Problem:
Balls and Boxes 5

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 not 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

3 5

Sample Output 1

10

Sample Input 2

5 10

Sample Output 2

252

Sample Input 3

100 200

Sample Output 3

407336795

Solution in C# (judge verdict: Accepted)

using System;
using System.Linq;//リストの使用
using System.Collections.Generic;
class Program
{
static string[] input = Console.ReadLine().Split(' ');
static long n = long.Parse(input[1]);
static long k = long.Parse(input[0]);
static long p = 1000000007;
static long[] factorials = new long[n+1];//i!(mod p)を先にメモ
static long[] factorialRs = new long[n+1];//i!^-1(mod p), pは素数

	static void Main()
	{
		factorials[0] = 1;
		factorialRs[n] = DivideModFactorial(n,p);
		for(long i = 1; i <= n; i++)
		{
			factorials[i] = (factorials[i-1]*i)%p;//i!(mod p)
			factorialRs[n-i] = (factorialRs[n+1-i]*(n+1-i))%p;//逆元も先にメモ
		}

		if(n >= k) Console.WriteLine(Comb(n, k, p));//nCk(mod p),pは素数
		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;
    answer *= factorials[a];
    answer %= p;
    answer *= factorialRs[a-b];//引数a-bは負にならないようにする
    answer %= p;
    answer *= factorialRs[b];
    answer %= p;
    return answer;
  }
  
}