I am having a bit of trouble understanding which combinatorial methods to use for this problem. I've actually resorted to listing some of these scenarios out (brute force) to get my solution. I would like some advice on counting methods (and possibly references to texts on counting methods) in order to solve problems like this WITHOUT using brute force.
Say I have 6 similar balls and 9 distinguishable boxes labeled A - I. I want to distribute my balls as follows:
- box A gets 1 ball
- box B gets 2 balls
- the remaining three balls can be distributed any way amongst the leftover boxes
There are 6 ways I can give box A a ball. There are $_{5}C_{2}=10$ ways for box B to get two balls. Now I have 3 balls left and 7 boxes (C through I) left. I can put all three balls in box C or box D, etc in 7 different ways. I can chose to put one ball in one box and two balls in one box in 42 (and I literally listed each way on paper to figure this one out). Finally, I can place 1 ball in each of three boxes in $_{7}C_{3}$ ways.
This leads me to my solution of: $6 \cdot 10 \cdot 35 + 6 \cdot 10 \cdot 42 + 6 \cdot 10 \cdot 7 = 5,040$ ways for box A to get 1 ball, box B to get 2 balls and the remaining C, D, E, F, G, H, I boxes to get the remaining three balls in any way.
My questions:
(1) Am I correct that there are 42 ways for 3 balls to be split amongst 7 boxes if one box gets one ball and another box (any box) gets two balls? If so, how can I enumerate this WITHOUT using brute force?
(2) Is it correct that I added three scenarios? That is, A getting 1 ball, B getting 2 balls, and one of the remaining 7 boxes getting all three can be done (6)(10)(7) ways. I added this amount to the other two possible outcomes for the remaining 7 boxes.