# 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: 3
The 3 graphs are {1-2, 3}, {2-3, 1}, {1-3, 2}.

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

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

Approach: The N vertices are numbered from 1 to N. As there is 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 are NC2 which is equal to (N * (N – 1)) / 2. Assume it P.
Now M edges must be used with these pair 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 Code ` `int` `main() ` `{ ` `    ``int` `N = 5, M = 1; ` ` `  `    ``int` `P = (N * (N - 1)) / 2; ` ` `  `    ``cout << binomialCoeff(P, M); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `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 Code ` `public` `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 Code ` `if` `__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 approach ` `using` `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 Code ` `public` `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 ` `?> `

Output:

```10
```

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