Marbles


Problem Statement :


Rohit dreams he is in a shop with an infinite amount of marbles. He is allowed to select n marbles. There are marbles of k different colors. From each color there are also infinitely many marbles. Rohit wants to have at least one marble of each color, but still there are a lot of possibilities for his selection. In his effort to make a decision he wakes up. Now he asks you how many possibilities for his selection he would have had. Assume that marbles of equal color can't be distinguished, and the order of the marbles is irrelevant.

Input

The first line of input contains a number T <= 100 that indicates the number of test cases to follow. Each test case consists of one line containing n and k, where n is the number of marbles Rohit selects and k is the number of different colors of the marbles. You can assume that 1<=k<=n<=1000000.

Output

For each test case print the number of possibilities that Rohit would have had. You can assume that this number fits into a signed 64 bit integer.

Example

Input:

2
10 10
30 7

Output:

1
475020



Solution :


                            Solution in C :

#include <stdio.h>
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		long long int n,k;scanf("%lld %lld",&n,&k);
		long long int d=n-k,sum=1;
		if(n==k) {
		sum=1;
		}
		else
		{
			for(long long int i=1;i<k;i++)
			{
				sum=sum*(d+i);
				sum=sum/i;
			}
		}
		printf("%lld\n",sum);
	}
	return 0;
}
                        

                        Solution in C++ :

#include<bits/stdc++.h>
using namespace std;

#define ll long long 
#define  ld long double 
#define vl vector<ll> 
#define vi vector<int> 
#define vs vector<string> 
#define vvl vector<vl> 
#define vvvl vector<vvl> 
#define  vvs vector<vs> 
#define   vvi vector<vi> 
#define  vs vector<string> 
#define  vvs vector<vs>
using namespace std;


ll comb(ll n, ll k){
    ll ans = 1;
    if (2*k > n) k = n-k;

    for(ll i=1; i<=k; i++){
        ans *= n;
        ans /= i;
        n--;
    }

    return ans;
}

void solve(){
    ll n,k; cin>>n>>k;

    cout << comb(n-1, k-1) << endl;
}


int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll t; 
    cin>>t;
   while(t--)
        solve();
    
    return 0;
}
                    

                        Solution in Java :

/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
    public static long fact(int n)
    {
        long res=0;
        if(n==1||n==0)
        {
            res= 1;
        }
        else{
            res =  n*fact(n-1);
        }
        return res;
    }
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		int n,k,t;
		long poss;
		Scanner sc = new Scanner(System.in);
		t = sc.nextInt();
		while(t>0)
		{
		    t--;
		    n = sc.nextInt();
		    k = sc.nextInt();
		    if(n==k)
		    {
		        poss =1;
		    }
		    else 
		    {
		        int m = n-k; poss = 1;
		        for(int i=1;i<k;i++)
		        {
		            poss = poss*(m+i)/i;
		        }
		    }
		    System.out.println(poss);
		}
	}
}
                    

                        Solution in Python : 
                            
n=int(input())
for i in range(n):
    n,b=map(int,input().split())
    r1=1
    for i in range(1,b):
        r1=r1*(n-b+i)//(i)
    print(int(r1))
                    

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