Given two positive integers **N** and **K**. The task is to find the number of arrays of size N that can be formed such that elements of the array should be positive integers and the sum of elements is equal to K.

Examples:

Input : N = 2, K = 3 Output : 2Explanation:[1, 2] and [2, 1] are the only arrays of size 2 whose sum is 3. Input : n = 3, k = 7 Output : 15

**Prerequisite:** Stars and Bars

Suppose there are K identical objects which needs to be placed in N bins (N indices of the array) such that each bin have at least one object. Instead of starting to place objects into bins, we start placing the objects on a line, where the object for the first bin will be taken from the left, followed by the objects for the second bin, and so forth. Thus, the configuration will be determined once one knows what is the first object going to the second bin, and the first object going to the third bin, and so on. We can indicate this by placing N X 1 separating bars at some places between two objects; since no bin is allowed to be empty, there can be at most one bar between a given pair of objects. So, we have K objects in a line with K – 1 gaps. Now we have to choose N – 1 gaps to place bars from K – 1 gaps. This can be chosen by ^{K – 1}C_{N – 1}.

Below is C++ implementation of this approach:

// CPP Program to find the number of arrays of // size N whose elements are positive integers // and sum is K #include <bits/stdc++.h> using namespace std; // Return nCr int binomialCoeff(int n, int k) { int C[k + 1]; memset(C, 0, sizeof(C)); C[0] = 1; // nC0 is 1 for (int i = 1; i <= n; i++) { // Compute next row of pascal triangle using // the previous row for (int j = min(i, k); j > 0; j--) C[j] = C[j] + C[j - 1]; } return C[k]; } // Return the number of array that can be // formed of size n and sum equals to k. int countArray(int N, int K) { return binomialCoeff(K - 1, N - 1); } // Driver Code int main() { int N = 2, K = 3; cout << countArray(N, K) << endl; return 0; }

**Output**

2

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