<p>Given an integer <code>n</code>, return the count of all numbers with unique digits, <code>x</code>, where <code>0 &lt;= x &lt; 10<sup>n</sup></code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> 91
<strong>Explanation:</strong> The answer should be the total numbers in the range of 0 &le; x &lt; 100, excluding 11,22,33,44,55,66,77,88,99
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 0
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>0 &lt;= n &lt;= 8</code></li>
</ul>

class Solution:
    def countNumbersWithUniqueDigits(self, n: int) -> int:
        arr = []
        def dp(n):
            if n == 0:
                return 1
            if n == 1:
                return 10
            res = 9
            j = 9
            for i in range(n-1):
                res *= j
                j -= 1
            s = self.countNumbersWithUniqueDigits(n-1)
            arr.append(res+s)
            return res + s
        a = dp(n)
        return a
            
15900