给定两个整数X和N。任务是找到从N个男人中选择X个男人的总数,包括或不包括某个特定男人。
例子:
Input: N = 3 X = 2
Output: 3
Including a man say M1, the ways can be (M1, M2) and (M1, M3).
Excluding a man say M1, the only way is (M2, M3).
Total ways = 2 + 1 = 3.
Input: N = 5 X = 3
Output: 10
方法:从N个人中选择X个人的总数为N C X
- 包括一个特殊的人:我们可以从N – 1 C X – 1中的(N – 1)个中选择(X – 1)个男人。
- 扣除特定的人:我们可以选择从(N – 1)X战警中的N -图1C X
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the value of nCr
int nCr(int n, int r)
{
// Initialize the answer
int ans = 1;
for (int i = 1; i <= r; i += 1) {
// Divide simultaneously by
// i to avoid overflow
ans *= (n - r + i);
ans /= i;
}
return ans;
}
// Function to return the count of ways
int total_ways(int N, int X)
{
return (nCr(N - 1, X - 1) + nCr(N - 1, X));
}
// Driver code
int main()
{
int N = 5, X = 3;
cout << total_ways(N, X);
return 0;
}
Java
// Java implementation of the approach
import java.io.*;
class GFG
{
// Function to return the value of nCr
static int nCr(int n, int r)
{
// Initialize the answer
int ans = 1;
for (int i = 1; i <= r; i += 1)
{
// Divide simultaneously by
// i to avoid overflow
ans *= (n - r + i);
ans /= i;
}
return ans;
}
// Function to return the count of ways
static int total_ways(int N, int X)
{
return (nCr(N - 1, X - 1) + nCr(N - 1, X));
}
// Driver code
public static void main (String[] args)
{
int N = 5, X = 3;
System.out.println (total_ways(N, X));
}
}
// This code is contributed by Sachin
Python3
# Python3 implementation of the approach
# Function to return the value of nCr
def nCr(n, r) :
# Initialize the answer
ans = 1;
for i in range(1, r + 1) :
# Divide simultaneously by
# i to avoid overflow
ans *= (n - r + i);
ans //= i;
return ans;
# Function to return the count of ways
def total_ways(N, X) :
return (nCr(N - 1, X - 1) + nCr(N - 1, X));
# Driver code
if __name__ == "__main__" :
N = 5; X = 3;
print(total_ways(N, X));
# This code is contributed by AnkitRai01
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the value of nCr
static int nCr(int n, int r)
{
// Initialize the answer
int ans = 1;
for (int i = 1; i <= r; i += 1)
{
// Divide simultaneously by
// i to avoid overflow
ans *= (n - r + i);
ans /= i;
}
return ans;
}
// Function to return the count of ways
static int total_ways(int N, int X)
{
return (nCr(N - 1, X - 1) + nCr(N - 1, X));
}
// Driver code
public static void Main (String[] args)
{
int N = 5, X = 3;
Console.WriteLine(total_ways(N, X));
}
}
// This code is contributed by 29AjayKumar
Javascript
输出:
10