#include <stdio.h>  
#include <stdlib.h>  
double x = 1;  
int i = 0;  
int n, r; 

void a(){      
	if(i < r){          
    	    x *= (double) (n - i) / (double) (r - i);    
            i ++;   
            a();  
	}
}  

int main(int argc, char *argv[]) {      
    scanf("%d %d", &n, &r);      
    a();      
    printf("%.lf", x);       
    return 0; 
}

#include <stdio.h>
#include <stdlib.h>
#define Max 1000    // maximum value of 'r'
// Note that n C r = n-1 C r-1 + n-1 C r 

long long int min(long long int a, long long int b){
    if(a < b)   return a;     
    else    return b;  
}  

int main(int argc, char *argv[]) {     
    long long int n, r, i, j;      
    scanf("%lld %lld", &n, &r);    
    r = min(r, n - r);  // for optimization  
    long long int C[Max + 1][2] = { {0, }, };      
    C[0][0] = 1;     
    for(i = 1; i <= n; i ++){          
    	C[0][1] = 1;         
        for(j = 1; j <= min(i, r); j ++)        C[j][1] = C[j-1][0] + C[j][0];        
        // C[j][0] = i-1 C j  and  C[j][1] = i C j        
        for(j = 0; j <= min(i, r); j ++)        C[j][0] = C[j][1];     
    }    
    printf("%d", C[r][0]);   
    return 0; 
}

#include <stdio.h>
#include <stdlib.h>
#define p 1999  // 'p' is 'p'rime 
#define g(p) 3  // g( ): primitive root  

int main(int argc, char *argv[]) {      
	int notepad = 1, exponent_set[p], remainder_set[p];     // 0 ~ p-1 (mod p)    
    long long int n, r, i, j;    
    int exponent_I = 0; // n!  ÷ (n - r)! ≡ (g(p)) exponent_I (mod p)    
    int exponent_II = 0;     // r! ≡(g(p)) exponent_II (mod p)    
    
    for(i = 0; i < p - 1; i ++){    // (g(p)) p - 1

 ≡1 (mod p) (Fermat's little Theorem) (①)
    	exponent_set[notepad] = i;  // remainder → exponent of g(p)       
        remainder_set[i] = notepad;     // exponent of g(p) → remainder 
        notepad *= g(p);    notepad %= p;   
    }    
    scanf("%lld %lld", &n, &r);   
    notepad = 0;    
    for(i = n; i > n - r; i --){    
    	j = i;      
        while(1){   
        	if(j % p != 0)  break;   
            notepad ++; 
            j /= p;      
        }        
        exponent_I += exponent_set[j % p];   
        exponent_I %= p - 1;  // (①)        
    }    
    for(i = r; i > 0; i --){        
    	j = i;     
        while(1){   
        	if(j % p != 0)  break;      
            notepad --;    
            j /= p;            
        }          
        exponent_II += exponent_set[j % p];    
        exponent_II %= p - 1;    // (①)          
    }           
    
    // n C r ≡ (g(p)) exponent_I - exponent_II (mod p)          
    if(notepad > 0) printf("0");    // notepad is always over -1      
    else    {          
    	if(exponent_I < exponent_II)    exponent_II -= p - 1;   // (①)          
        printf("%d", remainder_set[exponent_I - exponent_II]);      
    }      
    return 0;
}