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?
Best Answer
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.