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Number of Permutations such that no Three Terms forms Increasing Subsequence
  • Difficulty Level : Expert
  • Last Updated : 13 May, 2021

Given a number N. The task is to find the number of permutations of 1 to N such that no three terms of the permutation form an increasing subsequence.

Examples:  

Input : N = 3
Output : 5
Valid permutations : 132, 213, 231, 312 and 321 and not 123

Input : N = 4
Output : 14 

The above problem is an application of Catalan numbers. So, the task is to only find the nth Catalan Number. First few Catalan numbers are 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, … (considered from 0th number)

Below is the program to find Nth Catalan Number:  

C++




// C++ program to find the
// nth catalan number
#include <bits/stdc++.h>
using namespace std;
 
// Returns value of Binomial Coefficient C(n, k)
unsigned long int binomialCoeff(unsigned int n,
                                unsigned int k)
{
    unsigned long int res = 1;
 
    // Since C(n, k) = C(n, n-k)
    if (k > n - k)
        k = n - k;
 
    // Calculate value of
    // [n*(n-1)*---*(n-k+1)] / [k*(k-1)*---*1]
    for (int i = 0; i < k; ++i) {
        res *= (n - i);
        res /= (i + 1);
    }
 
    return res;
}
 
// A Binomial coefficient based function
// to find nth catalan
// number in O(n) time
unsigned long int catalan(unsigned int n)
{
    // Calculate value of 2nCn
    unsigned long int c = binomialCoeff(2 * n, n);
 
    // return 2nCn/(n+1)
    return c / (n + 1);
}
 
// Driver code
int main()
{
    int n = 3;
 
    cout << catalan(n) << endl;
 
    return 0;
}

Java




// Java program to find the
// nth catalan number
import java.io.*;
 
class GFG
{
 
// Returns value of Binomial
// Coefficient C(n, k)
static long binomialCoeff(long n, long k)
{
    long res = 1;
 
    // Since C(n, k) = C(n, n-k)
    if (k > n - k)
        k = n - k;
 
    // Calculate value of
    // [n*(n-1)*---*(n-k+1)] /
    // [k*(k-1)*---*1]
    for (int i = 0; i < k; ++i)
    {
        res *= (n - i);
        res /= (i + 1);
    }
 
    return res;
}
 
// A Binomial coefficient based
// function to find nth catalan
// number in O(n) time
static long catalan(long n)
{
    // Calculate value of 2nCn
    long c = binomialCoeff(2 * n, n);
 
    // return 2nCn/(n+1)
    return c / (n + 1);
}
 
// Driver code
public static void main (String[] args)
{
    int n = 3;
     
    System.out.println(catalan(n));
}
}
 
// This code has been contributed
// by inder_verma.

Python3




# Python program to find the
# nth catalan number
 
# Returns value of Binomial
# Coefficient C(n, k)
def binomialCoeff(n, k):
    res = 1
     
    # Since C(n, k) = C(n, n-k)
    if k > n - k:
        k=n-k
    # Calculate value of
    # [n*(n-1)*---*(n-k+1)] //
    # [k*(k-1)*---*1]
 
    for i in range(k):
        res = res * (n - i)
        res = res // (i + 1)
    return res
     
# A Binomial coefficient based
# function to find nth catalan
# number in O(n) time
def catalan(n):
     
    # Calculate value of 2nCn
    c = binomialCoeff(2 * n, n)
     
    # return 2nCn/(n+1)
    return c // (n + 1)
     
# Driver code
n = 3
print(catalan(n))
 
# This code is contributed
# by sahil shelangia

C#




// C# program to find the
// nth catalan number
using System;
 
class GFG
{
 
// Returns value of Binomial
// Coefficient C(n, k)
static long binomialCoeff(long n,
                          long k)
{
    long res = 1;
 
    // Since C(n, k) = C(n, n-k)
    if (k > n - k)
        k = n - k;
 
    // Calculate value of
    // [n*(n-1)*---*(n-k+1)] /
    // [k*(k-1)*---*1]
    for (int i = 0; i < k; ++i)
    {
        res *= (n - i);
        res /= (i + 1);
    }
 
    return res;
}
 
// A Binomial coefficient based
// function to find nth catalan
// number in O(n) time
static long catalan(long n)
{
    // Calculate value of 2nCn
    long c = binomialCoeff(2 * n, n);
 
    // return 2nCn/(n+1)
    return c / (n + 1);
}
 
// Driver code
public static void Main (String[] args)
{
    int n = 3;
     
    Console.WriteLine(catalan(n));
}
}
 
// This code is contributed
// by Kirti_Mangal

PHP




<?php
// PHP program to find the
// nth catalan number
 
// Returns value of Binomial
// Coefficient C(n, k)
function binomialCoeff($n, $k)
{
    $res = 1;
     
    // Since C(n, k) = C(n, n-k)
    if($k >$n - $k)
        $k = $n - $k;
         
    // Calculate value of
    // [n*(n-1)*---*(n-k+1)] //
    // [k*(k-1)*---*1]
    for ($i = 0; $i < $k; $i++)
    {
        $res = $res * ($n - $i);
        $res = $res / ($i + 1);
    }
    return $res;
}
 
// A Binomial coefficient based
// function to find nth catalan
// number in O(n) time
function catalan($n)
{
    // Calculate value of 2nCn
    $c = binomialCoeff(2 * $n, $n);
     
    // return 2nCn/(n+1)
    return $c / ($n + 1);
}
 
// Driver code
$n = 3;
print(catalan($n));
 
// This code is contributed
// by mits
?>

Javascript




<script>
 
 
// Javascript program to find the
// nth catalan number
 
// Returns value of Binomial Coefficient C(n, k)
function binomialCoeff(n, k)
{
    var res = 1;
 
    // Since C(n, k) = C(n, n-k)
    if (k > n - k)
        k = n - k;
 
    // Calculate value of
    // [n*(n-1)*---*(n-k+1)] / [k*(k-1)*---*1]
    for (var i = 0; i < k; ++i) {
        res *= (n - i);
        res /= (i + 1);
    }
 
    return res;
}
 
// A Binomial coefficient based function
// to find nth catalan
// number in O(n) time
function catalan(n)
{
    // Calculate value of 2nCn
    var c = binomialCoeff(2 * n, n);
 
    // return 2nCn/(n+1)
    return c / (n + 1);
}
 
// Driver code
var n = 3;
document.write( catalan(n));
 
</script>
Output: 
5

 

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