0
$\begingroup$

I was given the next question: What is the number of the options to distribute $a$ distinguishable balls into $b$ distinguishable boxes leaving exactly $c$ boxes empty whereas $a,b,c$ are all natural numbers which equal to/bigger than $1.$ and $a$ => $b$ - $c$

I'd really like some hints/clues..I find it quite confusing, handling problems with distinguishable/un-distinguishable balls and boxes

2 Answers 2

2

First pick the empty boxes-how many ways to do that? Then each ball has $b-c$ places to go. Then you have to subtract the ways that leave another box empty, which will require inclusion/exclusion. Subtract the ways with $c-1$ boxes empty, but you have subtracted the ones with $c-2$ empty too may times, so add them back in, and so on.

  • 0
    @rogerl: You are right-brain glitch. Added. – 2017-01-08
  • 0
    How do I find the possible options for placing the distinct balls into the b-c distinct boxes? – 2017-01-08
  • 0
    As I said, each ball has $b-c$ places to go, and you make that choice $a$ times, so there are $(b-c)^a$ ways to place them. The problem comes in that you might leave some other boxes empty. – 2017-01-08
  • 0
    Sorry for the stupid question..but I don;t understand why it's (b−c)^a? Why their factorial product isn't the right answer? – 2017-01-08
  • 0
    Having chosen $c$ bins to be empty, there are $b-c$ you can fill. Take the first ball. You have $b-c$ places to put it. Now take the second. You again have $b-c$ places to put it. As the balls are distinguishable, putting the first in bin 1 and the second in bin 2 is different from the first in bin 2 and the second in bin 1, so the total number of ways to place them is $(b-c)^2$. The choices are completely intdependent, so you multiply. – 2017-01-08
  • 0
    Thank you so so much! I got confused for a moment – 2017-01-08
0

You can choose the empty boxes in ${b \choose c}$ ways. To place the $a$ balls into the $b-c$ non-emtpy labelled boxes, you have $ (b-c)! S_{a,b-c} $ ways, where $S_{i,j}$ is the Stirling number of the second kind. Hence the answer is

$$ {b \choose c} (b-c)! S_{a,b-c} $$