<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 {
public:
void findanswer(int& n, vector<int>f,string ans,int& count )
{
    if(ans.length()>=n)
    {
        return;
    }
    for(char i='A';i<='Z';i++)
    {
        if(f[i-'A']>0)
        {
            ans.push_back(i);
            count++;
            f[i-'A']--;
            findanswer(n,f,ans,count);
            f[i-'A']++;
            ans.pop_back();

        }
    }

}
    int numTilePossibilities(string tiles) {
        int count=0;
        vector<int>f(26);
        for(char i:tiles)
        {
            f[i-'A']++;
        }
        string ans;
        int n=tiles.length();

        findanswer(n,f,ans,count);
        return count;


    }
};
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