以下是二项式系数的常见定义。
- 可以将二项式系数C(n,k)定义为(1 + X)^ n展开中的X ^ k系数。
- 二项式系数C(n,k)还给出了可从n个对象中选择k个对象的方式数量,而无需考虑顺序。更正式地讲,n个元素集的k个元素子集(或k个组合)的数量。
给定两个数字n和r,求n C r的值
例子 :
Input : n = 5, r = 2
Output : 10
The value of 5C2 is 10
Input : n = 3, r = 1
Output : 3
这个想法只是基于以下公式。
nCr = (n!) / (r! * (n-r)!)
C++
// CPP program To calculate The Value Of nCr
#include
using namespace std;
int fact(int n);
int nCr(int n, int r)
{
return fact(n) / (fact(r) * fact(n - r));
}
// Returns factorial of n
int fact(int n)
{
int res = 1;
for (int i = 2; i <= n; i++)
res = res * i;
return res;
}
// Driver code
int main()
{
int n = 5, r = 3;
cout << nCr(n, r);
return 0;
}
Java
// Java program To calculate
// The Value Of nCr
class GFG {
static int nCr(int n, int r)
{
return fact(n) / (fact(r) *
fact(n - r));
}
// Returns factorial of n
static int fact(int n)
{
int res = 1;
for (int i = 2; i <= n; i++)
res = res * i;
return res;
}
// Driver code
public static void main(String[] args)
{
int n = 5, r = 3;
System.out.println(nCr(n, r));
}
}
// This code is Contributed by
// Smitha Dinesh Semwal.
Python 3
# Python 3 program To calculate
# The Value Of nCr
def nCr(n, r):
return (fact(n) / (fact(r)
* fact(n - r)))
# Returns factorial of n
def fact(n):
res = 1
for i in range(2, n+1):
res = res * i
return res
# Driver code
n = 5
r = 3
print(int(nCr(n, r)))
# This code is contributed
# by Smitha
C#
// C# program To calculate
// The Value Of nCr
using System;
class GFG {
static int nCr(int n, int r)
{
return fact(n) / (fact(r) *
fact(n - r));
}
// Returns factorial of n
static int fact(int n)
{
int res = 1;
for (int i = 2; i <= n; i++)
res = res * i;
return res;
}
// Driver code
public static void Main()
{
int n = 5, r = 3;
Console.Write(nCr(n, r));
}
}
// This code is Contributed by nitin mittal.
PHP
Javascript
输出:
10
更高效的解决方案:
动态编程|集合9(二项式系数)
时空有效的二项式系数
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