Count numbers having N 0’s and M 1’s with no leading zeros

Given two integer N and M, the task is to find the number of distinct numbers having N 0’s and M 1’s with no leading zeros and N + M total digits.

Examples:

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. 

Approach: The problem can be easily solved by finding the total permutation of N similar items and M – 1 similar items. Since no leading zeros are allowed, one 1 is always fixed at the start of the number. The remaining M – 1 1’s and N 0’s are arranged in different permutations. Now, the number of permutations of (A + B) objects where A objects of the same type and B objects of other type is given by (A + B)! / (A! * B!). Therefore, the number of distinct numbers is given by (N + M -1)! / (N! * (M – 1)!).

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
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  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  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  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  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

PHP

<?php

// PHP implementation of the approach
// Function to return the factorial of a number
function factorial($f)
{
    $fact = 1;

    for ($i = 2; $i <= $f; $i++)
        $fact *= $i;

    return $fact;
}

// Function to return the count of distinct
// (N + M) digit numbers having N 0's
// and  M 1's with no leading zeros
function findPermutation($N,$M)
{
    $permutation = factorial($N + $M - 1)
                    / (factorial($N) * factorial($M - 1));
    return $permutation;
}

    // Driver code
    $N = 3;
    $M = 3;
    echo findPermutation($N, $M);

// This code is contributed by ajit..
?>

Javascript

<script>

// Javascript implementation of the approach

// Function to return the factorial of a number
function factorial(f)
{
    var fact = 1;

    for(var i = 2; i <= f; i++)
        fact *= i;

    return fact;
}

// Function to return the count of distinct
// (N + M) digit numbers having N 0's
// and  M 1's with no leading zeros
function findPermutation(N, M)
{
    var permutation = factorial(N + M - 1) /
                    (factorial(N) * factorial(M - 1));
    return permutation;
}

// Driver code
var N = 3, M = 3;

document.write(findPermutation(N, M));

// This code is contributed by noob2000

</script>

Output:

10

Time Complexity: O(N+M)
Auxiliary Space: O(1), as no extra space is used