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.)
 A: 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}  $ 
A: You can select 4 red balls from 10 white balls in $10\choose4$ ways. Similarly, for white and black balls. 
So, we can select 4 red, 4 black and 4 white balls in ${{10}\choose {4}}* {{10}\choose {4}}*{{10}\choose {4}} $ ways. 
We now have $5$ balls to select, from $30-12=18$ balls. This can be done in $18\choose5$ ways.
The answer thus becomes ${{10}\choose {4}}* {{10}\choose {4}}*{{10}\choose {4}} *{{18\choose5}}$ ways. 
A: If the balls are identical you want the number of compositions of $17$ from three parts with each part being at least $4$ and at most $10$.  In this case you can subtract $3$ from each number and ask the number of  strong compositions of $8$ from three parts of at most $7$ and we note that the upper limit is no problem.  You can use stars and bars to put the eight stars in a row and choose two of the seven gaps to change colors.  You have $7 \choose 2$ ways to pick the balls.
