## Rock and Box Question

Posted August 7th, 2006 @ 06:10pm by Erik J. Barzeski

You have three boxes. One is red, one is yellow, one is green. You also have six unique rocks. How many ways can you put the six rocks in the three boxes if you must put two rocks in each box? Explain how you arrive at your answer.

### 4 Responses to "Rock and Box Question"

Posted 08 Aug 2006 at 11:09am #The obvious answer is that you're choosing four things from a set of six (because the you don't have any choice about the last two). That would lead you to think that there should be (6*5)*(4*3)=360 possibilities. That's wrong, though, because we don't care what order the rocks were put into the boxes. A box that contains rock A and rock B is the same as a box that contains rock B and rock A.

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.

Posted 08 Aug 2006 at 11:39am #I arrived to the same answer: 90. Here's how I got it:

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.

Posted 08 Aug 2006 at 4:43pm #Olivier, I like your second answer the best. And yes, 90 is the correct answer.

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.

Posted 09 Aug 2006 at 10:02am #7th graders are 11 to 13 years old, is that right? Then yes, that seems way too tough. Two boxes and four rocks would be more appropriate to make them understand the concept of indistinct pairs, assuming the teacher helps with the reasoning.

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".