Total number of Subsets of size at most K

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Given a number N which is the size of the set and a number K, the task is to find the count of subsets, of the set of N elements, having at most K elements in it, i.e. the size of subset is less than or equal to K.
Examples:

Input: N = 3, K = 2
Output: 6
Subsets with 1 element in it = {1}, {2}, {3}
Subsets with 2 elements in it = {1, 2}, {1, 3}, {1, 2}
Since K = 2, therefore only the above subsets will be considered for length atmost K. Therefore the count is 6.
Input: N = 5, K = 2
Output: 15

Approach:

Since the number of subsets of exactly K elements that can be made from N items is (^NC_K). Therefore for “at most”, the required count will be
$\LARGE \sum _{i = 1}^{K} \text{ } ^{N}\textrm{C}_{i} =\text{ } ^{N}\textrm{C}_{1} + ^{N}\textrm{C}_{2} + ^{N}\textrm{C}_{3} + ...+ ^{N}\textrm{C}_{K}$
Inorder to calculate the value of ^NC_K, Binomial Coefficient is used. Please refer this article to see how it works.
So to get the required subsets for length atmost K, run a loop from 1 to K and add the ^NC_i for each value of i.

Below is the implementation of the above approach:

C++

// C++ code to find total number of
// Subsets of size at most K
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to compute the value
// of Binomial Coefficient C(n, k)
int binomialCoeff(int n, int k)
{
    int C[n + 1][k + 1];
    int i, j;
 
    // Calculate value of Binomial Coefficient
    // in bottom up manner
    for (i = 0; i <= n; i++) {
        for (j = 0; j <= min(i, k); j++) {
 
            // Base Cases
            if (j == 0 || j == i)
                C[i][j] = 1;
 
            // Calculate value using previously
            // stored values
            else
                C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
        }
    }
 
    return C[n][k];
}
 
// Function to calculate sum of
// nCj from j = 1 to k
int count(int n, int k)
{
    int sum = 0;
    for (int j = 1; j <= k; j++) {
 
        // Calling the nCr function
        // for each value of j
        sum = sum + binomialCoeff(n, j);
    }
 
    return sum;
}
 
// Driver code
int main()
{
    int n = 3, k = 2;
    cout << count(n, k) << endl;
 
    n = 5, k = 2;
    cout << count(n, k) << endl;
    return 0;
}

Java

// Java code to find total number of
// Subsets of size at most K
import java.lang.*;
class GFG
{
 
// Function to compute the value
// of Binomial Coefficient C(n, k)
public static int binomialCoeff(int n, int k)
{
    int[][] C = new int[n + 1][k + 1];
    int i, j;
 
    // Calculate value of Binomial Coefficient
    // in bottom up manner
    for (i = 0; i <= n; i++) 
    {
        for (j = 0; j <= Math.min(i, k); j++)
        {
 
            // Base Cases
            if (j == 0 || j == i)
                C[i][j] = 1;
 
            // Calculate value using previously
            // stored values
            else
                C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
        }
    }
 
    return C[n][k];
}
 
// Function to calculate sum of
// nCj from j = 1 to k
public static int count(int n, int k)
{
    int sum = 0;
    for (int j = 1; j <= k; j++)
    {
 
        // Calling the nCr function
        // for each value of j
        sum = sum + binomialCoeff(n, j);
    }
 
    return sum;
}
 
// Driver code
public static void main(String args[])
{
    GFG g = new GFG();
    int n = 3, k = 2;
    System.out.print(count(n, k));
 
    int n1 = 5, k1 = 2;
    System.out.print(count(n1, k1));
}
}
 
// This code is contributed by SoumikMondal

Python3

# Python code to find total number of
# Subsets of size at most K
 
# Function to compute the value
# of Binomial Coefficient C(n, k)
def binomialCoeff(n, k):
    C = [[0 for i in range(k + 1)] for j in range(n + 1)];
    i, j = 0, 0;
 
    # Calculate value of Binomial Coefficient
    # in bottom up manner
    for i in range(n + 1):
        for j in range( min(i, k) + 1):
 
            # Base Cases
            if (j == 0 or j == i):
                C[i][j] = 1;
 
            # Calculate value using previously
            # stored values
            else:
                C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
    return C[n][k];
 
# Function to calculate sum of
# nCj from j = 1 to k
def count(n, k):
    sum = 0;
    for j in range(1, k+1):
 
        # Calling the nCr function
        # for each value of j
        sum = sum + binomialCoeff(n, j);
    return sum;
 
# Driver code
if __name__ == '__main__':
    n = 3;
    k = 2;
    print(count(n, k), end="");
 
    n1 = 5;
    k1 = 2;
    print(count(n1, k1));
 
# This code is contributed by 29AjayKumar

C#

// C# code to find total number of
// Subsets of size at most K
using System;
 
class GFG
{
 
    // Function to compute the value
    // of Binomial Coefficient C(n, k)
    public static int binomialCoeff(int n, int k)
    {
        int[,] C = new int[n + 1, k + 1];
        int i, j;
     
        // Calculate value of Binomial Coefficient
        // in bottom up manner
        for (i = 0; i <= n; i++) 
        {
            for (j = 0; j <= Math.Min(i, k); j++)
            {
     
                // Base Cases
                if (j == 0 || j == i)
                    C[i, j] = 1;
     
                // Calculate value using previously
                // stored values
                else
                    C[i, j] = C[i - 1, j - 1] + C[i - 1, j];
            }
        }
     
        return C[n, k];
    }
     
    // Function to calculate sum of
    // nCj from j = 1 to k
    public static int count(int n, int k)
    {
        int sum = 0;
        for (int j = 1; j <= k; j++)
        {
     
            // Calling the nCr function
            // for each value of j
            sum = sum + binomialCoeff(n, j);
        }
     
        return sum;
    }
     
    // Driver code
    public static void Main()
    {
 
        int n = 3, k = 2;
        Console.Write(count(n, k));
     
        int n1 = 5, k1 = 2;
        Console.Write(count(n1, k1));
    }
}
 
// This code is contributed by AnkitRai01

Javascript

<script>
// Javascript implementation of the
// above approach
 
// Function for the binomial coefficient
function binomialCoeff(n, k)
{
 
    var C = new Array(n + 1);
    // Loop to create 2D array using 1D array
    for (var i = 0; i < C.length; i++) {
        C[i] = new Array(k + 1);
    }
    var i, j;
   
    // Calculate value of Binomial Coefficient
    // in bottom up manner
    for (i = 0; i <= n; i++) {
        for (j = 0; j <= Math.min(i, k); j++) {
            // Base Cases
            if (j == 0 || j == i)
                C[i][j] = 1;
   
            // Calculate value using previously
            // stored values
            else
                C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
        }
    }
   
    return C[n][k];
}
   
// Function to calculate sum of
// nCj from j = 1 to k
function count(n, k)
{
    var sum = 0;
    for (var j = 1; j <= k; j++) {
   
        // Calling the nCr function
        // for each value of j
        sum = sum + binomialCoeff(n, j);
    }
   
    return sum;
}
 
// Driver code
var n = 3;
var k = 2;
document.write(count(n, k));
 
var n = 5;
var k = 2;
document.write(count(n, k));
 
// This code is contributed by ShubhamSingh10
</script>

Output

6
15

Time Complexity: O(n² * k)

Auxiliary Space: O(n + k)

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