<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 subseq(int index,string &tiles,int n,string temp,unordered_set<string> &res){
        if(index==n){
            res.insert(temp);
            return;
        }
        subseq(index+1,tiles,n,temp+tiles[index],res);
        subseq(index+1,tiles,n,temp,res);
    }
    int fact(int num){
        int res=1;
        for(int i=1;i<=num;i++){
            res=res*i;
        }
        return res;
    }
    int numTilePossibilities(string tiles) {
        int n=tiles.length();
        sort(tiles.begin(),tiles.end());
        unordered_set<string> res;
        subseq(0,tiles,n,"",res);
        int count=0;
        for(auto r:res){
            unordered_map<char,int> ump;
            for(auto r1:r)
               ump[r1]++;
            int total=r.length();
            int f=1;
            for(auto m:ump){
                f*=fact(m.second);}
            count+=fact(total)/f; 
        }
        

          
        
        return count-1;

    }
};
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