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copied!<p>Looks like a good problem for recursion.</p> <p>Define a function <code>f(prefix, iMin, iMax, n)</code>, that prints all combinations of <code>n</code> digits in the range [<code>iMin</code>, <code>iMax</code>] and returns the total number of combinations. For <code>n</code> = 1, it should print every digit from <code>iMin</code> to <code>iMax</code> and return <code>iMax - iMin + 1</code>.</p> <p>For your <code>iRange = 5</code> and <code>n = 3</code> case, you call <code>f("", 1, 5, 3)</code>. The output should be <code>123 - 124 - 125 - 134 - 135 - 145 - 234 - 235 - 245 - 345</code>.</p> <p>Notice that the first group of outputs are simply <code>1</code> prefixed onto the outputs of <code>f("", 2, 5, 2)</code>, i.e. <code>f("1", 2, 5, 2)</code>, followed by <code>f("2", 3, 5, 2)</code> and <code>f("3", 4, 5, 2)</code>. See how you would do that with a loop. Between this, the case for <code>n</code> = 1 above, and traps for bad inputs (best if they print nothing and return 0, it should simplify your loop), you should be able to write <code>f()</code>.</p> <p>I'm stopping short because this looks like a homework assignment. Is this enough to get you started?</p> <p>EDIT: Just for giggles, I wrote a Python version. Python has an easier time throwing around sets and lists of things and staying legible.</p> <pre><code>#!/usr/bin/env python def Combos(items, n): if n <= 0 or len(items) == 0: return [] if n == 1: return [[x] for x in items] result = [] for k in range(len(items) - n + 1): for s in Combos(items[k+1:], n - 1): result.append([items[k]] + s) return result comb = Combos([str(x) for x in range(1, 6)], 3) print len(comb), " - ".join(["".join(c) for c in comb]) </code></pre> <p>Note that <code>Combos()</code> doesn't care about the types of the items in the <code>items</code> list.</p>

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