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

Input : N = 15
        K = 5
Output : 1001

Solving this question is equivalent to solving the mathematics equation x₁ + x₂ + ….. + 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 ^{(N – 1)}C_{(K – 1)}, where ^NC_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 10⁷ + 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 pieces
    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 pieces.
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 pieces.
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.
?>

Javascript

<script>
//Javascript Implementation
// 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

    var numerator = 1;
    for (var i = n; i > n - r; i--)
        numerator = (numerator * i);

    var denominator = 1;
    for (var i = 1; i < r + 1; i++)
        denominator = (denominator * i);

    return Math.floor(numerator / denominator);
}

// Returns number of ways to cut
// a rod of length N into K pieces.
function countWays(N,K)
{
    return nCr(N - 1, K - 1);
}

// Driver code
var N = 5;
var K = 2;
document.write(countWays(N, K));

// This code is contributed by shubhamsingh10
</script>

Output

4

Time complexity: O(N)
Auxiliary space: O(1)

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 y₁ + y₂ + ….. + y_K = N + K. The number of solutions to this equation is ^{(N + K – 1)}C_{(K – 1)}
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