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  1. PO
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    <p>I think it is best to handle this with math.</p> <pre><code>$range = array('a','b','c'); $length = 3; // looking for 3 digit results $minRep = 2; // with &gt;=2 consecutive characters $rangeLength = count($range); $count = (pow($rangeLength,$length-$minRep+1) * ($length-$minRep+1)) - ($rangeLength * ($length-$minRep)); // is the result </code></pre> <p>Now, $count is getting true result for three situation. But it may not be general formula and need to improve.</p> <p>Try to explain it:</p> <pre><code>pow($rangeLength,$length-$minRep+1) </code></pre> <p>in this, we count repetitive characters like as one. For instance, in second example that you gave, we think in aab, aa is a one character. Because, two characters need to change together. We think now there is two character like xy. So there is same possibilities for both character a, b, and c namely 3 ($rangeLength) possible value for two characters($length-$minRep+1). So 3^2=9 is possible situations for second example.</p> <p>We calculate 9 is for just xy not yx. For this, we multiply length of xy ($length-$minRep+1). And then we have 18.</p> <p>It can be seemed that we calculated the result, but there is a repeat in our calculation. We didn't reckon with this situation: xy => aaa and yx => aaa. For this, we calculate and substract repeated results</p> <pre><code>- ($rangeLength * ($length-$minRep)) </code></pre> <p>So after this, we get result. As i said begining of the description, this formula may need to improve.</p>
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    1. POCounting possibilities in a char. combination using a consecutive repetition criterion
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    1. COThanks for the explanation. I'm thinking on it - whether it needs a tweak or not.
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    2. CODoesn't work. For `$range = array('a', 'b'), $length = 5, $minRep = 2` it returns 58, which is greater than the size of the set defined by `$range` and `$length` (which is 32).
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    3. CO@Amine Yes, you are right. There is a problem with calculation of repeated results.
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