两个数的二项式系数的所有乘积之和，最高为K

给定三个整数N ， M和K ，任务是计算二项式系数C(N，i)和C(M，K – i)的乘积之和，其中i介于[0，K]之间。

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

例子：

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

天真的方法：解决此问题的最简单方法是简单地在[0，K]范围内进行迭代，并为每个i计算C(N，i)和C(M，K – 1) ，并通过相加它们的乘积来更新总和。
下面是上述方法的实现：

C++

// C++ implementation of
// the above approach
 
#include 
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

输出：

10

时间复杂度： O(K ² )
辅助空间： O(1)
高效方法：
可以使用Vandermonde的Identity对上述方法进行优化。

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.

因此，给定的表达式简化为以下形式：

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

下面是上述方法的实现：

C++

// C++ implementation of
// the above approach
 
#include 
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

Java脚本

 
// 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.

输出：

10

时间复杂度： O(K)
辅助空间： O(1)