Sum of products of all combination taken (1 to n) at a time

Namachila October 13, 2025

0 0 12 minutes read

Given N, we have to find the sum of products of all combinations taken 1 to N at a time. In simple words, we have to find the sum of products of all combinations taken 1 at a time, then 2 at a time, then 3 at a time till N at a time.
If you think closely about the problem, a large value of N could result in producing many combinations.

Examples:

Input :  N = 3
Output : f(1) = 6
         f(2) = 11
         f(3) = 6

Explanation: f(x) is sum of products of all 
             combination taken x at a time
             1 + 2 + 3 = 6
             f(2) = (1*2) + (1*3) + (2*3) = 11
             f(3) = (1*2*3) 

Input :  N = 4
Output : f(1) = 10
         f(2) = 35
         f(3) = 50
         f(4) = 24

Explanation: f(1) = 1 + 2 + 3 + 4 = 10
             f(2) = (1*2) + (1*3) + (1*4) + 
                    (2*3) + (2*4) + (3*4) 
                  = 35
             f(3) = (1*2*3) + (1*2*4) +(1*3*4) + 
                    (2*3*4) 
                 = 50
             f(4) = (1*2*3*4) = 24

A Brute force approach would be to produce all the combinations and then find their products and sum.
Recursion would do the trick to produce the combinations taken x at a time.
Example: N = 4 taken 3 at a time

C++

// Program to find SOP of all combination taken
// (1 to N) at a time using brute force
#include <iostream>
using namespace std;
 
// to store sum of every combination
int sum = 0;
 
void Combination(int a[], int combi[], int n, 
                int r, int depth, int index) {
 
  // if we have reached sufficient depth
  if (index == r) {
     
    // find the product of combination
    int product = 1;
    for (int i = 0; i < r; i++)
      product = product * combi[i];
 
    // add the product into sum
    sum += product;
    return;
  }
 
  // recursion to produce different combination
  for (int i = depth; i < n; i++) {
    combi[index] = a[i];
    Combination(a, combi, n, r, i + 1, index + 1);
  }
}
 
// function to print sum of products of
// all combination taken 1-N at a time
void allCombination(int a[], int n) {
  for (int i = 1; i <= n; i++) {
 
    // creating temporary array for storing
    // combination
    int *combi = new int[i];
 
    // call combination with r = i
    // for combination taken i at a time
    Combination(a, combi, n, i, 0, 0);
 
    // displaying sum
    cout << "f(" << i << ") --> " << sum << "\n";
    sum = 0;
 
    // free from heap area
    free(combi);
  }
}
 
// Driver's code
int main() {
  int n = 5;
  int *a = new int[n];
 
  // storing numbers from 1-N in array
  for (int i = 0; i < n; i++)
    a[i] = i + 1;
 
  // calling allCombination
  allCombination(a, n);
 
  return 0;
}

Java

// Program to find SOP of 
// all combination taken
// (1 to N) at a time using
// brute force
import java.io.*;
 
class GFG
{
    // to store sum of
    // every combination
    static int sum = 0;
     
    static void Combination(int []a, int []combi, 
                            int n, int r, 
                            int depth, int index) 
    {
     
    // if we have reached
    // sufficient depth
    if (index == r) 
    {
         
        // find the product
        // of combination
        int product = 1;
        for (int i = 0; i < r; i++)
        product = product * combi[i];
     
        // add the product into sum
        sum += product;
        return;
    }
     
    // recursion to produce 
    // different combination
    for (int i = depth; i < n; i++) 
    {
        combi[index] = a[i];
        Combination(a, combi, n, r, 
                    i + 1, index + 1);
    }
    }
     
    // function to print sum of 
    // products of all combination
    // taken 1-N at a time
    static void allCombination(int []a, 
                               int n)
    {
        for (int i = 1; i <= n; i++) 
        {
         
            // creating temporary array 
            // for storing combination
            int []combi = new int[i];
         
            // call combination with 
            // r = i for combination 
            // taken i at a time
            Combination(a, combi, n,
                        i, 0, 0);
         
            // displaying sum
            System.out.print("f(" + i + ") --> " + 
                                      sum + "\n");
            sum = 0;
        }
    }
     
    // Driver code
    public static void main(String args[]) 
    {
        int n = 5;
        int []a = new int[n];
         
        // storing numbers 
        // from 1-N in array
        for (int i = 0; i < n; i++)
            a[i] = i + 1;
         
        // calling allCombination
        allCombination(a, n);
    }
} 
 
// This code is contributed by 
// Manish Shaw(manishshaw1)

Python3

# Python3 Program to find SOP of all combination 
# taken (1 to N) at a time using brute force 
 
# to store sum of every combination 
def Combination(a, combi, n, r, depth, index):
    global Sum
     
    # if we have reached sufficient depth 
    if index == r: 
     
        # find the product of combination 
        product = 1
        for i in range(r):
            product = product * combi[i] 
     
        # add the product into sum 
        Sum += product 
        return
 
    # recursion to produce different
    # combination 
    for i in range(depth, n):
        combi[index] = a[i] 
        Combination(a, combi, n, r, 
                    i + 1, index + 1)
 
# function to print sum of products of 
# all combination taken 1-N at a time 
def allCombination(a, n):
    global Sum
    for i in range(1, n + 1):
         
        # creating temporary array for 
        # storing combination 
        combi = [0] * i 
     
        # call combination with r = i 
        # for combination taken i at a time 
        Combination(a, combi, n, i, 0, 0) 
     
        # displaying sum 
        print("f(", i, ") --> ", Sum) 
        Sum = 0
 
# Driver Code 
Sum = 0
n = 5
a = [0] * n 
 
# storing numbers from 1-N in array
for i in range(n):
    a[i] = i + 1
 
# calling allCombination 
allCombination(a, n) 
 
# This code is contributed by PranchalK

C#

// Program to find SOP of 
// all combination taken
// (1 to N) at a time using
// brute force
using System;
 
class GFG
{
    // to store sum of
    // every combination
    static int sum = 0;
     
    static void Combination(int []a, int []combi, 
                            int n, int r, 
                            int depth, int index) 
    {
     
    // if we have reached
    // sufficient depth
    if (index == r) 
    {
         
        // find the product
        // of combination
        int product = 1;
        for (int i = 0; i < r; i++)
        product = product * combi[i];
     
        // add the product into sum
        sum += product;
        return;
    }
     
    // recursion to produce 
    // different combination
    for (int i = depth; i < n; i++) 
    {
        combi[index] = a[i];
        Combination(a, combi, n, r, 
                    i + 1, index + 1);
    }
    }
     
    // function to print sum of 
    // products of all combination
    // taken 1-N at a time
    static void allCombination(int []a, 
                               int n)
    {
    for (int i = 1; i <= n; i++) 
    {
     
        // creating temporary array 
        // for storing combination
        int []combi = new int[i];
     
        // call combination with 
        // r = i for combination 
        // taken i at a time
        Combination(a, combi, n,
                    i, 0, 0);
     
        // displaying sum
        Console.Write("f(" + i + ") --> " + 
                               sum + "\n");
        sum = 0;
    }
    }
     
    // Driver code
    static void Main() 
    {
        int n = 5;
        int []a = new int[n];
         
        // storing numbers 
        // from 1-N in array
        for (int i = 0; i < n; i++)
            a[i] = i + 1;
         
        // calling allCombination
        allCombination(a, n);
    }
}
 
// This code is contributed by 
// Manish Shaw(manishshaw1)

Javascript

<script>
 
// JavaScript program to find sum of all combination taken
// (1 to N) at a time using brute force
  
    // to store sum of
    // every combination
    let sum = 0;
      
    function Combination(a, combi, n,  r, depth,  index)
    {
      
    // if we have reached
    // sufficient depth
    if (index == r)
    {
          
        // find the product
        // of combination
        let product = 1;
        for (let i = 0; i < r; i++)
        product = product * combi[i];
      
        // add the product into sum
        sum += product;
        return;
    }
      
    // recursion to produce
    // different combination
    for (let i = depth; i < n; i++)
    {
        combi[index] = a[i];
        Combination(a, combi, n, r,
                    i + 1, index + 1);
    }
    }
      
    // function to print sum of
    // products of all combination
    // taken 1-N at a time
    function allCombination(a, n)
    {
        for (let i = 1; i <= n; i++)
        {
          
            // creating temporary array
            // for storing combination
            let combi = [];
          
            // call combination with
            // r = i for combination
            // taken i at a time
            Combination(a, combi, n,
                        i, 0, 0);
          
            // displaying sum
            document.write("f(" + i + ") --> " +
                                      sum + "<br/>");
            sum = 0;
        }
    }
       
  
// Driver code
 
        let n = 5;
        let a = [];
          
        // storing numbers
        // from 1-N in array
        for (let i = 0; i < n; i++)
            a[i] = i + 1;
          
        // calling allCombination
        allCombination(a, n);
 
</script>

Output

f(1) --> 15
f(2) --> 85
f(3) --> 225
f(4) --> 274
f(5) --> 120

Time complexity of the above code is exponential when the value of N is large.

An Efficient Method is to use the concept of dynamic programming. We don’t have to find the sum of products every time. We can make use of previous results.
Let’s take an example: N = 4

C++

// CPP Program to find sum of all combination taken
// (1 to N) at a time using dynamic programming
#include <iostream>
using namespace std;
 
// find the postfix sum array
void postfix(int a[], int n) {
  for (int i = n - 1; i > 0; i--)
    a[i - 1] = a[i - 1] + a[i];
}
 
// modify the array such that we don't have to
// compute the products which are obtained before
void modify(int a[], int n) {
  for (int i = 1; i < n; i++)
    a[i - 1] = i * a[i];
}
 
// finding sum of all combination taken 1 to N at a time
void allCombination(int a[], int n) {
 
  int sum = 0;
 
  // sum taken 1 at time is simply sum of 1 - N
  for (int i = 1; i <= n; i++)
    sum += i;
  cout << "f(1) --> " << sum << "\n";
 
  // for sum of products for all combination
  for (int i = 1; i < n; i++) {
 
    // finding postfix array
    postfix(a, n - i + 1);
     
    // sum of products taken i+1 at a time
    sum = 0;
    for (int j = 1; j <= n - i; j++) {
      sum += (j * a[j]);
    }
    cout << "f(" << i + 1 << ") --> " << sum << "\n";
 
    // modify the array for overlapping problem
    modify(a, n);
  }
}
 
// Driver's Code
int main() {
  int n = 5;
  int *a = new int[n];
 
  // storing numbers from 1 to N
  for (int i = 0; i < n; i++)
    a[i] = i + 1;
 
  // calling allCombination
  allCombination(a, n);
 
  return 0;
}

Java

// Java Program to find sum of all combination taken
// (1 to N) at a time using dynamic programming
import java.util.*;
 
class GFG 
{
 
    // find the postfix sum array
    static void postfix(int a[], int n) 
    {
        for (int i = n - 1; i > 0; i--) 
        {
            a[i - 1] = a[i - 1] + a[i];
        }
    }
 
    // modify the array such that we don't 
    // have to compute the products which
    // are obtained before
    static void modify(int a[], int n)
    {
        for (int i = 1; i < n; i++) 
        {
            a[i - 1] = i * a[i];
        }
    }
 
    // finding sum of all combination 
    // taken 1 to N at a time
    static void allCombination(int a[], int n) 
    {
        int sum = 0;
 
        // sum taken 1 at time is simply sum of 1 - N
        for (int i = 1; i <= n; i++) 
        {
            sum += i;
        }
        System.out.println("f(1) --> " + sum);
 
        // for sum of products for all combination
        for (int i = 1; i < n; i++) 
        {
 
            // finding postfix array
            postfix(a, n - i + 1);
 
            // sum of products taken i+1 at a time
            sum = 0;
            for (int j = 1; j <= n - i; j++) 
            {
                sum += (j * a[j]);
            }
            System.out.println("f(" + (i + 1) + 
                               ") --> " + sum);
 
            // modify the array for overlapping problem
            modify(a, n);
        }
    }
 
    // Driver's Code
    public static void main(String[] args) 
    {
        int n = 5;
        int[] a = new int[n];
 
        // storing numbers from 1 to N
        for (int i = 0; i < n; i++) 
        {
            a[i] = i + 1;
        }
 
        // calling allCombination
        allCombination(a, n);
    }
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 Program to find 
# sum of all combination taken
# (1 to N) at a time using 
# dynamic programming
 
# Find the postfix sum array
def postfix(a, n):
   
    for i in range (n - 1, 1, -1):
        a[i - 1] = a[i - 1] + a[i]
 
# Modify the array such 
# that we don't have to
# compute the products 
# which are obtained before
def modify(a, n):
   
    for i in range (1, n):
        a[i - 1] = i * a[i];
 
# Finding sum of all combination 
# taken 1 to N at a time
def allCombination(a, n):
 
    sum = 0
 
    # sum taken 1 at time is 
    # simply sum of 1 - N
    for i in range (1, n + 1):
       
        sum += i
    print ("f(1) --> ", sum )
 
    # for sum of products for 
    # all combination
    for i in range (1, n):
 
        # finding postfix array
        postfix(a, n - i + 1)
     
        # sum of products taken 
        # i+1 at a time
        sum = 0
         
        for j in range(1, n - i + 1):
            sum += (j * a[j])
     
        print ("f(", i + 1, ") --> ", sum)
 
        # modify the array for 
        # overlapping problem
        modify(a, n)
 
# Driver's Code
if __name__ == "__main__":
 
    n = 5
    a = [0] * n
 
    # storing numbers 
    # from 1 to N
    for i in range(n):
        a[i] = i + 1
 
    # calling allCombination
    allCombination(a, n)
 
# This code is contributed by Chitranayal

C#

// C# Program to find sum of all combination taken
// (1 to N) at a time using dynamic programming
using System;
     
class GFG 
{
 
    // find the postfix sum array
    static void postfix(int []a, int n) 
    {
        for (int i = n - 1; i > 0; i--) 
        {
            a[i - 1] = a[i - 1] + a[i];
        }
    }
 
    // modify the array such that we don't 
    // have to compute the products which
    // are obtained before
    static void modify(int []a, int n)
    {
        for (int i = 1; i < n; i++) 
        {
            a[i - 1] = i * a[i];
        }
    }
 
    // finding sum of all combination 
    // taken 1 to N at a time
    static void allCombination(int []a, int n) 
    {
        int sum = 0;
 
        // sum taken 1 at time is simply sum of 1 - N
        for (int i = 1; i <= n; i++) 
        {
            sum += i;
        }
        Console.WriteLine("f(1) --> " + sum);
 
        // for sum of products for all combination
        for (int i = 1; i < n; i++) 
        {
 
            // finding postfix array
            postfix(a, n - i + 1);
 
            // sum of products taken i+1 at a time
            sum = 0;
            for (int j = 1; j <= n - i; j++) 
            {
                sum += (j * a[j]);
            }
            Console.WriteLine("f(" + (i + 1) + 
                            ") --> " + sum);
 
            // modify the array for overlapping problem
            modify(a, n);
        }
    }
 
    // Driver's Code
    public static void Main(String[] args) 
    {
        int n = 5;
        int[] a = new int[n];
 
        // storing numbers from 1 to N
        for (int i = 0; i < n; i++) 
        {
            a[i] = i + 1;
        }
 
        // calling allCombination
        allCombination(a, n);
    }
}
 
// This code is contributed by Rajput-Ji

Javascript

<script>
// Javascript Program to find sum of all combination taken
// (1 to N) at a time using dynamic programming
     
    // find the postfix sum array
    function postfix(a,n)
    {
        for (let i = n - 1; i > 0; i--)
        {
            a[i - 1] = a[i - 1] + a[i];
        }
    }
     
    // modify the array such that we don't
    // have to compute the products which
    // are obtained before
    function modify(a,n)
    {
        for (let i = 1; i < n; i++)
        {
            a[i - 1] = i * a[i];
        }
    }
     
    // finding sum of all combination
    // taken 1 to N at a time
    function allCombination(a,n)
    {
        let sum = 0;
  
        // sum taken 1 at time is simply sum of 1 - N
        for (let i = 1; i <= n; i++)
        {
            sum += i;
        }
        document.write("f(1) --> " + sum+"<br>");
  
        // for sum of products for all combination
        for (let i = 1; i < n; i++)
        {
  
            // finding postfix array
            postfix(a, n - i + 1);
  
            // sum of products taken i+1 at a time
            sum = 0;
            for (let j = 1; j <= n - i; j++)
            {
                sum += (j * a[j]);
            }
            document.write("f(" + (i + 1) +
                               ") --> " + sum+"<br>");
  
            // modify the array for overlapping problem
            modify(a, n);
        }
    }
     
    // Driver's Code
    let n = 5;
    let a = new Array(n);
    // storing numbers from 1 to N
    for (let i = 0; i < n; i++)
    {
        a[i] = i + 1;
    }
  
    // calling allCombination
    allCombination(a, n);
     
    // This code is contributed by avanitrachhadiya2155
</script>

Output

f(1) --> 15
f(2) --> 85
f(3) --> 225
f(4) --> 274
f(5) --> 120

Time Complexity: O(n^2)
Auxiliary Space: O(1)

You can also find the execution time of both the method for a large value of N and can see the difference for yourself.

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