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

03 Dec, 2021
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 + ..... + 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++
Java
Python3
C#
PHP
Javascript
// 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 nChoosekunsigned 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 codeint main(){    unsigned long long N = 5;    unsigned long long K = 2;    cout << countWays(N, K);    return 0;}
// Java program to find the number of ways in which// a stick of length n can be divided into K piecesimport 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));    }}
# 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 nChoosekdef 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 codeN = 5K = 2print(countWays(N, K))
// C# program to find the number of// ways in which a stick of length n// can be divided into K piecesusing 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 program to calculate the// number of ways to divide a// stick of length n into k pieces  // function to generate nCk or nChoosekfunction 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.?>
<script>//Javascript Implementation// to calculate the number of ways// to divide a stick of length n into k pieces // function to generate nCk or nChoosekfunction 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 codevar N = 5;var K = 2;document.write(countWays(N, K)); // This code is contributed by shubhamsingh10</script>
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 write.geeksforgeeks.org or mail your article to review-team@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. 
nitin mittal
Akanksha_Rai
SHUBHAMSINGH10
simmytarika5
Combinatorial
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