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Number of compositions of a natural number

  • Difficulty Level : Hard
  • Last Updated : 13 Apr, 2021

Given a natural number n, find the number of ways in which n can be expressed as a sum of natural numbers when order is taken into consideration. Two sequences that differ in the order of their terms define different compositions of their sum. 
Examples: 
 

Input :  4
Output : 8
Explanation  
All 8 position composition are:
4, 1+3, 3+1, 2+2, 1+1+2, 1+2+1, 2+1+1
and 1+1+1+1

Input :  8
Output : 128

 

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A Simple Solution is to generate all compositions and count them.
Using the concept of combinatorics, it can be proved that any natural number n will have 2^(n-1) distinct compositions when order is taken into consideration.
 



One way to see why the answer is 2^(n-1) directly is to write n as a sum of 1s: 
n = 1 + 1 + 1 +…+ 1 (n times).
There are (n-1) plus signs between all 1s. For every plus sign we can choose to split ( by putting a bracket) at the point or not split. Therefore answer is 2^(n-1).
For example, n = 4 
No Split 
4 = 1 + 1 + 1 + 1 [We write as single 4]
Different ways to split once 
4 = (1) + (1 + 1 + 1) [We write as 1 + 3] 
4 = (1 + 1) + (1 + 1) [We write as 2 + 2] 
4 = (1 + 1 + 1) + (1) [We write as 3 + 1]
Different ways to split twice 
4 = (1) + (1 + 1) + (1) [We write as 1 + 2 + 1] 
4 = (1 + 1) + (1) + (1) [We write as 2 + 1 + 1] 
4 = (1) + (1) + (1 + 1) [We write as 1 + 1 + 2]
Different ways to split three times 
4 = (1) + (1) + (1) + (1) [We write as 1 + 1 + 1 + 1]
Since there are (n-1) plus signs between the n 1s, there are 2^(n-1) ways of choosing where to split the sum, and hence 2^(n-1) possible sums .

 

C++




// C++ program to find the total number of
// compositions of a natural number
#include<iostream>
using namespace std;
 
#define ull unsigned long long
 
ull countCompositions(ull n)
{
    // Return 2 raised to power (n-1)
    return (1L) << (n-1);
}
 
// Driver Code
int main()
{
    ull n = 4;
    cout << countCompositions(n) << "\n";
    return 0;
}

Java




// Java program to find
// the total number of
// compositions of a
// natural number
import java.io.*;
import java.util.*;
 
class GFG
{
public static int countCompositions(int n)
{
    // Return 2 raised
    // to power (n-1)
    return 1 << (n - 1);
}
 
// Driver Code
public static void main(String args[])
{
    int n = 4;
    System.out.print(countCompositions(n));
}
}
 
// This code is contributed by
// Akanksha Rai(Abby_akku)

Python




# Python code to find the total number of
# compositions of a natural number
def countCompositions(n):
 
    # function to return the total number
    # of composition of n
    return (2**(n-1))
 
# Driver Code
print(countCompositions(4))

C#




// C# program to find the
// total number of compositions
// of a natural number
using System;
 
class GFG
{
public static int countCompositions(int n)
{
    // Return 2 raised
    // to power (n-1)
    return 1 << (n - 1);
}
 
// Driver Code
public static void Main()
{
    int n = 4;
    Console.Write(countCompositions(n));
}
}
 
// This code is contributed by mits

PHP




<?php
// PHP program to find the
// total number of compositions
// of a natural number
 
function countCompositions($n)
{
    // Return 2 raised
    // to power (n-1)
    return ((1) << ($n - 1));
}
 
// Driver Code
$n = 4;
echo countCompositions($n), "\n";
     
// This code is contributed
// by ajit
?>

Javascript




<script>
 
// javascript program to find
// the total number of
// compositions of a
// natural number
 
function countCompositions(n)
{
    // Return 2 raised
    // to power (n-1)
    return 1 << (n - 1);
}
 
// Driver Code
 
    var n = 4;
    document.write(countCompositions(n));
 
 
 
// This code is contributed by 29AjayKumar
</script>

Output: 
 

8

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