# 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. – Duncan Jul 28 '14 at 14:30
• @Duncan Edited the question. – Vishwa Iyer 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 $\binom{3}{2}$). – angryavian 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. – Vishwa Iyer 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? – Vishwa Iyer Jul 28 '14 at 14:52

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 $\binom{3}{1}\binom{5}{1}\binom{4}{2}$.
• 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. – André Nicolas Jul 28 '14 at 14:45