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copied!<p>This sounds like homework (if it is, please tag it as such), so I'll only give a few hints:</p> <ol> <li><code>HashSet</code> should do the job. However, note that in order for <code>HashSet</code> to work like a mathematical set (i.e. it cannot contain multiple occurrences of the same element), the element type (in this case, <code>Set<Integer></code>) must have <code>equals()</code> and <code>hashCode()</code> implementations that tell the <code>HashSet</code> when two elements are equal. Most <code>Set</code> implementations do not have these methods implemented the way you'd expect, but you probably don't need it here. Just remember to not add the same set several times.</li> <li><code>new HashSet<Set<Integer>>()</code> <em>is</em> an empty set. If you add a set to it, it is no longer empty.</li> <li>A set can be considered to be a binary number. If your range is 2-5, the elements you can select from are <code>{2, 3, 4, 5}</code>. Any subset of this set can be represented as a binary number with four digits, each digit corresponding to an element in the main set, and being 1 if the element is in the subset, and 0 if it is not. So the subset <code>{3, 5}</code> is <code>0101</code> (2 is absent, 3 is present, 4 is absent, 5 is present), and <code>{2, 3, 4}</code> is <code>1110</code>. So if you can find a way of creating all binary numbers with the correct number of digits, and how to create a subset based on each number, you have your algorithm.</li> </ol>

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