Ways to multiply n elements with an associative operation

Given a number n, find the number of ways to multiply n elements with an associative operation.

Examples :

Input : 2
Output : 2
For a and b there are two ways to multiply them.
1. (a * b)        
2. (b * a)

Input : 3
Output : 12

Explanation(Example 2) :

For a, b and c there are 12 ways to multiply them.
1.  ((a * b) * c)     2.  (a * (b * c))
3.  ((a * c) * b)     4.  (a * (c * b))
5.  ((b * a) * c)     6.  (b * (a * c))
7.  ((b * c) * a)     8.  (b * (c * a))
9.  ((c * a) * b)     10.  (c * (a * b))
11.  ((c * b) * a)    12.  (c * (b * a))

Approach : First, we try to find out the recurrence relation. From above examples, we can see h(1) = 1, h(2) = 2, h(3) = 12 . Now, for n elements there will be n – 1 multiplications and n – 1 parentheses. And, (a1, a2, …, an ) can be obtained from (a1, a2, …, a(n – 1)) in exactly one of the two ways :

  1. Take a multiplication (a1, a2, …, a(n – 1))(which has n – 2 multiplications and n – 2 parentheses) and insert the nth element ‘an’ on either side of either factor in one of the n – 2 multiplications. Thus, for each scheme for n – 1 numbers gives 2 * 2 * (n – 2) = 4 * (n – 2) schemes for n numbers in this way.
  2. Take a multiplication scheme for (a1, a2, .., a(n-1)) and multiply on left or right by (‘an’). Thus, for each each scheme for n – 1 numbers gives two schemes for n numbers in this way.

So after adding above two, we get, h(n) = (4 * n – 8 + 2) * h(n – 1), h(n) = (4 * n – 6) * h(n – 1). This recurrence relation with same initial value is satisfied by the pseudo-Catalan number. Hence, h(n) = (2 * n – 2)! / (n – 1)!

C++

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// CPP code to find number of ways to multiply n 
// elements with an associative operation
# include <bits/stdc++.h>
using namespace std;
  
// Function to find the required factorial
int fact(int n)
{
    if (n == 0 || n == 1)    
        return 1 ;
  
    int ans = 1;   
    for (int i = 1 ; i <= n; i++)    
        ans = ans * i ; 
  
    return ans ;
}
  
// Function to find nCr
int nCr(int n, int r)
{
    int Nr = n , Dr = 1 , ans = 1;
    for (int i = 1 ; i <= r ; i++ ) {
        ans = ( ans * Nr ) / ( Dr ) ;
        Nr-- ;
        Dr++ ;
    }
    return ans ;
}
  
// function to find the number of ways
int solve ( int n )
{
    int N = 2*n - 2 ;
    int R = n - 1 ;    
    return nCr (N, R) * fact(n - 1) ;
}
  
// Driver code
int main()
{
    int n = 6 ;
    cout << solve (n) ;    
    return 0 ;
}

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Java

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// Java code to find number of 
// ways to multiply n elements 
// with an associative operation
import java.io.*;
  
class GFG 
{
// Function to find the
// required factorial
static int fact(int n)
{
    if (n == 0 || n == 1
        return 1 ;
  
    int ans = 1
    for (int i = 1 ; i <= n; i++) 
        ans = ans * i ; 
  
    return ans ;
}
  
// Function to find nCr
static int nCr(int n, int r)
{
    int Nr = n , Dr = 1 , ans = 1;
    for (int i = 1 ; i <= r ; i++ ) 
    {
        ans = ( ans * Nr ) / ( Dr ) ;
        Nr-- ;
        Dr++ ;
    }
    return ans ;
}
  
// function to find
// the number of ways
static int solve ( int n )
{
    int N = 2 * n - 2 ;
    int R = n - 1
    return nCr (N, R) * fact(n - 1) ;
}
  
// Driver Code
public static void main (String[] args) 
{
int n = 6 ;
System.out.println( solve (n)) ; 
}
}
  
// This code is contributed by anuj_67.

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Python3

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# Python3 code to find number
# of ways to multiply n 
# elements with an
# associative operation
  
# Function to find the
# required factorial
def fact(n):
    if (n == 0 or n == 1): 
        return 1;
  
    ans = 1
    for i in range(1, n + 1): 
        ans = ans * i; 
  
    return ans;
  
# Function to find nCr
def nCr(n, r):
    Nr = n ; Dr = 1 ; ans = 1;
    for i in range(1, r + 1):
        ans = int((ans * Nr) / (Dr));
        Nr = Nr - 1;
        Dr = Dr + 1;
    return ans;
  
# function to find 
# the number of ways
def solve ( n ):
    N = 2* n - 2;
    R = n - 1
    return (nCr (N, R) * 
            fact(n - 1));
  
# Driver code
n = 6 ;
print(solve (n) ); 
  
# This code is contributed
# by mits

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C#

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// C# code to find number of 
// ways to multiply n elements 
// with an associative operation
using System;
  
class GFG {
      
    // Function to find the
    // required factorial
    static int fact(int n)
    {
        if (n == 0 || n == 1) 
            return 1 ;
      
        int ans = 1; 
        for (int i = 1 ; i <= n; i++) 
            ans = ans * i ; 
      
        return ans ;
    }
      
    // Function to find nCr
    static int nCr(int n, int r)
    {
        int Nr = n , Dr = 1 , ans = 1;
        for (int i = 1 ; i <= r ; i++ ) 
        {
            ans = ( ans * Nr ) / ( Dr ) ;
            Nr-- ;
            Dr++ ;
        }
        return ans ;
    }
      
    // function to find
    // the number of ways
    static int solve ( int n )
    {
        int N = 2 * n - 2 ;
        int R = n - 1 ; 
        return nCr (N, R) * fact(n - 1) ;
    }
      
    // Driver Code
    public static void Main () 
    {
        int n = 6 ;
        Console.WriteLine( solve (n)) ; 
    }
}
  
// This code is contributed by anuj_67.

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PHP

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<?php
// PHP code to find number
// of ways to multiply n 
// elements with an
// associative operation
  
// Function to find the
// required factorial
function fact($n)
{
    if ($n == 0 || $n == 1) 
        return 1;
  
    $ans = 1; 
    for ($i = 1 ; $i <= $n; $i++) 
        $ans = $ans * $i
  
    return $ans;
}
  
// Function to find nCr
function nCr($n, $r)
{
    $Nr = $n ; $Dr = 1 ; $ans = 1;
    for ($i = 1 ; $i <= $r ; $i++ )
    {
        $ans = ($ans * $Nr) / 
                      ($Dr);
        $Nr--;
        $Dr++;
    }
    return $ans ;
}
  
// function to find 
// the number of ways
function solve ( $n )
{
    $N = 2* $n - 2 ;
    $R = $n - 1 ; 
    return nCr ($N, $R) * 
           fact($n - 1) ;
}
  
// Driver code
$n = 6 ;
echo solve ($n) ; 
  
// This code is contributed
// by ajit
?>

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Output :

30240


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