Given a stick of size N, find the number of ways in which it can be cut into K pieces such that length of every piece is greater than 0.

**Examples :**

Input : N = 5 K = 2 Output : 4 Input : N = 15 K = 5 Output : 1001

Solving this question is equivalent to solving the mathematics equation **x _{1} + x_{2} + ….. + x_{K} = N**

We can solve this by using the

**bars and stars**method in Combinatorics, from which we obtain the fact that the number of positive integral solutions to this equation is

**, where**

^{(N – 1)}C_{(K – 1)}^{N}C

_{K}is N! / ((N – K) ! * (K!)), where ! stands for factorial.

In C++ and Java, for large values of factorials, there might be overflow errors. In that case we can introduce a large prime number such as 10^{7} + 7 to mod the answer. We can calculate nCr % p by using Lucas Theorem.

However, python can handle large values without overflow.

## C++

`// C++ program to calculate the number of ways` `// to divide a stick of length n into k pieces` `#include <bits/stdc++.h>` `using` `namespace` `std;` ` ` `// function to generate nCk or nChoosek` `unsigned ` `long` `long` `nCr(unsigned ` `long` `long` `n,` ` ` `unsigned ` `long` `long` `r)` `{` ` ` `if` `(n < r)` ` ` `return` `0;` ` ` ` ` `// Reduces to the form n! / n!` ` ` `if` `(r == 0)` ` ` `return` `1;` ` ` ` ` `// nCr has been simplified to this form by` ` ` `// expanding numerator and denominator to ` ` ` `// the form n(n - 1)(n - 2)...(n - r + 1)` ` ` `// -----------------------------` ` ` `// (r!)` ` ` `// in the above equation, (n - r)! is cancelled ` ` ` `// out in the numerator and denominator` ` ` ` ` `unsigned ` `long` `long` `numerator = 1;` ` ` `for` `(` `int` `i = n; i > n - r; i--)` ` ` `numerator = (numerator * i);` ` ` ` ` `unsigned ` `long` `long` `denominator = 1;` ` ` `for` `(` `int` `i = 1; i < r + 1; i++)` ` ` `denominator = (denominator * i);` ` ` ` ` `return` `(numerator / denominator);` `}` ` ` `// Returns number of ways to cut ` `// a rod of length N into K pieces.` `unsigned ` `long` `long` `countWays(unsigned ` `long` `long` `N,` ` ` `unsigned ` `long` `long` `K)` `{` ` ` `return` `nCr(N - 1, K - 1);` `}` ` ` `// Driver code` `int` `main()` `{` ` ` `unsigned ` `long` `long` `N = 5;` ` ` `unsigned ` `long` `long` `K = 2;` ` ` `cout << countWays(N, K);` ` ` `return` `0;` `}` |

## Java

`// Java program to find the number of ways in which` `// a stick of length n can be divided into K pieces` `import` `java.io.*;` `import` `java.util.*;` ` ` `class` `GFG` `{` ` ` `// function to generate nCk or nChoosek` ` ` `public` `static` `int` `nCr(` `int` `n, ` `int` `r)` ` ` `{` ` ` `if` `(n < r)` ` ` `return` `0` `;` ` ` ` ` `// Reduces to the form n! / n!` ` ` `if` `(r == ` `0` `)` ` ` `return` `1` `;` ` ` ` ` `// nCr has been simplified to this form by` ` ` `// expanding numerator and denominator to ` ` ` `// the form n(n - 1)(n - 2)...(n - r + 1)` ` ` `// -----------------------------` ` ` `// (r!)` ` ` `// in the above equation, (n-r)! is cancelled ` ` ` `// out in the numerator and denominator` ` ` ` ` `int` `numerator = ` `1` `;` ` ` `for` `(` `int` `i = n ; i > n - r ; i--)` ` ` `numerator = (numerator * i);` ` ` ` ` `int` `denominator = ` `1` `;` ` ` `for` `(` `int` `i = ` `1` `; i < r + ` `1` `; i++)` ` ` `denominator = (denominator * i);` ` ` ` ` `return` `(numerator / denominator);` ` ` `}` ` ` ` ` `// Returns number of ways to cut ` ` ` `// a rod of length N into K peices` ` ` `public` `static` `int` `countWays(` `int` `N, ` `int` `K)` ` ` `{` ` ` `return` `nCr(N - ` `1` `, K - ` `1` `);` ` ` `}` ` ` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `N = ` `5` `;` ` ` `int` `K = ` `2` `;` ` ` `System.out.println(countWays(N, K));` ` ` `}` `}` |

## Python3

`# Python program to find the number ` `# of ways in which a stick of length ` `# n can be divided into K pieces` ` ` `# function to generate nCk or nChoosek` `def` `nCr(n, r):` ` ` ` ` `if` `(n < r):` ` ` `return` `0` ` ` ` ` `# reduces to the form n! / n!` ` ` `if` `(r ` `=` `=` `0` `):` ` ` `return` `1` ` ` ` ` `# nCr has been simplified to this form by` ` ` `# expanding numerator and denominator to ` ` ` `# the form n(n - 1)(n - 2)...(n - r + 1)` ` ` `# -----------------------------` ` ` `# (r!)` ` ` `# in the above equation, (n-r)! is cancelled ` ` ` `# out in the numerator and denominator` ` ` ` ` `numerator ` `=` `1` ` ` `for` `i ` `in` `range` `(n, n ` `-` `r, ` `-` `1` `):` ` ` `numerator ` `=` `numerator ` `*` `i` ` ` ` ` `denominator ` `=` `1` ` ` `for` `i ` `in` `range` `(` `1` `, r ` `+` `1` `):` ` ` `denominator ` `=` `denominator ` `*` `i` ` ` ` ` `return` `(numerator ` `/` `/` `denominator)` ` ` `# Returns number of ways to cut ` `# a rod of length N into K peices.` `def` `countWays(N, K) :` ` ` `return` `nCr(N ` `-` `1` `, K ` `-` `1` `);` ` ` `# Driver code` `N ` `=` `5` `K ` `=` `2` `print` `(countWays(N, K))` |

## C#

`// C# program to find the number of ` `// ways in which a stick of length n ` `// can be divided into K pieces` `using` `System;` ` ` `class` `GFG` `{` ` ` `// function to generate nCk or nChoosek` ` ` `public` `static` `int` `nCr(` `int` `n, ` `int` `r)` ` ` `{` ` ` `if` `(n < r)` ` ` `return` `0;` ` ` ` ` `// Reduces to the form n! / n!` ` ` `if` `(r == 0)` ` ` `return` `1;` ` ` ` ` `// nCr has been simplified to this form by` ` ` `// expanding numerator and denominator to ` ` ` `// the form n(n - 1)(n - 2)...(n - r + 1)` ` ` `// -----------------------------` ` ` `// (r!)` ` ` `// in the above equation, (n-r)! is cancelled` ` ` `// out in the numerator and denominator` ` ` ` ` `int` `numerator = 1;` ` ` `for` `(` `int` `i = n; i > n - r; i--)` ` ` `numerator = (numerator * i);` ` ` ` ` `int` `denominator = 1;` ` ` `for` `(` `int` `i = 1; i < r + 1; i++)` ` ` `denominator = (denominator * i);` ` ` ` ` `return` `(numerator / denominator);` ` ` `}` ` ` ` ` `// Returns number of ways to cut ` ` ` `// a rod of length N into K pieces` ` ` `public` `static` `int` `countWays(` `int` `N, ` `int` `K)` ` ` `{` ` ` `return` `nCr(N - 1, K - 1);` ` ` `}` ` ` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `N = 5;` ` ` `int` `K = 2;` ` ` `Console.Write(countWays(N, K));` ` ` ` ` `}` `}` ` ` `// This code is contributed by nitin mittal.` |

## PHP

`<?php` `// PHP program to calculate the` `// number of ways to divide a ` `// stick of length n into k pieces` ` ` ` ` `// function to generate nCk or nChoosek` `function` `nCr(` `$n` `, ` `$r` `)` `{` ` ` `if` `(` `$n` `< ` `$r` `)` ` ` `return` `0;` ` ` ` ` `// Reduces to the form n! / n!` ` ` `if` `(` `$r` `== 0)` ` ` `return` `1;` ` ` ` ` `// nCr has been simplified to this form by` ` ` `// expanding numerator and denominator to ` ` ` `// the form n(n - 1)(n - 2)...(n - r + 1)` ` ` `// -----------------------------` ` ` `// (r!)` ` ` `// in the above equation, (n - r)! is cancelled ` ` ` `// out in the numerator and denominator` ` ` ` ` `$numerator` `= 1;` ` ` `for` `(` `$i` `= ` `$n` `; ` `$i` `> ` `$n` `- ` `$r` `; ` `$i` `--)` ` ` `$numerator` `= (` `$numerator` `* ` `$i` `);` ` ` ` ` `$denominator` `= 1;` ` ` `for` `(` `$i` `= 1; ` `$i` `< ` `$r` `+ 1; ` `$i` `++)` ` ` `$denominator` `= (` `$denominator` `* ` `$i` `);` ` ` ` ` `return` `(` `floor` `(` `$numerator` `/ ` `$denominator` `));` `}` ` ` `// Returns number of ways to cut ` `// a rod of length N into K peices.` `function` `countWays(` `$N` `, ` `$K` `)` `{` ` ` `return` `nCr(` `$N` `- 1, ` `$K` `- 1);` `}` ` ` `// Driver code` `$N` `= 5;` `$K` `= 2;` `echo` `countWays(` `$N` `, ` `$K` `);` `return` `0;` ` ` `// This code is contributed by nitin mittal.` `?>` |

**Output :**

4

**Exercise :**

Extend the above problem with 0 length pieces allowed. Hint : The number of solutions can similarly be found by writing each x_{i} as y_{i} – 1, and we get an equation **y _{1} + y_{2} + ….. + y_{K} = N + K**. The number of solutions to this equation is

^{(N + K – 1)}C_{(K – 1)}This article is contributed by **Deepak Srivatsav**. 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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