# Number of integral solutions of the equation x1 + x2 +…. + xN = k

Given N and K. The task is to count the number of the integral solutions of a linear equation having N variable as given below:

x1 + x2+ x3…+ xN-1+…+xN = K

Examples

`Input: N = 3, K = 3Output: 10Explanation: Possible solutions are: (1,1,1),(1,0,2),(2,0,1),(1,2,0),(2,1,0),(0,1,2)(0,2,1),(3,0,0),(0,3,0),(0,0,3).Input: N = 2, K = 2Output: 3`

Approach: This problem can be solved using the concept of Permutation and Combination. Below are the direct formulas for finding non-negative and positive integral solutions respectively.

Number of non-negative integral solutions of equation x1 + x2 + …… + xn = k is given by (n+k-1)! / (n-1)!*k!.
Number of positive integral solutions of equation x1 + x2 + ….. + xn = k is given by (k-1)! / (n-1)! * (k-n)!.

Note  here that the non-negative integral solutions already include the positive integral solutions. Therefore, there is no need to add the number of positive integral solutions to the answer.
Below is the implementation of above approach:

## c++

 `// C++ program for above implementation ``#include``#define MAX 100``using` `namespace` `std ;` `int` `nCr(``int` `n, ``int` `r)``{``    ``int` `fac[MAX] = {1} ;``    ` `    ``for` `(``int` `i = 1 ; i < n + 1 ; i++)``    ``{``        ``fac[i] = fac[i - 1] * i ;``    ``}``    ` `    ``int` `ans = fac[n] / (fac[n - r] * ``                        ``fac[r]) ;``    ``return` `ans ;``}` `// Driver Code``int` `main()``{``    ``int` `n = 3 ;``    ``int` `k = 3 ;``    ` `    ``int` `ans = nCr(n + k - 1 , k); ``    ``cout << ans ;``    ` `    ``return` `0 ; ``}` `// This code is contributed``// by ANKITRAI1`

## Java

 `// Java program for above implementation``import` `java.io.*;` `class` `GFG ``{``static` `int` `nCr(``int` `n, ``int` `r)``{``    ``int` `fac[] = ``new` `int``[``100``] ;``    ``for``(``int` `i = ``0``; i < n; i++)``    ``fac[i] = ``1``;``    ` `    ``for` `(``int` `i = ``1` `; i < n + ``1` `; i++)``    ``{``        ``fac[i] = fac[i - ``1``] * i ;``    ``}``    ` `    ``int` `ans = fac[n] / (fac[n - r] * ``                        ``fac[r]);``    ``return` `ans ;``}` `// Driver Code``public` `static` `void` `main (String[] args) ``{``    ``int` `n = ``3` `;``    ``int` `k = ``3` `;``    ` `    ``int` `ans = nCr(n + k - ``1` `, k);``    ` `    ``System.out.println(ans) ;``}``}` `// This code is contributed``// by anuj_67`

## Python3

 `# Python implementation of ``# above approach` `# Calculate nCr i.e binomial ``# coefficient nCr = n !/(r !*(n-r)!)``def` `nCr(n, r):` `    ``# first find factorial ``    ``# upto n``    ``fac ``=` `list``()``    ``fac.append(``1``)``    ``for` `i ``in` `range``(``1``, n ``+` `1``):``        ``fac.append(fac[i ``-` `1``] ``*` `i)` `    ``# use nCr formula``    ``ans ``=` `fac[n] ``/` `(fac[n ``-` `r] ``*` `fac[r])``    ``return` `ans` `# n = number of variables``n ``=` `3` `# sum of n variables = k``k ``=` `3` `# find number of solutions``ans ``=` `nCr(n ``+` `k ``-` `1``, k)` `print``(ans)` `# This code is contributed``# by ChitraNayal`

## C#

 `// C# program for above implementation``using` `System; ` `class` `GFG ``{``static` `int` `nCr(``int` `n, ``int` `r)``{``    ``int``[] fac = ``new` `int``[100] ;``    ``for``(``int` `i = 0; i < n; i++)``    ``fac[i] = 1;``    ` `    ``for` `(``int` `i = 1 ; i < n + 1 ; i++)``    ``{``        ``fac[i] = fac[i - 1] * i ;``    ``}``    ` `    ``int` `ans = fac[n] / (fac[n - r] * ``                        ``fac[r]);``    ``return` `ans ;``}` `// Driver Code``public` `static` `void` `Main () ``{``    ``int` `n = 3 ;``    ``int` `k = 3 ;``    ` `    ``int` `ans = nCr(n + k - 1 , k);``    ` `    ``Console.Write(ans) ;``}``}` `// This code is contributed ``// by ChitraNayal`

## Javascript

 ``

## PHP

 ``

Output
```10
```

Time Complexity: O(n)
Auxiliary Space: O(MAX)

Applications of the above concepts:

1. Number of non-negative integral solutions of equation x1 + x2 +…+ xn = k is equal to the number of ways in which k identical balls can be distributed into N unique boxes.
2. Number of positive integral solutions of equation x1 + x2 + … + xn = k is equal to the number of ways in which k identical balls can be distributed into N unique boxes such that each box must contain at-least 1 ball.

Previous
Next