<p>You have <code>n</code>&nbsp;&nbsp;<code>tiles</code>, where each tile has one letter <code>tiles[i]</code> printed on it.</p>

<p>Return <em>the number of possible non-empty sequences of letters</em> you can make using the letters printed on those <code>tiles</code>.</p>

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

<pre>
<strong>Input:</strong> tiles = &quot;AAB&quot;
<strong>Output:</strong> 8
<strong>Explanation: </strong>The possible sequences are &quot;A&quot;, &quot;B&quot;, &quot;AA&quot;, &quot;AB&quot;, &quot;BA&quot;, &quot;AAB&quot;, &quot;ABA&quot;, &quot;BAA&quot;.
</pre>

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

<pre>
<strong>Input:</strong> tiles = &quot;AAABBC&quot;
<strong>Output:</strong> 188
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> tiles = &quot;V&quot;
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>1 &lt;= tiles.length &lt;= 7</code></li>
	<li><code>tiles</code> consists of uppercase English letters.</li>
</ul>

class Solution:
    def numTilePossibilities(self, tiles: str) -> int:
        self.ans = 0
        def permutation(cur):
            count = Counter(cur)
            res = factorial(len(cur))
            for k, v in count.items():
                res /= factorial(v)
            return int(res)

        seen = set()
        tiles = sorted(tiles)
        def backtracking(cur_idx, cur_string):
            for i in range(cur_idx, len(tiles)):
                cur_string.append(tiles[i])
                tmp_string = ''.join(cur_string)
                if tmp_string not in seen:
                    self.ans += permutation(tmp_string)
                    seen.add(tmp_string)
                backtracking(i+1, cur_string.copy())
                cur_string.pop()

        backtracking(0, [])
        return self.ans
16700