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    <p>You're going to have to be more specific on exactly WHAT you want your function to get. There are many different definitions of "combinations" and you haven't specified whether you want ordered or unordered combinations.</p> <p>Mathematically, if you have n elements and want a LIST of k of them (ordered with repeats), that gives you</p> <pre><code>n ^ k </code></pre> <p>combinations. (6 ^ 4 = 1296 combinations in your original example, which is a lot!). However, if you have n elements and want a MULTISET of k of them (unordered with repeats), that gives you</p> <pre><code>(n + k - 1)! / (k! * (n - 1)!) </code></pre> <p>combinations and is a much harder enumeration.</p> <p>If k is small, you can generate the first one with a limited number of for loops but this becomes cumbersome very quickly as k grows. This strongly hints at the need for a RECURSIVE method:</p> <pre><code>public static String[] getAllLists(String[] elements, int lengthOfList) { //initialize our returned list with the number of elements calculated above String[] allLists = new String[(int)Math.pow(elements.length, lengthOfList)]; //lists of length 1 are just the original elements if(lengthOfList == 1) return elements; else { //the recursion--get all lists of length 3, length 2, all the way up to 1 String[] allSublists = getAllLists(elements, lengthOfList - 1); //append the sublists to each element int arrayIndex = 0; for(int i = 0; i &lt; elements.length; i++) { for(int j = 0; j &lt; allSublists.length; j++) { //add the newly appended combination to the list allLists[arrayIndex] = elements[i] + allSublists[j]; arrayIndex++; } } return allLists; } } </code></pre> <p>Not only will this method generate all the lists, but it will enumerate them in order. That is, the output will be</p> <pre><code>aaaa aaab aaac aaa1 aaa2 aaa3 aaba aabb aabc aab1 ... 3323 333a 333b 333c 3331 3332 3333 </code></pre> <p>using your original input. It can also generate any length of words (be very careful with this! Just with words of length 8 I wound up with 1,679,616 combinations!).</p> <p>If the method confuses you (it's a recursive method, so it's a bit hard to follow) or if you want a solution to the second combination problem, feel free to ask. Also, this method is somewhat inefficient because it recalculates the combinations for all the sublists, so it's not viable for really long lists. If you really wanted efficiency you would store the already-calculated tuples in a global list.</p>
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    1. POGetting every possible permutation of a string or combination including repeated characters in Java
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    1. COThank you very very much! This solution has solved my problem! I understand the idea of recursion but I'm still trying to wrap my head around the solution:( This is fantastic though and has given me a lot to think about, I would love to ask for some help understanding the recursion and exactly how it works, I'm going to try and understand it now but I think I will be back soon haha!
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