计算具有N 0和M 1且无前导零的数字

给定两个整数N和M ，任务是查找具有N 0和M 1且无前导零和N + M总位数的不同数字的数量。

例子：

Input: N = 2, M = 2
Output: 3
The numbers are 1001, 1010 and 1100.

Input: N = 2, M = 3
Output: 6
The numbers are 10011, 10101, 10110, 11001, 11010 and 11100.

方法：通过找到N个相似项和M – 1个相似项的总排列，可以轻松解决该问题。由于不允许前导零，所以数字1始终固定为1 。其余的M – 1 1和N 0以不同的排列方式排列。现在，排列的(A + B)对象的数量，其中A的对象相同的类型和其它类型的对象乙通过(A + B)中给出！ /(A！* B！) 。因此，不同数字的数量由(N + M -1)给出！ /(N！*(M – 1)！) 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
#define ll long long int
  
// Function to return the factorial of a number
ll factorial(int f)
{
    ll fact = 1;
  
    for (int i = 2; i <= f; i++)
        fact *= (ll)i;
  
    return fact;
}
  
// Function to return the count of distinct
// (N + M) digit numbers having N 0's
// and and M 1's with no leading zeros
ll findPermutation(int N, int M)
{
    ll permutation = factorial(N + M - 1)
                     / (factorial(N) * factorial(M - 1));
    return permutation;
}
  
// Driver code
int main()
{
    int N = 3, M = 3;
    cout << findPermutation(N, M);
  
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
  
class GFG
{
// Function to return the factorial of a number
static int factorial(int f)
{
    int fact = 1;
  
    for (int i = 2; i <= f; i++)
        fact *= (int)i;
  
    return fact;
}
  
// Function to return the count of distinct
// (N + M) digit numbers having N 0's
// and and M 1's with no leading zeros
static int findPermutation(int N, int M)
{
    int permutation = factorial(N + M - 1)
                    / (factorial(N) * factorial(M - 1));
    return permutation;
}
  
// Driver code
public static void main (String[] args)
{
  
    int N = 3, M = 3;
    System.out.println(findPermutation(N, M));
}
}
  
// This code is contributed
// by ajit

Python3

# Python3 implementation of the approach
  
# Function to return the factorial
# of a number
def factorial(f):
    fact = 1
    for i in range(2, f + 1):
        fact *= i
    return fact
  
# Function to return the count of distinct
# (N + M) digit numbers having N 0's
# and and M 1's with no leading zeros
def findPermuatation(N, M):
    permutation = (factorial(N + M - 1) // 
                  (factorial(N) * factorial(M - 1)))
    return permutation
  
# Driver code
N = 3; M = 3
print(findPermuatation(N, M))
  
# This code is contributed 
# by Shrikant13

C#

// C# implementation of the approach
using System;
  
class GFG
{
// Function to return the factorial of a number
static int factorial(int f)
{
    int fact = 1;
  
    for (int i = 2; i <= f; i++)
        fact *= (int)i;
  
    return fact;
}
  
// Function to return the count of distinct
// (N + M) digit numbers having N 0's
// and and M 1's with no leading zeros
static int findPermutation(int N, int M)
{
    int permutation = factorial(N + M - 1)
                    / (factorial(N) * factorial(M - 1));
    return permutation;
}
  
// Driver code
public static void Main()
{
    int N = 3, M = 3;
    Console.Write(findPermutation(N, M));
}
}
  
// This code is contributed
// by Akanksha Rai

输出：

10