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# Number of non-negative integral solutions of a + b + c = n

• Difficulty Level : Medium
• Last Updated : 24 Jun, 2022

Given a number n, find the number of ways in which  we can add 3 non-negative integers so that their sum is n.
Examples :

```Input : n = 1
Output : 3
There are three ways to get sum 1.
(1, 0, 0), (0, 1, 0) and (0, 0, 1)

Input : n = 2
Output : 6
There are six ways to get sum 2.
(2, 0, 0), (0, 2, 0), (0, 0, 2), (1, 1, 0)
(1, 0, 1) and (0, 1, 1)

Input : n = 3
Output : 10
There are ten ways to get sum 3.
(3, 0, 0), (0, 3, 0), (0, 0, 3), (1, 2, 0)
(1, 0, 2), (0, 1, 2), (2, 1, 0), (2, 0, 1),
(0, 2, 1) and (1, 1, 1)```

Recommended Practice

Method 1 [ Brute Force : O(n3) ]
A simple solution is to consider all triplets using three loops. For every triplet, check if its sum is equal to n. If the sum is n, increment the count of solutions.
Below is the implementation.

## C++

 `// A naive C++ solution to count solutions of` `// a + b + c = n` `#include` `using` `namespace` `std;`   `// Returns count of solutions of a + b + c = n` `int` `countIntegralSolutions(``int` `n)` `{` `    ``// Initialize result` `    ``int` `result = 0;`   `    ``// Consider all triplets and increment` `    ``// result whenever sum of a triplet is n.` `    ``for` `(``int` `i=0; i<=n; i++)` `      ``for` `(``int` `j=0; j<=n-i; j++)` `          ``for` `(``int` `k=0; k<=(n-i-j); k++)` `             ``if` `(i + j + k == n)` `              ``result++;`   `    ``return` `result;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `n = 3;` `    ``cout <<  countIntegralSolutions(n);` `    ``return` `0;` `}`

## Java

 `// A naive Java solution to count` `// solutions of a + b + c = n` `import` `java.io.*;`   `class` `GFG ` `{` `    ``// Returns count of solutions of a + b + c = n` `    ``static` `int` `countIntegralSolutions(``int` `n)` `    ``{` `        ``// Initialize result` `        ``int` `result = ``0``;` `    `  `        ``// Consider all triplets and increment` `        ``// result whenever sum of a triplet is n.` `        ``for` `(``int` `i = ``0``; i <= n; i++)` `        ``for` `(``int` `j = ``0``; j <= n - i; j++)` `            ``for` `(``int` `k = ``0``; k <= (n - i - j); k++)` `                ``if` `(i + j + k == n)` `                ``result++;` `    `  `        ``return` `result;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main (String[] args) ` `    ``{` `        ``int` `n = ``3``;` `        ``System.out.println( countIntegralSolutions(n));` `    `  `    ``}` `}`   `// This article is contributed by vt_m`

## Python3

 `# Python3 code to count ` `# solutions of a + b + c = n`   `# Returns count of solutions` `# of a + b + c = n` `def` `countIntegralSolutions (n):`   `    ``# Initialize result` `    ``result ``=` `0` `    `  `    ``# Consider all triplets and ` `    ``# increment result whenever ` `    ``# sum of a triplet is n.` `    ``for` `i ``in` `range``(n ``+` `1``):` `        ``for` `j ``in` `range``(n ``+` `1``):` `            ``for` `k ``in` `range``(n ``+` `1``):` `                ``if` `i ``+` `j ``+` `k ``=``=` `n:` `                    ``result ``+``=` `1` `    `  `    ``return` `result` `    `  `# Driver code` `n ``=` `3` `print``(countIntegralSolutions(n))`   `# This code is contributed by "Sharad_Bhardwaj".`

## C#

 `// A naive C# solution to count` `// solutions of a + b + c = n` `using` `System;`   `class` `GFG {` `    `  `    ``// Returns count of solutions` `    ``// of a + b + c = n` `    ``static` `int` `countIntegralSolutions(``int` `n)` `    ``{` `        `  `        ``// Initialize result` `        ``int` `result = 0;` `    `  `        ``// Consider all triplets and increment` `        ``// result whenever sum of a triplet is n.` `        ``for` `(``int` `i = 0; i <= n; i++)` `            ``for` `(``int` `j = 0; j <= n - i; j++)` `                ``for` `(``int` `k = 0; k <= (n - i - j); k++)` `                    ``if` `(i + j + k == n)` `                    ``result++;` `    `  `        ``return` `result;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main () ` `    ``{` `        ``int` `n = 3;` `        ``Console.Write(countIntegralSolutions(n));` `    `  `    ``}` `}`   `// This article is contributed by Smitha.`

## PHP

 ``

## Javascript

 ``

Output:

`10`

Time complexity: O(n3)

Auxiliary Space: O(1)

Method 2 [ Direct Formula : O(1) ]
If we take a closer look at the pattern, we can find that the count of solutions is ((n+1) * (n+2)) / 2. The problem is equivalent to distributing n identical balls  in three boxes and the solution is n+2C2. In general, if there are m variables (or boxes) and n balls , the formula becomes n+m-1Cm-1

## C++

 `// A naive C++ solution to count solutions of` `// a + b + c = n` `#include` `using` `namespace` `std;`   `// Returns count of solutions of a + b + c = n` `int` `countIntegralSolutions(``int` `n)` `{` `    ``return` `((n+1)*(n+2))/2;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `n = 3;` `    ``cout <<  countIntegralSolutions(n);` `    ``return` `0;` `}`

## Java

 `// Java solution to count ` `// solutions of a + b + c = n` `import` `java.io.*;`   `class` `GFG ` `{` `    ``// Returns count of solutions ` `    ``// of a + b + c = n` `    ``static` `int` `countIntegralSolutions(``int` `n)` `    ``{` `    ``return` `((n + ``1``) * (n + ``2``)) / ``2``;` `        `  `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `main (String[] args) ` `    ``{` `        ``int` `n = ``3``;` `        ``System.out.println ( countIntegralSolutions(n));` `        `  `    ``}` `}` `// This article is contributed by vt_m`

## Python3

 `# Python3 solution to count ` `# solutions of a + b + c = n`   `# Returns count of solutions` `# of a + b + c = n` `def` `countIntegralSolutions (n):` `    ``return` `int``(((n ``+` `1``) ``*` `(n ``+` `2``)) ``/` `2``)` `    `  `# Driver code` `n ``=` `3` `print``(countIntegralSolutions(n))`   `# This code is contributed by "Sharad_Bhardwaj".`

## C#

 `// C# solution to count ` `// solutions of a + b + c = n` `using` `System;`   `class` `GFG {` `    `  `    ``// Returns count of solutions ` `    ``// of a + b + c = n` `    ``static` `int` `countIntegralSolutions(``int` `n)` `    ``{` `        ``return` `((n + 1) * (n + 2)) / 2;` `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `Main (String[] args) ` `    ``{` `        ``int` `n = 3;` `        ``Console.Write ( countIntegralSolutions(n));` `        `  `    ``}` `}`   `// This code is contributed by parashar.`

## PHP

 ``

## Javascript

 ``

Output :

`10`

Time complexity: O(1)

Auxiliary Space: O(1)
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