# Combinatorics Question about balls in boxes

There are 5 balls numbered 1 to 5, and there are 3 boxes numbered 1 to 3. The question asks in how many distinct ways can the balls be put into the boxes if 2 boxes have 2 balls each and the other box has the remaining ball?

My try: We choose 2 balls to go in one box, 2 to go in another, and the last one goes in the last box for a total of $${5 \choose 2} {3\choose 2}$$ ways. So one arrangement would be having balls 1 and 2 in box 1, balls 3 and 4 in box 2, and ball 5 in box 3. But we can put balls 1 and 2 in box 2 or 3, so we multiply the expression above by $3! = 6$ to get a total of 180 ways.

The solution manual states that they divide by 2 because the 2 groups of 2 balls are indistinguishable, what does mean, and why? Because from my understanding, we're using combinations, so aren't the balls considered indistinguishable? And we multiplied by 6 to account for each group of balls being in a different box. I don't really understand the reasoning behind indistinguishable, so any help is greatly appreciated.

EDIT: This is the actual question:

Five balls are numbered 1 to 5. Three boxes are numbered 1 to 3. How many distinct ways can the balls be put in the boxes if two boxes have two balls each and the other has the remaining ball?

• Can you post the exact wording to the question? By the way you've posed it above (numbered balls and numbered boxes), I would agree with your solution, but "indistinguishable" in a combinations problem generally means that they are interchangeable and you can't tell the difference, so you would have to divide. But the exact wording will help. Jul 28 '14 at 14:30
• @Duncan Edited the question. Jul 28 '14 at 14:32
• I think André's solution below is a nicer way to think of it. With regard to your question about dividing by $2$, this is because you have double-counted: if you take the example you give of $12|34|5$, note that $34|12|5$ is counted not only as one of the $3!$ permutations of the boxes, but also as choosing $3$ and $4$ first (in $\binom{5}{2}$) and then $1$ and $2$ (in $\binom{3}{2}$). Jul 28 '14 at 14:44
• Based on all of your comments, you're saying that the boxes with two balls is indistinguishable and the box with one ball isn't? Therefore, it's like having a word "AAB" and finding the distinct permutations of it. Jul 28 '14 at 14:49
• @angryavian The reason I was confused was that I thought $12|34|5$ and $34|12|5$ were different because the boxes are labeled, so it's like having the two balls in different boxes, resulting in distinct ways. Why is this wrong? Jul 28 '14 at 14:52

The balls are distinguishable, as are the boxes: they all have labels.

I would solve the problem like this. There are $\binom{3}{1}$ ways to choose the box that will have a singleton, and for each such way there are $\binom{5}{1}$ ways to choose the lonely ball.

Now we have two boxes left, a lower-numbered one and a higher-numbered one. Choose the two balls that will go into the lower-numbered one. This can be done in $\binom{4}{2}$ ways, for a total of $\binom{3}{1}\binom{5}{1}\binom{4}{2}$.

Remark: If the boxes are indistinguishable, but the balls are distinguishable, then something like what you quote is reasonable. However, both are labelled, and there is no sense as interpreting numbering as conferring distinguishability for balls, but not for boxes.

• Makes sense, but why are you not multiplying by 2? Initially you choose 1 box out of the 3, and then you have to choose either the lower-numbered box or the higher-numbered box. Are you assuming that the if balls 1 and 2 are in the lower-numbered or the higher numbered one, that's it's indistinguishable? My overall question is why do the balls have to go into the lower-numbered one? Can't they go into the higher-numbered one? Jul 28 '14 at 14:40
• I am trying to avoid mysterious divisions by $2$. We have two boxes left, say Box 1 and Box 3. As soon as we decide which of the remaining balls go into Box 1, we will be finished. Jul 28 '14 at 14:45
• Yes, but you have to choose one of the two boxes for the two balls to go in correct? Then you assume that the other two balls go into the last box. Jul 28 '14 at 14:46
• If we try to have choice of balls and choice of boxes, we will inevitably double-count. That is why I gave a specific procedure (fill the lower ranking box first), because this guarantees against double-counting. Jul 28 '14 at 14:49
• @AndréNicolas Actually the answer is $\binom{3}{1}\binom{5}{1}\binom{4}{2}\binom{2}{2}$. Isn't it?
– snr
Apr 25 at 9:13