# 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``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

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## Javascript

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Output:

`8`

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