated. The number of ways to distribute the question paper in this way is M!. For example, 

N = 6, M = 3 
A, B, C, A, B, C 
A, C, B, A, C, B 
B, C, A, B, C, A 
B, A, C, B, A, C 
C, A, B, C, A, B 
C, B, A, C, B, A 

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

const int MOD = 1000000007;

// Function to return n! % 1000000007
int factMod(int n)
{

    // To store the factorial
    long fact = 1;

    // Find the factorial
    for (int i = 2; i <= n; i++) {
        fact *= (i % MOD);
        fact %= MOD;
    }

    return fact;
}

// Function to return the
// count of possible ways
int countWays(int n, int m)
{
    return factMod(m);
}

// Driver code
int main()
{
    int n = 2, m = 2;

    cout << countWays(n, m);

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG
{
static int MOD = 1000000007;

// Function to return n! % 1000000007
static int factMod(int n)
{

    // To store the factorial
    long fact = 1;

    // Find the factorial
    for (int i = 2; i <= n; i++)
    {
        fact *= (i % MOD);
        fact %= MOD;
    }
    return (int)fact;
}

// Function to return the
// count of possible ways
static int countWays(int n, int m)
{
    return factMod(m);
}

// Driver code
public static void main(String args[])
{
    int n = 2, m = 2;

    System.out.print(countWays(n, m));
}
}

// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of the approach
MOD = 1000000007;

# Function to return n! % 1000000007
def factMod(n) :

    # To store the factorial
    fact = 1;

    # Find the factorial
    for i in range(2, n + 1) :
        fact *= (i % MOD);
        fact %= MOD;

    return fact;

# Function to return the
# count of possible ways
def countWays(n, m) :

    return factMod(m);

# Driver code
if __name__ == "__main__" :

    n = 2; m = 2;

    print(countWays(n, m));

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;

class GFG
{
    static int MOD = 1000000007;

    // Function to return n! % 1000000007
    static int factMod(int n)
    {
        // To store the factorial
        int fact = 1;

        // Find the factorial
        for (int i = 2; i <= n; i++)
        {
            fact *= (i % MOD);
            fact %= MOD;
        }
        return fact;
    }

    // Function to return the
    // count of possible ways
    static int countWays(int n, int m)
    {
        return factMod(m);
    }

    // Driver code
    public static void Main()
    {
        int n = 2, m = 2;
        Console.Write(countWays(n, m));
    }
}

// This code is contributed by Sanjit Prasad

Javascript

<script>
// javascript implementation of the approach
     MOD = 1000000007;

    // Function to return n! % 1000000007
    function factMod(n) {

        // To store the factorial
        var fact = 1;

        // Find the factorial
        for (i = 2; i <= n; i++) {
            fact *= (i % MOD);
            fact %= MOD;
        }
        return parseInt( fact);
    }

    // Function to return the
    // count of possible ways
    function countWays(n , m) {
        return factMod(m);
    }

    // Driver code

        var n = 2, m = 2;

        document.write(countWays(n, m));

// This code contributed by aashish1995
</script>

Output:

2

Time Complexity: O(n)

Auxiliary Space: O(1)