What is the number of ways in which I can select 17 balls from an urn containing 30 balls? I have an urn that contains 30 balls with 10 coloured white, 10 coloured black and the remaining coloured red. Each ball is numbered, from 1 to 10: that is I have red balls numbered 1 to 10, black balls numbered 1 to 10 and white balls numbered 1 to 10. In how many ways can I select 17 balls such that I select a minimum of 4 red balls, 4 black balls and 4 white balls?
I need the quickest method to find the above out. There are long methods of solving it where I find the number of scenarios that violate the condition, such as those scenarios with 3 red balls or no black balls; these I subtract from the total number of ways of choosing 17 balls from the urn.
But is there a quicker way to solve this question?
(EDIT: All balls are now numbered. Please accept my apologies.)
(I will say this: The above problem is not a home-work problem from school.)

There are 21 composition of 17:
$17 = 4+4+9 , 17 = 4+5+8, .... , 17=9+4+4$
However, some of have the same structure are equivalent and we may take 3 or 6 of them at a time:
$ 3  \binom{10}{4} \binom{10}{4} \binom{10}{9} + 
 3  \binom{10}{5} \binom{10}{5} \binom{10}{7}  + 
 3  \binom{10}{6} \binom{10}{6} \binom{10}{5}  + 
 6  \binom{10}{4} \binom{10}{5} \binom{10}{8}  + 
 6  \binom{10}{4} \binom{10}{6} \binom{10}{7}  $ 
