Sum of all products of the Binomial Coefficients of two numbers up to K

Given three integers N, M and K, the task is to calculate the sum of products of Binomial Coefficients C(N, i) and C(M, K – i), where i ranges between [0, K].

\begin{*align} \sum_{i=0}^{k}C(n, i)*C(m, k-i) \label{sum} \end{*align}

Examples:

Input: N = 2, M = 2, K = 2 
Output:
Explanation: 
C(2, 0) * C(2, 2) + C(2, 1) * C(2, 1) + C(2, 2) * C(2, 0) = 1*1 + 2*2 +1*1 = 6

Input: N = 2, M = 3, K = 1 
Output:
Explanation: 
C(2, 0) * C(3, 1) + C(2, 1) * C(3, 0) = 1*3 + 2*1 = 5



Naive Approach:The simplest approach to solve this problem is to simply iterate over the range [0, K] and calculate C(N, i) and C(M, K – 1) for every i and update sum by adding their product. 

Below is the implementation of the above approach:

C++

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// C++ implementation of
// the above approach
  
#include <bits/stdc++.h>
using namespace std;
  
// Function returns nCr
// i.e. Binomial Coefficient
int nCr(int n, int r)
{
  
    // Initialize res with 1
    int res = 1;
  
    // Since C(n, r) = C(n, n-r)
    if (r > n - r)
        r = n - r;
  
    // Evaluating expression
    for (int i = 0; i < r; ++i) {
  
        res *= (n - i);
        res /= (i + 1);
    }
  
    return res;
}
  
// Function to calculate and
// return the sum of the products
int solve(int n, int m, int k)
{
  
    // Initialize sum to 0
    int sum = 0;
  
    // Traverse from 0 to k
    for (int i = 0; i <= k; i++)
        sum += nCr(n, i)
               * nCr(m, k - i);
  
    return sum;
}
  
// Driver Code
int main()
{
    int n = 3, m = 2, k = 2;
  
    cout << solve(n, m, k);
    return 0;
}

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Java

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// Java implementation of
// the above approach
import java.util.*;
class GFG{
  
// Function returns nCr
// i.e. Binomial Coefficient
static int nCr(int n, int r)
{
  
    // Initialize res with 1
    int res = 1;
  
    // Since C(n, r) = C(n, n-r)
    if (r > n - r)
        r = n - r;
  
    // Evaluating expression
    for (int i = 0; i < r; ++i) 
    {
        res *= (n - i);
        res /= (i + 1);
    }
  
    return res;
}
  
// Function to calculate and
// return the sum of the products
static int solve(int n, int m, int k)
{
  
    // Initialize sum to 0
    int sum = 0;
  
    // Traverse from 0 to k
    for (int i = 0; i <= k; i++)
        sum += nCr(n, i)
               * nCr(m, k - i);
  
    return sum;
}
  
// Driver Code
public static void main(String[] args)
{
    int n = 3, m = 2, k = 2;
  
    System.out.print(solve(n, m, k));
}
}
  
// This code is contributed by Rohit_ranjan

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Python3

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# Python3 implementation of
# the above approach
  
# Function returns nCr
# i.e. Binomial Coefficient
def nCr(n, r):
      
    # Initialize res with 1
    res = 1
      
    # Since C(n, r) = C(n, n-r)
    if r > n - r:
        r = n - r
      
    # Evaluating expression
    for i in range(r):
        res *= (n - i)
        res /= (i + 1)
      
    return res;
      
# Function to calculate and
# return the sum of the products
def solve(n, m, k):
      
    # Initialize sum to 0
    sum = 0;
      
    # Traverse from 0 to k
    for i in range(k + 1):
        sum += nCr(n, i) * nCr(m, k - i)
      
    return int(sum)
      
# Driver code 
if __name__ == '__main__'
      
    n = 3
    m = 2
    k = 2;
      
    print(solve(n, m, k))
  
# This code is contributed by jana_sayantan    

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C#

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// C# implementation of
// the above approach
using System;
class GFG{
  
// Function returns nCr
// i.e. Binomial Coefficient
static int nCr(int n, int r)
{
  
    // Initialize res with 1
    int res = 1;
  
    // Since C(n, r) = C(n, n-r)
    if (r > n - r)
        r = n - r;
  
    // Evaluating expression
    for (int i = 0; i < r; ++i) 
    {
        res *= (n - i);
        res /= (i + 1);
    }
  
    return res;
}
  
// Function to calculate and
// return the sum of the products
static int solve(int n, int m, int k)
{
  
    // Initialize sum to 0
    int sum = 0;
  
    // Traverse from 0 to k
    for (int i = 0; i <= k; i++)
        sum += nCr(n, i)
            * nCr(m, k - i);
  
    return sum;
}
  
// Driver Code
public static void Main(String[] args)
{
    int n = 3, m = 2, k = 2;
  
    Console.Write(solve(n, m, k));
}
}
  
// This code is contributed by Rajput-Ji 

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Output: 

10

Time complexity: O(K2) 
Auxiliary Space: O(1)

Efficient Approach: 
The above approach can be optimized using Vandermonde’s Identity.

According to Vandermonde’s Identity, any combination of K items from a total of (N + M) items should have r items from M and (K – r) items from N items.

Therefore, the given expression is reduced to the following:

\begin{*align} \sum_{i=0}^{k}C(n, i)*C(m, k-i) = C(n+m, k) \label{sum} \end{*align}

Below is the implementation of the above approach:

C++

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code

// C++ implementation of
// the above approach
  
#include <bits/stdc++.h>
using namespace std;
  
// Function returns nCr
// i.e. Binomial Coefficient
int nCr(int n, int r)
{
  
    // Initialize res with 1
    int res = 1;
  
    // Since C(n, r) = C(n, n-r)
    if (r > n - r)
        r = n - r;
  
    // Evaluating expression
    for (int i = 0; i < r; ++i) {
  
        res *= (n - i);
        res /= (i + 1);
    }
  
    return res;
}
  
// Driver Code
int main()
{
    int n = 3, m = 2, k = 2;
  
    cout << nCr(n + m, k);
    return 0;
}

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Java

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// Java implementation of
// the above approach
import java.util.*;
class GFG{
  
// Function returns nCr
// i.e. Binomial Coefficient
static int nCr(int n, int r)
{
  
    // Initialize res with 1
    int res = 1;
  
    // Since C(n, r) = C(n, n-r)
    if (r > n - r)
        r = n - r;
  
    // Evaluating expression
    for (int i = 0; i < r; ++i) 
    {
        res *= (n - i);
        res /= (i + 1);
    }
    return res;
}
  
// Driver Code
public static void main(String[] args)
{
    int n = 3, m = 2, k = 2;
  
    System.out.print(nCr(n + m, k));
}
}
  
// This code is contributed by sapnasingh4991

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Python3

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# Python3 implementation of
# the above approach
  
# Function returns nCr
# i.e. Binomial Coefficient 
def nCr(n, r):
  
    # Initialize res with 1
    res = 1
  
    # Since C(n, r) = C(n, n-r)
    if(r > n - r):
        r = n - r
  
    # Evaluating expression
    for i in range(r):
        res *= (n - i)
        res //= (i + 1)
  
    return res
  
# Driver Code
if __name__ == '__main__':
  
    n = 3
    m = 2
    k = 2
  
    # Function call
    print(nCr(n + m, k))
  
# This code is contributed by Shivam Singh

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C#

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code

// C# implementation of
// the above approach
using System;
class GFG{
  
// Function returns nCr
// i.e. Binomial Coefficient
static int nCr(int n, int r)
{
  
    // Initialize res with 1
    int res = 1;
  
    // Since C(n, r) = C(n, n-r)
    if (r > n - r)
        r = n - r;
  
    // Evaluating expression
    for (int i = 0; i < r; ++i)
    {
        res *= (n - i);
        res /= (i + 1);
    }
  
    return res;
}
  
// Driver Code
public static void Main()
{
    int n = 3, m = 2, k = 2;
    Console.Write(nCr(n + m, k));
}
}
  
// This code is contributed by Code_Mech

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Output: 

10

Time Complexity: O(K) 
Auxiliary Space: O(1)
 

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