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Posts ## Rock and Box Question

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"

1. 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.

2. 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.