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 Codeint 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 implementationimport 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 Codepublic 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 variablesn = 3 # sum of n variables = kk = 3 # find number of solutionsans = nCr(n + k - 1, k) + nCr(k - 1, n - 1) print(ans) # This code is contributed# by ChitraNayal |
C#
// C# program for above implementationusing 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 Codepublic 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?> |
11.0
Applications of the above concepts:
- 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.
- 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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