# C Programming: Count all the numbers with unique digits in a range

## C Programming Mathematics: Exercise-30 with Solution

Write a C program that accepts a number (n) and counts all numbers with unique digits of length x within a specified range.

Range: 0 <= x < 10^{n}

Example:

When n = 1, numbers with unique digits (10) between 0 and 9 are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

When n = 2, numbers with unique digits (91) between 0 and 100 are 0,1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15 …..99 except 11, 22, 33, 44, 55, 66, 77, 88 and 99.

**Test Data:**

(1) -> 10

(2) -> 91

**Sample Solution:**

**C Code:**

```
#include <stdio.h>
int min(int a, int b){
return (a > b) ? b : a;
}
int test(int n) {
if(n == 0)
return 1;
n = min(10, n);
if(n == 1)
return 10;
int flag = 9;
int result = 10;
for(int i = 2; i<= n; i++){
flag *= (9 - i + 2);
result += flag;
}
return result;
}
int main(void) {
int n = 1;
printf("n = %d",n);
printf("\nNumbers with unique digits in the range 0, 10: %d", test(n));
n = 2;
printf("\n\nn = %d",n);
printf("\nNumbers with unique digits in the range 0, 100: %d", test(n));
n = 3;
printf("\n\nn = %d",n);
printf("\nNumbers with unique digits in the range 0, 1000: %d", test(n));
}
```

Sample Output:

n = 1 Numbers with unique digits in the range 0, 10: 10 n = 2 Numbers with unique digits in the range 0, 100: 91 n = 3 Numbers with unique digits in the range 0, 1000: 739

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