Solving Stats Problem with Physics: 3 Eggs & 5 Boxes

[GoGoGadget]

Homework Statement

You have 5 identical boxes that stand one after another. How many different ways can you put your 3 identical eggs into these boxes?


For whatever reason, my physics professor decided to give us this as a problem to think about in how to solve. However, from my understanding, this is something you would normally see in Stats. I'm not sure how you go about incorporating physics to solve this but I believe the statistical property n! could be used to solve this problem? However, upon going to try to answer it on my class website, I seem to be misunderstanding how to solve it. I know you have 120 total combinations with the boxes in shifting them in different rows. But I'm not sure on how to figure out how to properly calculate how you would rearrange the eggs? Any input would be appreciated.

 

[HallsofIvy]

I don't understand why you are concerned about "incorporating physics to solve this". Solve it however you can! I imagine that you are given this question because determininjg how many ways something can happen is important in quantum physics. And this is a relatively simple question. How many choices do you have for where to put the first egg? How many for the second egg? How many for the third? And then use the "fundamental principle of counting": "if one thing can happen in m ways and then another in n ways, the two can happen in mn ways".

Your problem allows you to put any number of eggs in a box. You would get a different answer if you were only allowed to put one egg in anyone box. In fact, since those are both important problems in quantum physics (Bose-Einstein statistics versus Fermi statistics), I wouldn't be surprised if your teacher next gave you the "only one egg per box" question.

 

[GoGoGadget]

I wasn't sure how it related to physics principles in general as I haven't taken Stats before and was just intrigued if it was at all but that would make sense with regard to quantum physics then. I've otherwise have tried multiple different calculations for it and haven't been able to get it right. I would have 5 different choices for each egg on where I can place it, a total of 15. So in adding it together for three eggs and 5 boxes, I have a total of three situations. So I'd take three eggs times 5 boxes and then find the total for 3 eggs which be 45. (5 boxes x 3 eggs = 15 arrangements x 4 possible options.) Then repeat for 2 eggs for each box and 1 egg for each box. And then adding everything together, I'd get 45 + 40 + 20 = 105 ways. But this also was incorrect so I'm still uncertain about it. I also tried taking the scientific notation of 5! to get a total 120 arrangements for the boxes. But I didn't know what to do with the different combinations for eggs. Any further input would be great.

 

[haruspex]

You're told the boxes are identical and the eggs are identical. Normally this would be taken to mean that which box is which doesn't matter, etc. But then you're told the boxes are in a line, which means they're not really identical - you do know which is which. So I find the question ambiguous.
Going with my first interpretation, all that matters is the pattern of numbers per box: how many boxes get 3 eggs, how many 2 eggs, and so on. In maths, this is the partitioning problem, and in general it's quite tough. But with only 3 eggs it's not hard to figure it out by going through the cases.
With the second interpretation, there's a neat trick. Imagine laying E eggs in a line and placing sticks between them to indicate which go in the first box, which in the second, and so forth. If there are B boxes you'll need B-1 sticks. Now think of it as just a line of E+B-1 things, of which E are eggs and B-1 are sticks. The number of arrangements is the number of ways of choosing which are eggs and which are sticks. Do you know a formula for the number of ways of choosing M things from N things?

 

[Nickie]

I would approach this problem by first understanding the physical properties involved. In this case, we have 3 identical eggs and 5 identical boxes. The eggs can be placed in any of the 5 boxes, but the order in which they are placed does not matter. Therefore, we can use the statistical property of combinations to solve this problem.

The formula for combinations is nCr = n! / (r!(n-r)!), where n is the total number of items and r is the number of items chosen. In this case, n = 5 (number of boxes) and r = 3 (number of eggs). So, the number of ways to put the eggs in the boxes is 5C3 = 5! / (3!(5-3)!) = 10.

Alternatively, we can think of this problem as a permutation, where the order matters. In this case, the formula is nPr = n! / (n-r)!. Since the eggs are identical, the order does not matter. Therefore, we need to divide the result by the number of ways the eggs can be arranged, which is 3! (permutations of 3 items). So, the number of ways to put the eggs in the boxes is 5P3 / 3! = 10.

In conclusion, using the statistical properties of combinations or permutations, we can determine that there are 10 different ways to put 3 identical eggs in 5 identical boxes. This problem may seem more suited for statistics, but understanding the physical properties involved can help us apply statistical concepts to solve it.