I remember solving a lot of logic problems at school when I was that age, but we were doing them during class, with help from the teacher and only half of the class headcount (about 10 kids). Nothing as difficult as this and we had chosen the scientific option.

The puzzles I preferred were the ones where there are N people living in N houses, each with different characteristics, and you had to tell who lived where using incomplete propositions such as "the doctor lives next to the house of the man who has a cat".

]]>I asked the question because I don't think that a 7th grader would know how to solve this problem. I thought I might have overlooked some glaringly obvious solution.

]]>There are 6! = 720 ways to permute six rocks. You then take the first two and put them in the first box, the next two in the second box and the last two in the third box. Now you get the same arrangement if you swap the two rocks in any of the boxes. There are 2^3 = 8 ways to swap or not the rocks in the three boxes, so the final answer is 720 / 8 = 90.

The second way I tried was to figure out that there are 15 distinct ways to make three pairs out of 6 rocks (if you arbitrarily put rock #1 in the first pair, there are 5 ways to make the first pair and then 3 ways to arrange the remaining 4 rocks into 2 pairs) and then 3! = 6 ways to permute these pairs into 3 boxes. 6 x 15 = 90.

]]>The easiest way to get to the right answer is to take the simple answer above and adjust it to remove duplicates.

There are 6*5=30 possible ways to choose two rocks in order from a set of six, and 2*1=2 ways to arrange two things, so there are (6*5)/2=15 ways to choose two rocks from a set of six irrespective of order. Continuing, we have:

((6*5)/2)*((4*3)/2)*((2*1)/2) = 15*6*1 = 90

You can put the rocks in the boxes 90 different ways.

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