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: 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++ 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 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 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# 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
<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:
\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++ 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 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 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# 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
<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)
