0
$\begingroup$

my math teacher wrote a problem today:

In how many ways can you put 9 similar balls into 3 bins stacked on top of each other, so that the top bin will have at least 4 balls.

His answer was: let's put 4 balls into the top bin. then we're left with 5 balls into 3 bins, without order being important, and repetition (putting into the same bin) is allowed. Therefore the answer is $5+3-1 \choose 3-1$ = $7 \choose 2$ = 21

What I don't understand is why is it $5+3-1 \choose 3-1$ and not $5+3-1 \choose 3$? we're choosing 3 bins, not 2.

$\endgroup$
  • $\begingroup$ I am confused: Don't you have $4$ bins in the first sentence? $\endgroup$ – zoli Dec 18 '16 at 10:57
  • $\begingroup$ @zoli There are 3 bins, the top bin have at least 4 balls $\endgroup$ – blz Dec 18 '16 at 10:59
  • $\begingroup$ Is there any problem with my answer? $\endgroup$ – zoli Dec 21 '16 at 22:31
2
$\begingroup$

OK. So we have three bins and five balls. (We can forget about the first four balls in the top bin.)

So the question is: How many ways there are to arrange five indistinguishable balls in three distinguishable bins?

The answer is

$$\binom{7}{2},$$

because we have $5+2$ abstract objects ($2$ walls between bins and $5$ balls) and we have to select two of the $7$ objects to play the role of the walls.

So, to answer your question explicitly: Three bins are separated by $2$ (and not $3$) walls [floors]. $\color{red}{\text{That is, we are choosing two separators and not three bins.}}$

$\endgroup$

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.