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Not sure this can be done. Certainly 3 combinations can be accommodated by just putting condom #1 nested inside condom #2 and then going through the scenarios. But two couples means 4 different combinations, and I can't see how to get the 4th one. Any takers? -Nick

Alternative Solutions to A Problem Hi. This is a great forum!! The interview questions are very useful. I have a method to solve two of the problems that may have been overlooked. 1. Find x in x^(x^(x^ ... ) = 2 2. Calculate \sqrt{ 2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + ...} }}} In...