c - find all the possible combination of coins, implementing by recursion -

solve problem recursion:

using 3 type coins include 1 yuan, 2 yuan , 5 yuan, plus 10 yuan, how many combinations?

the following code :

int coinnum(int num){    if(num>=0){        if(num==0)            return 1;        else             return coinnum(num-5)+coinnum(num-2)+coinnum(num-1);    }else        return 0;  }  int main(){    int num=coinnum(10);    printf("%d\n",num);//the result 128    system("pause");    return 0; } 

what's error of recursion algorithm or what's right code ?
question supplement :
1. (5,2,2,1) , (2,5,2,1) should counted 1 combination .
2. following code of enumeration algorithm .

void coin(){   int i,j,k,count=0;   for(i=0;i<=10;i++)     for(j=0;j<=5;j++)         for(k=0;k<=2;k++)             if((i+2*j+5*k)==10){                 count++;                 printf("one yuan :%d，2 yuan :%d，5 yuan :%d\n",i,j,k);             }     printf("总方法数%d\n",count);//the result 10  } 

your code counting number of permutations add 10. want combinations. means (5,2,2,1) , (2,5,2,1) should counted 1 combination.

in case, answer should 10: (5,5), (5,2,2,1), (5,2,1,1,1), (5,1,..1), (2,2,2,2,2), (2,2,2,2,1,1), (2,2,2,1,1,1,1), (2,2,1,..1), (2,1,..1), , (1,..1).

try code:

int coinnum(int num, int *coins){   if (num == 0) return 1;   if (num < 0 || !*coins) return 0;   return coinnum(num - *coins, coins) + coinnum(num, coins+1); }  int main(){   int coins[] = {5,2,1,0}; // don't forget 0 or program won't end    int num=coinnum(10,coins);   printf("%d\n",num); // result 10   system("pause");   return 0; } 

the code above tries combinations until sum equals or exceeds desired sum. note not efficient algorithm solve problem, simple one. better algorithms, should @ computer science stack exchange.