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].

Examples:

Input: N = 2, M = 2, K = 2 
Output: 6 
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: 5 
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++

// 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;
}

Java

// 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

Python3

# 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

C#

// 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

Javascript

<script>
// JavaScript program for the above approach

// Function returns nCr
// i.e. Binomial Coefficient
function nCr(n, r)
{

    // Initialize res with 1
    let res = 1;

    // Since C(n, r) = C(n, n-r)
    if (r > n - r)
        r = n - r;

    // Evaluating expression
    for (let i = 0; i < r; ++i)
    {
        res *= (n - i);
        res /= (i + 1);
    }

    return res;
}

// Function to calculate and
// return the sum of the products
function solve(n, m, k)
{

    // Initialize sum to 0
    let sum = 0;

    // Traverse from 0 to k
    for (let i = 0; i <= k; i++)
        sum += nCr(n, i)
               * nCr(m, k - i);

    return sum;
}

// Driver Code

    let n = 3, m = 2, k = 2;

    document.write(solve(n, m, k));

</script>

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:

Below is the implementation of the above approach:

C++

// 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;
}

Java

// 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

Python3

# 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

C#

// 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

Javascript

<script>

// JavaScript implementation of the above approach

// Function returns nCr
// i.e. Binomial Coefficient
function nCr(n, r)
{

    // Initialize res with 1
    let res = 1;

    // Since C(n, r) = C(n, n-r)
    if (r > n - r)
        r = n - r;

    // Evaluating expression
    for (let i = 0; i < r; ++i)
    {
        res *= (n - i);
        res /= (i + 1);
    }
    return res;
}

// Driver code

        let n = 3, m = 2, k = 2;

        document.write(nCr(n + m, k));

    // This code is contributed by code_hunt.
</script>

Output:

10

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