Do you know the formula for the number of ways to put K identical objects into N different cells: Combin(K+N-1,K)=Combin(K+N-1,N-1).Originally Posted by dcapdogg
If we were to select any twenty balls from the piles without restriction, the number of possible selections is the number of non-negative integer solutions to R+G+B=20.
That how many ways can we put twenty ones into three difference places? For example: R=6, G=9 & B=5; R=0, G=7 & B=13. From the above the answer is Combin(20+3-1,20)=Combin(22,2).
Now, how can we answer your question? We need R to be at least 2, G to be at least 3 and B to be at least 4. So using the 20, just go ahead and put 2 into the R cell, 3 into the G cell and 4 into the B cell. Thus we have 11 left to put into any of the three cells. How many ways can that be done?