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

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  • Difficulty Level : Easy
  • Last Updated : 31 Aug, 2022
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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 = 3
Output: 10
Explaination: 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 = 2
Output: 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<iostream>
 
using namespace std ;
 
int nCr(int n, int r)
{
    int fac[100] = {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
# cofficent 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

PHP




<?php
// PHP implementation of above approach
 
// Calculate nCr i.e binomial
// cofficent nCr = n !/(r !*(n-r)!)
function nCr($n, $r)
{
    // first find factorial
    // upto n
    $fac = array();
    array_push($fac, 1);
    for($i = 1; $i < $n + 1; $i++)
        array_push($fac, $fac[$i - 1] * $i);
 
    // use nCr formula
    $ans = $fac[$n] / ($fac[$n - $r] *
                       $fac[$r]);
    return $ans;
}
 
// Driver Code
 
// 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 mits
?>

Javascript




<script>
 
// Javascript program for above implementation
function nCr(n, r)
{
    var fac = Array(100).fill(1);
     
    for (var i = 1 ; i < n + 1 ; i++)
    {
        fac[i] = fac[i - 1] * i ;
    }
     
    var ans = fac[n] / (fac[n - r] *
                        fac[r]) ;
    return ans ;
}
 
// Driver Code
var n = 3 ;
var k = 3 ;
 
var ans = nCr(n + k - 1 , k);
 
document.write(ans );
 
// This code is contributed by noob2000.
</script>

Output

10

Time Complexity: O(1)

Auxiliary Space: O(1)

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.

 


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