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    <p>Let <code>N = cardsNumber</code>, <code>M = required[0] * required[1] * ... * required[maxColor]</code>. Then, actually, your problem could be easily solved in O(N * M) time. In your very case, that is <code>12 * 2 * 3 * 4 = 288</code> operations. :)</p> <p>One of possible ways to do this is to use a recurrence relation. Consider a function <code>logColours f(n, required)</code>. Let <code>n</code> be the current number of already considered cards; <code>required</code> is a vector from your example. Function returns the answer in a vector <code>logColours</code>. You are interested in <code>f(12, {2,3,4})</code>. Brief recurrent calculation inside a function <code>f</code> could be written like this:</p> <pre><code>std::vector&lt;int&gt; f(int n, std::vector&lt;int&gt; require) { if (cache[n].count(require)) { // we have already calculated function with same arguments, do not recalculate it again return cache[n][require]; } std::vector&lt;int&gt; logColours(maxColor, 0); // maxColor = 3 in your example for (int putColor=0; putColor&lt;maxColor; ++putColor) { if (/* there is still at least one card with color 'putColor'*/) { // put a card of color 'putColor' on place 'n' if (require[putColor] == 1) { // means we've reached needed amount of cards of color 'putColor' ++logColours[putColor]; } else { --require[putColor]; std::vector&lt;int&gt; logColoursRec = f(n+1, require); ++require[putColor]; // merge child array into your own. for (int i=0; i&lt;maxColor; ++i) logColours[i] += logColoursRec[i]; } } } // store logColours in a cache corresponding to this function arguments cache[n][required] = std::move(logColours); return cache[n][required]; } </code></pre> <p>Cache could be implemented as an <code>std::unordered_map&lt;int, std::unordered_map&lt;std::vector&lt;int&gt;, std::vector&lt;int&gt;&gt;&gt;</code>.</p> <p>Once you understand the main idea, you'll be able to implement it in even more efficient code.</p>
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    1. COhmm.. I don't understand the idea, but I can see that you don't care about the cards that I have. How can this work if you don't know the cards? in my example I have 4 cards of each colour, and 3 colours total, so 12 cards. I edited my question to indicate that..
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    2. COI do care. There is an `if` inside an `f` function, which checks for the number of cards of needed color left. I thought you'd be able to fill that code yourself.
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    3. COPlease, have a look at the complete implementation: http://ideone.com/9ibXaa
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