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    <p>You don't need recursion, or heavily nested loops, or even to generate/store the whole array of permutations in memory.</p> <p>Since the number of permutations is the product of the lengths of each of the arrays (call this <code>numPerms</code>), you can create a function <code>getPermutation(n)</code> that returns a unique permutation between index <code>0</code> and <code>numPerms - 1</code> by calculating the indices it needs to retrieve its characters from, based on <code>n</code>.</p> <p>How is this done? If you think of creating permutations on arrays each containing: [0, 1, 2, ... 9] it's very simple... the 245th permutation (n=245) is "245", rather intuitively, or:</p> <pre><code>arrayHundreds[Math.floor(n / 100) % 10] + arrayTens[Math.floor(n / 10) % 10] + arrayOnes[Math.floor(n / 1) % 10] </code></pre> <p>The complication in your problem is that array sizes differ. We can work around this by replacing the <code>n/100</code>, <code>n/10</code>, etc... with other divisors. We can easily pre-calculate an array of divisors for this purpose. In the above example, the divisor of 100 was equal to <code>arrayTens.length * arrayOnes.length</code>. Therefore we can calculate the divisor for a given array to be the product of the lengths of the remaining arrays. The very last array always has a divisor of 1. Also, instead of modding by 10, we mod by the length of the current array.</p> <p>Example code is below:</p> <pre><code>var allArrays = [first, second, third, ...]; // Pre-calculate divisors var divisors = []; for (var i = allArrays.length - 1; i &gt;= 0; i--) { divisors[i] = divisors[i + 1] ? divisors[i + 1] * allArrays[i + 1].length : 1; } function getPermutation(n) { var result = "", curArray; for (var i = 0; i &lt; allArrays.length; i++) { curArray = allArrays[i]; result += curArray[Math.floor(n / divisors[i]) % curArray.length]; } return result; } </code></pre>
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    1. POFinding All Combinations of JavaScript array values
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    1. USDavid Tang
    1. USDavid Tang
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    1. COVery nice. There is a typo here though, `results` should show `result` -- I notice you loop backwards for calculating the divisors, I assume the position of the divisor in the array is important?
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      1. USGary Green
    2. CO@Gary, thanks for picking that up. The order of the divisors matters because the first one depends on the second, the second depends on the third, etc... So by looping backwards I can build this up more easily.
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    3. CO@Box9: Does this function works with 1 array? Isn't it (n*n) - (n-1)?
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