# Counting — Placing 5 balls into 3 boxes

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 box has the remaining ball?

I approached this as you first choose which of the five balls to place in any of the three boxes, giving you $$(5 \cdot 3)$$ choices.

Then you choose which box to place the 2 out of four remaining balls, giving $$\left(\begin{array}{c} 4 \\ 2 \end{array}\right)\cdot 2$$ choices.

Finally, there is only one remaining choice left for the rest, giving a total of

$$\left( 5\cdot 3 \right)\cdot \left(\begin{array}{c} 4 \\ 2 \end{array}\right)\cdot 2$$ choices.

However, this gives the answer of $$180$$ whereas the solution to this is $$90$$.

I need to divide by 2 somewhere, but I cannot figure out why I would need to.

You need to divide by two because the following two procedures have the same effect:

• Place ball A in box X
• Place ball B,C in box Y
• Place ball D,E in box Z

and

• Place ball A in box X
• Place ball D,E in box Z
• Place ball B,C in box Y.

Your formula treats these as distinct, but they, of course, yield the same result.