Problem:
Balls and Boxes 2

Balls

Boxes

Any way

At most one ball

At least one ball

Distinguishable

Distinguishable

1

2

3

Indistinguishable

Distinguishable

4

5

6

Distinguishable

Indistinguishable

7

8

9

Indistinguishable

Indistinguishable

10

11

12

Problem

You have $n$ balls and $k$ boxes. You want to put these balls into the boxes.

Find the number of ways to put the balls under the following conditions:

Each ball is distinguished from the other.

Each box is distinguished from the other.

Each ball can go into only one box and no one remains outside of the boxes.

Each box can contain at most one ball.

Note that you must print this count modulo $10^9+7$.

Input

$n$ $k$

The first line will contain two integers $n$ and $k$.

Output

Print the number of ways modulo $10^9+7$ in a line.

Constraints

$1 \le n \le 1000$

$1 \le k \le 1000$

Sample Input 1

2 3

Sample Output 1

6

Sample Input 2

3 2

Sample Output 2

0

Sample Input 3

100 100

Sample Output 3

437918130

Solution in C# (judge verdict: Accepted)

using System;
using System.Linq;//リストの使用
using System.Collections.Generic;
class Program
{
	static void Main()
	{
    long mod = 1000000007;
    string[] input = Console.ReadLine().Split(' ');//Splitで区切り文字を指定して複数個受け取る。
		long n = long.Parse(input[0]);
		long k = long.Parse(input[1]);
	  if(k >= n) 
    {
      long answer = 1;
      for(long i = 2; i <= k; i++)
      {
        answer *= i;
        answer %= mod;
      }
      answer *= DivideModFactorial(k-n, mod);
      answer %= mod;
      Console.WriteLine(answer);
    }
    else Console.WriteLine(0);
	}

  static long DivideMod(long x, long a, long p)//戻り値はx^a(mod p)
  {
    long num = 2;
    long answer = 1;
    long check = a;
    long memo = x%p;
    
    while(check > 0)
    {
      if(check % num == num / 2)
      {
        check -= num / 2;
        answer *= memo;
        answer %= p;
      }
    memo *= memo;
    memo %= p;
    num *= 2;
    }
    return answer;
  }

  static long DivideModReverse(long x, long p)//戻り値はx^-1(mod p), pは素数
  {
    long answer = DivideMod(x, p-2, p);
    return answer;
  }

  static long DivideModFactorial(long x, long p)//戻り値はx!^-1(mod p), pは素数
  {
    long answer = 1;
    for(long i = x; i >= 2; i--)
    {
      answer *= DivideModReverse(i, p);
      answer %= p;
    }
    return answer;
  }

  static long Comb(long a, long b, long p)//戻り値は組み合わせC(a,b)のmod p
  {
    long answer = 1;
    for(long i = a; i > a-b; i--)
    {
      answer *= i;
      answer %= p;
    }
    answer *= DivideModFactorial(b, p);
    answer %= p;
    return answer;
  }
  
}