I would be extremely grateful for any help with this problem:
Let be numbers such that and .
Find the number of ways we can put indistinguishable balls into distinguishable boxes, so that:
-exactly three boxes remain empty,
-two boxes contain exactly balls, and that
-each of the remaining boxes has at least balls.
For example, I tried to sketch one of the possible distributions in case that and :
-the second, third and sixth boxes are empty;
-the first and fourth box contain exactly 2 balls;
-the fifth box contains three balls (which is > 2).
Now, how could this problem be solved for any as defined above? I know that we can find the number of partitions of a set containing n elements into k non-empty parts by Stirling numbers of the second kind, -- but in this problem, three of the parts must be empty! Should we first break this problem down into several disjoint cases where we first determine where would the empty boxes be, and then fill the remaining boxes with balls? There are, I believe, ways we can choose three of n distinguishable boxes to be empty, there then remain boxes in each case to be filled with balls. Of those boxes, 2 must contain exactly balls. We can choose those boxes in ways, and in those boxes we put balls. There now remain boxes to be filled with balls, but under the condition that there are at least balls in each of the remaining boxes...
Is this the right way to approach this problem? I'm not certain that the above argument is entirely correct, and would really appreciate some pointers in the right direction.