Can anyone solve this crazy combination problem? There are 9 blocks, 3 red, 3 yellow and 3 blue. Each set of colours is numbered 1 to 3. How many distinct sets of 4 towers can be made? The towers can have a maximum height of three blocks, and we must use all of the blocks.
Some clarifications:


*

*The blocks we have available are: R1, R2, R3, Y1, Y2, Y3, B1, B2, B3. 

*Towers are distinct if: they contain different blocks; or they contain the same blocks but in a different order. For example, R1Y1B2 is distinct from R1Y1B3 and from Y1B2R1. 

*Towers can be any height from 0 to 3

*The order of the towers matters. For example, for given towers T1, T2, T3, T4, T1T2T3T4 counts as a different set to T4T3T2T1. 


My answer was 8,709,160, but I am very likely wrong. I'd be interested to see a method, any ways to simplify the problem, and of course any answers! 
Thanks! 
My method: 
We could have the following combinations of tower heights:



*

*3,3,3,0 

*3,3,2,1 

*3,2,2,2. 


We can arrange 1. in 4 ways, 
2. in 18 ways (3x3x2)
3. in 4 ways.
Now for 1. I worked out the permutations for each tower, and multiplied by the number of ways of arranging them:
9P3 * 6P3 * 3P3 * 4 = 1,451,520
And the same for 2.:
9P3 * 6P3 * 3P2 * 1 * 18 = 6531840
And for 3.:
9P3 * 6P2 * 4P2 * 2P2 * 2 = 725760 
Adding these, I got my answer of 8,709,160. 
P.s. One error I think I've made is counting the number of ways to arrange the towers in 2. 
 A: We can arrange the blocks in a line in $9!$ ways.  For each such arrangement, we have to break it into $4$ towers, with from $0$ to $4$ blocks in a tower.  This is the number of ways to put $9$ indistinguishable balls into $4$ buckets, with no more than $3$ balls in a bucket.  One way to compute this is with a combination of stars and bars and inclusion-exclusion.
There are $\binom{12}{3}=220$ ways to put the balls in the buckets with no restrictions.  There are $4$ ways to choose a bucket in which to place $4$ balls.  Then there are $\binom{8}{3}=56$ ways to distribute the remaining $5$ balls.  But distributions with $4$ balls in two buckets have been subtracted twice, so we have to add them back in.  There are $\binom42=6$ ways to choose the two buckets, and $4$ ways to distribute the last ball, giving $$220-4\cdot56+6\cdot4=20$$ 
A: Yes, you have miscounted the number of ways to arrange the towers in (2). We have $4$ choices for the one-block tower, then $3$ for the two-block tower. The remaining towers will each have three blocks. This gives us $4\times 3=12$ ways to arrange the towers. The total number of ways to arrange the towers is $4+12+4=20$.
Now all we need to do is arrange the blocks. There are nine of them, so we have $9!=362880$ arrangements.
This gives us a total of $20\times 9!=7257600$.
