Count Numbers with Unique Digits

Given a non-negative integer n, count all numbers with unique digits, x, where 0 ≤ x < 10n.

Example:
Given n = 2, return 91. (The answer should be the total numbers in the range of 0 ≤ x < 100, excluding [11,22,33,44,55,66,77,88,99])

 public class Solution {
     public int countNumbersWithUniqueDigits(int n) {

         if (n < ) return ;
         if (n == ) return ;  // 0 - 10
         int count = countNumbersWithUniqueDigits(n - ); // 1 到 n-1 位数。
         int result = ;  // there are 9 choices in the first number. 9 in the second number, 8 in the third, 7 in the fourth number.......
         for (int i = ; i < n - ; i++) {  // n位数
             result = result * ( - i);
         }
         return count + result;
     }
 }