Problem Statement: How many 5 digit number can be formed from the integers 1,2, ..., 9 if no digit can appear more than twice? Approach: 18!/((2!*9)(18-5)!) Reason: Using the combination formula, we were able to approach this problem by saying that there are 18 numbers from which only 5 are chosen and the order does not matter for individual groups of two numbers. Is this approach correct? I would love a very detailed explanation of this problem since many of the ones in class are about this same difficulty level.