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Find ways to arrange K green balls among N balls such that exactly i moves is needed to collect all K green balls
  • Last Updated : 03 Dec, 2019

Given two integers N and K. There are N balls placed in a row. K of them are green and N – K of them are black. The task is to find the number of ways to arrange N balls such that one will need exactly i ( 1 ≤ i ≤ K ) moves to collect all the green balls. In one move, we can collect any group of consecutive green balls. Note that the answer can be very large. So, output answer modulo 109 + 7.

Examples:

Input: N = 5, K = 3
Output: 3 6 1
There are three ways to arrange the balls so that
one will need exactly one move:
(G, G, G, B, B), (B, G, G, G, B), and (B, B, G, G, G).

There are six ways to arrange the balls so that
one will need exactly two moves:
(G, G, B, G, B), (G, G, B, B, G), (B, G, G, B, G), (B, G, B, G, G),
(G, B, G, G, B), and (G, B, B, G, G).

There is only one way to arrange the balls so that
one will need exactly three moves: (G, B, G, B, G).



Input: N = 100, K = 5
Output: 96 18240 857280 13287840 61124064

Approach: Only i moves have to be performed to collect K green balls, which means that K green balls are separated into i places by balck balls. Therefore, let’s consider the combination as follows.

  • First, arrange the N – K balck balls in a row.
  • In between these black balls, select i places from the left end to the right end and consider placing K green balls there. There are N – K + 1C i ways to choose these.
  • For each choice, consider how many green balls will be assigned to each gap. Since it is necessary to assign one or more to each, there are K – 1C i – 1 ways to determine this.

Therefore, for each i, the answer is N – K + 1C i * K – 1C i – 1 . Finding n C r is disscused here.

Below is the implementation of the above approach:

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define N 100005
#define mod (int)(1e9 + 7)
  
// To store the factorial and the
// factorial mod inverse of a number
int factorial[N], modinverse[N];
  
// Function to find (a ^ m1) % mod
int power(int a, int m1)
{
    if (m1 == 0)
        return 1;
    else if (m1 == 1)
        return a;
    else if (m1 == 2)
        return (1LL * a * a) % mod;
    else if (m1 & 1)
        return (1LL * a
                * power(power(a, m1 / 2), 2))
               % mod;
    else
        return power(power(a, m1 / 2), 2) % mod;
}
  
// Function to find factorial
// of all the numbers
void factorialfun()
{
    factorial[0] = 1;
    for (int i = 1; i < N; i++)
        factorial[i] = (1LL
                        * factorial[i - 1] * i)
                       % mod;
}
  
// Function to find the factorial
// mod inverse of all the numbers
void modinversefun()
{
    modinverse[N - 1]
        = power(factorial[N - 1], mod - 2) % mod;
  
    for (int i = N - 2; i >= 0; i--)
        modinverse[i] = (1LL * modinverse[i + 1]
                         * (i + 1))
                        % mod;
}
  
// Function to return nCr
int binomial(int n, int r)
{
    if (r > n)
        return 0;
  
    int a = (1LL * factorial[n]
             * modinverse[n - r])
            % mod;
  
    a = (1LL * a * modinverse[r]) % mod;
    return a;
}
  
// Function to find ways to arrange K green
// balls among N balls such that we need
// exactly i moves to collect all K green balls
void arrange_balls(int n, int k)
{
    factorialfun();
    modinversefun();
  
    for (int i = 1; i <= k; i++)
        cout << (1LL * binomial(n - k + 1, i)
                 * binomial(k - 1, i - 1))
                    % mod
             << " ";
}
  
// Driver code
int main()
{
    int n = 5, k = 3;
  
    // Function call
    arrange_balls(n, k);
  
    return 0;
}

Python3




# Python3 implementation of the approach 
N = 100005
mod = (int)(1e9 + 7
  
# To store the factorial and the 
# factorial mod inverse of a number 
factorial = [0] * N;
modinverse = [0] * N; 
  
# Function to find (a ^ m1) % mod 
def power(a, m1) : 
      
    if (m1 == 0) :
        return 1
    elif (m1 == 1) :
        return a; 
    elif (m1 == 2) :
        return (a * a) % mod; 
    elif (m1 & 1) :
        return (a * power(power(a, m1// 2), 2)) % mod; 
    else :
        return power(power(a, m1 // 2), 2) % mod; 
  
# Function to find factorial 
# of all the numbers 
def factorialfun() :
  
    factorial[0] = 1
    for i in range(1, N) :
        factorial[i] = (factorial[i - 1] * i) % mod; 
  
# Function to find the factorial 
# mod inverse of all the numbers 
def modinversefun() :
    modinverse[N - 1] = power(factorial[N - 1], 
                                mod - 2) % mod;
      
    for i in range(N - 2 , -1, -1) :
        modinverse[i] = (modinverse[i + 1] * 
                                   (i + 1)) % mod; 
  
# Function to return nCr 
def binomial(n, r) : 
  
    if (r > n) :
        return 0
  
    a = (factorial[n] * 
         modinverse[n - r]) % mod; 
  
    a = (a * modinverse[r]) % mod; 
    return a; 
  
# Function to find ways to arrange K green 
# balls among N balls such that we need 
# exactly i moves to collect all K green balls 
def arrange_balls(n, k) :
    factorialfun(); 
    modinversefun(); 
  
    for i in range(1, k + 1) :
        print((binomial(n - k + 1, i) * 
               binomial(k - 1, i - 1)) % mod, 
                                  end = " "); 
  
# Driver code 
if __name__ == "__main__"
  
    n = 5; k = 3
  
    # Function call 
    arrange_balls(n, k); 
  
# This code is contributed by AnkitRai01
Output:
3 6 1

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