# Number of Simple Graph with N Vertices and M Edges

Given two integers N and M, the task is to count the number of simple undirected graphs that can be drawn with N vertices and M edges. A simple graph is a graph that does not contain multiple edges and self-loops.

Examples:

Input: N = 3, M = 1
Output:
The 3 graphs are {1-2, 3}, {2-3, 1}, {1-3, 2}.

Input: N = 5, M = 1
Output: 10

Approach: The N vertices are numbered from 1 to N. As there are no self-loops or multiple edges, the edge must be present between two different vertices. So the number of ways we can choose two different vertices is NC2 which is equal to (N * (N – 1)) / 2. Assume it P

Now M edges must be used with these pairs of vertices, so the number of ways to choose M pairs of vertices between P pairs will be PCM

If P < M then the answer will be 0 as the extra edges can not be left alone.

Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach#include using namespace std; // Function to return the value of// Binomial Coefficient C(n, k)int binomialCoeff(int n, int k){     if (k > n)        return 0;     int res = 1;     // Since C(n, k) = C(n, n-k)    if (k > n - k)        k = n - k;     // Calculate the value of    // [n * (n - 1) *---* (n - k + 1)] / [k * (k - 1) * ... * 1]    for (int i = 0; i < k; ++i) {        res *= (n - i);        res /= (i + 1);    }     return res;} // Driver Codeint main(){    int N = 5, M = 1;     int P = (N * (N - 1)) / 2;     cout << binomialCoeff(P, M);     return 0;}

## Java

 // Java implementation of the approachclass GFG {     // Function to return the value of    // Binomial Coefficient C(n, k)    static int binomialCoeff(int n, int k)    {         if (k > n)            return 0;         int res = 1;         // Since C(n, k) = C(n, n-k)        if (k > n - k)            k = n - k;         // Calculate the value of        // [n * (n - 1) *---* (n - k + 1)] /        // [k * (k - 1) * ... * 1]        for (int i = 0; i < k; ++i)        {            res *= (n - i);            res /= (i + 1);        }        return res;    } // Driver Codepublic static void main(String[] args){    int N = 5, M = 1;    int P = (N * (N - 1)) / 2;     System.out.println(binomialCoeff(P, M));}}  // This code is contributed by Shivi_Aggarwal

## Python 3

 # Python 3 implementation of the approach # Function to return the value of# Binomial Coefficient C(n, k)def binomialCoeff(n, k):     if (k > n):        return 0     res = 1     # Since C(n, k) = C(n, n-k)    if (k > n - k):        k = n - k     # Calculate the value of    # [n * (n - 1) *---* (n - k + 1)] /     # [k * (k - 1) * ... * 1]    for i in range( k):        res *= (n - i)        res //= (i + 1)     return res # Driver Codeif __name__=="__main__":         N = 5    M = 1     P = (N * (N - 1)) // 2     print(binomialCoeff(P, M)) # This code is contributed by ita_c

## C#

 // C# implementation of the approachusing System; class GFG{// Function to return the value of// Binomial Coefficient C(n, k)static int binomialCoeff(int n, int k){     if (k > n)        return 0;     int res = 1;     // Since C(n, k) = C(n, n-k)    if (k > n - k)        k = n - k;     // Calculate the value of    // [n * (n - 1) *---* (n - k + 1)] /     // [k * (k - 1) * ... * 1]    for (int i = 0; i < k; ++i)    {        res *= (n - i);        res /= (i + 1);    }     return res;} // Driver Codepublic static void Main(){    int N = 5, M = 1;     int P = (N * (N - 1)) / 2;     Console.Write(binomialCoeff(P, M));}} // This code is contributed// by Akanksha Rai

## PHP

 \$n)         return 0;      \$res = 1;      // Since C(n, k) = C(n, n-k)     if (\$k > \$n - \$k)         \$k = \$n - \$k;      // Calculate the value of     // [n * (n - 1) *---* (n - k + 1)] /     // [k * (k - 1) * ... * 1]     for (\$i = 0; \$i < \$k; ++\$i)     {         \$res *= (\$n - \$i);         \$res /= (\$i + 1);     }      return \$res; }  // Driver Code \$N = 5;\$M = 1;  \$P = floor((\$N * (\$N - 1)) / 2);  echo binomialCoeff(\$P, \$M);  // This code is contributed by Ryuga?>

## Javascript



Output
10

Complexity Analysis:

• Time Complexity: O(M), where M is the number of edges.
• Space Complexity: O(1), since no extra space has been taken.

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