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))

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 :

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.
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++

// 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 ; 
} 

Java

// 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. 

Python3

# 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 

C#

// 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. 

PHP

<?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 
?> 

Output :

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2019	2020	2021

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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