==> arithmetic/digits/equations/123456789.s <==
 1-2 3+4 5+6 7-8 9 = 1
+1-2 3+4 5+6 7-8 9 = 1
 1+2 3+4-5+6 7-8 9 = 1
+1+2 3+4-5+6 7-8 9 = 1
-1+2 3-4+5+6 7-8 9 = 1
 1+2 3-4 5-6 7+8 9 = 1
+1+2 3-4 5-6 7+8 9 = 1
 1-2 3-4+5-6 7+8 9 = 1
+1-2 3-4+5-6 7+8 9 = 1
 1-2-3-4 5+6 7-8-9 = 1
+1-2-3-4 5+6 7-8-9 = 1
 1+2-3 4+5 6-7-8-9 = 1
+1+2-3 4+5 6-7-8-9 = 1
-1+2 3+4+5-6-7-8-9 = 1
-1 2+3 4-5-6+7-8-9 = 1
 1+2+3+4-5+6+7-8-9 = 1
+1+2+3+4-5+6+7-8-9 = 1
-1+2+3-4+5+6+7-8-9 = 1
 1-2-3+4+5+6+7-8-9 = 1
+1-2-3+4+5+6+7-8-9 = 1
 1+2 3+4 5-6 7+8-9 = 1
+1+2 3+4 5-6 7+8-9 = 1
 1+2 3-4-5-6-7+8-9 = 1
+1+2 3-4-5-6-7+8-9 = 1
 1+2+3+4+5-6-7+8-9 = 1
+1+2+3+4+5-6-7+8-9 = 1
-1+2+3+4-5+6-7+8-9 = 1
 1-2+3-4+5+6-7+8-9 = 1
+1-2+3-4+5+6-7+8-9 = 1
-1-2-3+4+5+6-7+8-9 = 1
 1-2+3+4-5-6+7+8-9 = 1
+1-2+3+4-5-6+7+8-9 = 1
 1+2-3-4+5-6+7+8-9 = 1
+1+2-3-4+5-6+7+8-9 = 1
-1-2+3-4+5-6+7+8-9 = 1
-1+2-3-4-5+6+7+8-9 = 1
-1+2 3+4 5-6 7-8+9 = 1
 1-2 3-4 5+6 7-8+9 = 1
+1-2 3-4 5+6 7-8+9 = 1
-1+2 3-4-5-6-7-8+9 = 1
-1+2+3+4+5-6-7-8+9 = 1
 1-2+3+4-5+6-7-8+9 = 1
+1-2+3+4-5+6-7-8+9 = 1
 1+2-3-4+5+6-7-8+9 = 1
+1+2-3-4+5+6-7-8+9 = 1
-1-2+3-4+5+6-7-8+9 = 1
 1+2-3+4-5-6+7-8+9 = 1
+1+2-3+4-5-6+7-8+9 = 1
-1-2+3+4-5-6+7-8+9 = 1
-1+2-3-4+5-6+7-8+9 = 1
 1-2-3-4-5+6+7-8+9 = 1
+1-2-3-4-5+6+7-8+9 = 1
 1-2 3+4+5+6+7-8+9 = 1
+1-2 3+4+5+6+7-8+9 = 1
 1+2+3+4 5-6 7+8+9 = 1
+1+2+3+4 5-6 7+8+9 = 1
 1 2+3 4+5-6 7+8+9 = 1
+1 2+3 4+5-6 7+8+9 = 1
 1+2+3-4-5-6-7+8+9 = 1
+1+2+3-4-5-6-7+8+9 = 1
-1+2-3+4-5-6-7+8+9 = 1
 1-2-3-4+5-6-7+8+9 = 1
+1-2-3-4+5-6-7+8+9 = 1
-1-2-3-4-5+6-7+8+9 = 1
-1-2 3+4+5+6-7+8+9 = 1
 1-2+3 4-5 6+7+8+9 = 1
+1-2+3 4-5 6+7+8+9 = 1
 1 2-3 4+5-6+7+8+9 = 1
+1 2-3 4+5-6+7+8+9 = 1
Total solutions  = 69 (26 of which have a leading "+", which is redundant)

69/19683 = 0.35 %

for those who care (it's not very elegant but it did the trick):

#include <stdio.h>
#include <math.h>

main (argc,argv)
     int argc;
     char *argv[];
{
  int sresult, result, operator[10],thisop;
  char buf[256],ops[3];
  int i,j,tot=0,temp;
  
  ops[0] = ' ';
  ops[1] = '-';
  ops[2] = '+';

  for (i=1; i<10; i++) operator[i] = 0;
  
  for (j=0; j < 19683; j++) {
    result = 0;
    sresult = 0;
    thisop = 1;
    for (i=1; i<10; i++) {
      switch (operator[i]) {
      case 0:
	sresult = sresult * 10 + i;
	break;
      case 1:
	result = result + sresult * thisop;
	sresult = i;
	thisop = -1;
	break;
      case 2:
	result = result + sresult * thisop;
	sresult = i;
	thisop = 1;
	break;
      }
    }
  
    result  = result + sresult * thisop;
    if (result == 1) {
      tot++;
      for  (i=1;i<10;i++) 
	printf("%c%d",ops[operator[i]],i);
      printf(" = %d\n",result);
    }
    temp = 0;
    operator[1] += 1;
    for (i=1;i<10;i++) {
      operator[i] += temp;
      if (operator[i] > 2) { operator[i] = 0; temp = 1;}
      else temp = 0;
    }
  
  }
  
  printf("Total solutions  = %d\n" , tot);
}

cwren@media.mit.edu (Christopher Wren)