Suppose 10 different red balls and 15 identical black balls will be distributed in 4 ballot boxes. Each urn must contain at least one red ball and the fourth urn must have at least 6 black balls. What is the total of ways in which can you make the distribution?
Attempt: I think it's complicate, so far I have: After selecting 1 red for each box, I can distribute the remaining 6 balls in any of the boxes, but I'm not sure how. In the case of the equal black balls, the limit is 6 for the fourth box, but can be 7, 8 ..., 15. How to cover all cases?
Determine the number of way to distribute the distinguishable red balls:
The Stirling Numbers of the Second Kind represent the number of ways to partition sets into nonempty partitions. Since the order of the partitions matters, the number of ways to distribute the red balls is given by:
$$\left\{\begin{array}{c}10 \\ 4\end{array}\right\}4!=818520$$
Multiply it by the number of ways to distribute the black balls. Assume $6$ balls are in the last urn. Then you have to distribute $9$ balls to $4$ urns:
$$\dbinom{9+4-1}{4-1} = \dbinom{12}{3} = 220$$
Total: $180074400$