Number of ways to cut a stick of length N into K pieces

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
divide_sticks

Input : N = 15
        K = 5
Output : 1001

Solving this question is equivalent to solving the mathematics equation x1 + x2 + ….. + xK = 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 (N – 1)C(K – 1), where NCK 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 107 + 7 to mod the answer. We can calculate nCr % p by using Lucas Theorem.
However, python can handle large values without overflow.

C++

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// 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;
}

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Java

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// 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));
    }
}

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Python3

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# 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))

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C#

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// 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.

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PHP

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<?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.
?>

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Output :

4 

Exercise :
Extend the above problem with 0 length pieces allowed. Hint : The number of solutions can similarly be found by writing each xi as yi – 1, and we get an equation y1 + y2 + ….. + yK = 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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Improved By : nitin mittal, Akanksha_Rai



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