0
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.