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Number of integral solutions of the equation x1 + x2 +…. + xN = k
  • Difficulty Level : Easy
  • Last Updated : 20 Apr, 2021

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

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)!. 
 

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) +
              nCr(k - 1, n - 1);
     
    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) +
              nCr(k - 1, n - 1);
     
    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) + nCr(k - 1, n - 1)
 
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) +
              nCr(k - 1, n - 1);
     
    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, $k) +
       nCr($k - 1, $n - 1);
 
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) +
        nCr(k - 1, n - 1);
 
document.write(ans );
 
// This code is contributed by noob2000.
</script>
Output: 
11.0

 

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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