# 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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**

In case you wish to attend **live classes **with experts, please refer **DSA Live Classes for Working Professionals **and **Competitive Programming Live for Students**.

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

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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.