Number of compositions of a natural number

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

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++

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// 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;
}

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Java

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// 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)

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Python

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# 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))

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C#

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// 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

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PHP

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<?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
?>

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Output:

8

This article is contributed by Sruti Rai . If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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