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Count the number of ways to fill K boxes with N distinct items

  • Difficulty Level : Medium
  • Last Updated : 23 Oct, 2020

Given two values N and K. Find the number of ways to arrange the N distinct items in the boxes such that exactly K (K<N) boxes are used from the N distinct boxes. The answer can be very large so return the answer modulo 109 + 7.
Note: 1 <= N <= K <= 105
Prerequisites: Factorial of a number, Compute nCr % p

Examples: 

Input: N = 5, k = 5 
Output: 120

Input: N = 5, k = 3 
Output: 1500 

Approach: We will use the inclusion-exclusion principle to count the ways. 



  1. Let us assume that the boxes are numbered 1 to N and now we have to choose any K boxes and use them. The number of ways to do this is NCK.
  2. Now any item can be put in any of the chosen boxes, hence the number of ways to arrange them is KN But here, we may count arrangements with some boxes empty. Hence, we will use the inclusion-exclusion principle to ensure that we count ways with all K boxes filled with at least one item.
  3. Let us understand the application of the inclusion-exclusion principle: 
    • So out of KN ways, we subtract the case when at least 1 box(out of K) is empty. Hence, subtract 
      (KC1)*((K-1)N).
    • Note that here, The cases where exactly two boxes are empty are subtracted twice(once when we choose the first element in (KC1) ways, and then when we choose the second element in (KC1) ways).
    • Hence, we add these ways one time to compensate. So we add (KC2)*((K – 2)N).
    • Similarly, here we need to add the number of ways when at least 3 boxes were empty, and so on…
  4. Hence, the total number of ways:
     


C++




// C++ program to calculate the
// above formula
#include <bits/stdc++.h>
#define mod 1000000007
#define int long long
 
using namespace std;
 
// To store the factorials
// of all numbers
int factorial[100005];
 
// Function to calculate factorial
// of all numbers
void StoreFactorials(int n)
{
    factorial[0] = 1;
    for (int i = 1; i <= n; i++)
    {
        factorial[i] =
          (i * factorial[i - 1])
            % mod;
         
    }
}
 
// Calculate x to the power y
// in O(log n) time
int Power(int x, int y)
{
    int ans = 1;
    while (y > 0) {
        if (y % 2 == 1) {
            ans = (ans * x) % mod;
        }
        x = (x * x) % mod;
        y /= 2;
    }
    return ans;
}
 
// Function to find inverse mod of
// a number x
int invmod(int x)
{
    return Power(x, mod - 2);
}
 
// Calculate (n C r)
int nCr(int n, int r)
{
    return (factorial[n]
            * invmod((factorial[r]
            * factorial[n - r]) % mod))
            % mod;
}
 
int CountWays(int n,int k)
{
    StoreFactorials(n);
 
     
    // Loop to compute the formula
    // evaluated
    int ans = 0;
    for (int i = k; i >= 0; i--)
    {
        if (i % 2 == k % 2)
        {
            // Add even power terms
            ans = (ans + (Power(i, n)
                  * nCr(k, i)) % mod)
                  % mod;
        }
        else
        {
            // Subtract odd power terms
            ans = (ans + mod - (Power(i, n)
                  * nCr(k, i)) % mod) % mod;
        }
    }
     
    // Choose the k boxes which
    // were used
    ans = (ans * nCr(n, k)) % mod;
 
    return ans;
}
 
// Driver code
signed main()
{
    int N = 5;
    int K = 5;
     
    cout << CountWays(N, K) << "\n";
     
    return 0;
}

Java




// Java program to calculate the
// above formula    
import java.util.*;
 
class GFG{        
     
static long mod = 1000000007;
 
// To store the factorials
// of all numbers
static long factorial[] = new long[100005];
 
// Function to calculate factorial
// of all numbers
static void StoreFactorials(int n)
{
    factorial[0] = 1;
     
    for(int i = 1; i <= n; i++)
    {
        factorial[i] = (i *
        factorial[i - 1]) % mod;
    }
}
 
// Calculate x to the power y
// in O(log n) time
static long Power(long x, long y)
{
    long ans = 1;
     
    while (y > 0)
    {
        if (y % 2 == 1)
        {
            ans = (ans * x) % mod;
        }
        x = (x * x) % mod;
        y /= 2;
    }
    return ans;
}
 
// Function to find inverse mod of
// a number x
static long invmod(long x)
{
    return Power(x, mod - 2);
}
 
// Calculate (n C r)
static long nCr(int n, int r)
{
    return (factorial[n] *
    invmod((factorial[r] *
            factorial[n - r]) % mod)) % mod;
}
 
static long CountWays(int n,int k)
{
    StoreFactorials(n);
 
    // Loop to compute the formula
    // evaluated
    long ans = 0;
    for(int i = k; i >= 0; i--)
    {
        if (i % 2 == k % 2)
        {
             
            // Add even power terms
            ans = (ans + (Power(i, n) *
                   nCr(k, i)) % mod) % mod;
        }
        else
        {
             
            // Subtract odd power terms
            ans = (ans + mod - (Power(i, n) *
                   nCr(k, i)) % mod) % mod;
        }
    }
     
    // Choose the k boxes which
    // were used
    ans = (ans * nCr(n, k)) % mod;
 
    return ans;
}    
         
// Driver Code        
public static void main (String[] args)
{        
    int N = 5;
    int K = 5;
     
    System.out.print(CountWays(N, K) + "\n");
}        
}
 
// This code is contributed by math_lover

Python3




# Python3 program to calculate the
# above formula
 
mod = 1000000007
 
# To store the factorials
# of all numbers
factorial = [0 for i in range(100005)]
 
# Function to calculate factorial
# of all numbers
def StoreFactorials(n):
     
    factorial[0] = 1
    for i in range(1, n + 1, 1):
        factorial[i] = (i * factorial[i - 1]) % mod
 
# Calculate x to the power y
# in O(log n) time
def Power(x, y):
     
    ans = 1
    while (y > 0):
         
        if (y % 2 == 1):
            ans = (ans * x) % mod
             
        x = (x * x) % mod
        y //= 2
         
    return ans
 
# Function to find inverse mod
# of a number x
def invmod(x):
     
    return Power(x, mod - 2)
 
# Calculate (n C r)
def nCr(n, r):
    return ((factorial[n] * invmod((factorial[r] *
                                    factorial[n - r]) %
                                    mod)) % mod)
 
def CountWays(n, k):
     
    StoreFactorials(n)
     
    # Loop to compute the formula
    # evaluated
    ans = 0
    i = k
     
    while(i >= 0):
        if (i % 2 == k % 2):
             
            # Add even power terms
            ans = ((ans + (Power(i, n) *
                           nCr(k, i)) % mod) % mod)
        else:
             
            # Subtract odd power terms
            ans = ((ans + mod - (Power(i, n) *
                                 nCr(k, i)) %
                                 mod) % mod)
        i -= 1
         
    # Choose the k boxes which
    # were used
    ans = (ans * nCr(n, k)) % mod
     
    return ans
 
# Driver code
if __name__ == '__main__':
     
    N = 5
    K = 5
     
    print(CountWays(N, K))
 
# This code is contributed by Surendra_Gangwar

C#




// C# program to calculate the
// above formula    
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG{        
     
static long mod = 1000000007;
 
// To store the factorials
// of all numbers
static long []factorial = new long[100005];
 
// Function to calculate factorial
// of all numbers
static void StoreFactorials(int n)
{
    factorial[0] = 1;
     
    for(int i = 1; i <= n; i++)
    {
        factorial[i] = (i *
        factorial[i - 1]) % mod;
    }
}
 
// Calculate x to the power y
// in O(log n) time
static long Power(long x, long y)
{
    long ans = 1;
     
    while (y > 0)
    {
        if (y % 2 == 1)
        {
            ans = (ans * x) % mod;
        }
        x = (x * x) % mod;
        y /= 2;
    }
    return ans;
}
 
// Function to find inverse mod of
// a number x
static long invmod(long x)
{
    return Power(x, mod - 2);
}
 
// Calculate (n C r)
static long nCr(int n, int r)
{
    return (factorial[n] *
    invmod((factorial[r] *
            factorial[n - r]) % mod)) % mod;
}
 
static long CountWays(int n,int k)
{
    StoreFactorials(n);
 
    // Loop to compute the formula
    // evaluated
    long ans = 0;
    for(int i = k; i >= 0; i--)
    {
        if (i % 2 == k % 2)
        {
             
            // Add even power terms
            ans = (ans + (Power(i, n) *
                  nCr(k, i)) % mod) % mod;
        }
        else
        {
             
            // Subtract odd power terms
            ans = (ans + mod - (Power(i, n) *
                  nCr(k, i)) % mod) % mod;
        }
    }
     
    // Choose the k boxes which
    // were used
    ans = (ans * nCr(n, k)) % mod;
 
    return ans;
}    
         
// Driver Code        
public static void Main (string[] args)
{        
    int N = 5;
    int K = 5;
     
    Console.Write(CountWays(N, K) + "\n");
}        
}
 
// This code is contributed by rutvik_56
Output: 
120




 

Time complexity: O(N*log N)
 

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