The question:
In a basket there are $20$ black balls, $15$ white balls and $18$ red balls. All balls with the same color are identical to each other. In how many ways it is possible to take out balls from the basket (when order doesn't matter), such that will take out at least one black ball, at least one white ball and at least one red ball (without using pascal triangle principles) ?
I though that the min number of balls we have to choose is $3$ (one from each color),and the max numbers of balls is $20 + 15 + 18 = 53$
So, for the min number of balls we'll have $1$ option (black, white and red), and from now we'll "glue" the subset {black, white and red} to all the other options.
For $4$ balls we have ${3 \choose 1}$ options in order to choose the ball which appears twice.
For the 5 balls we have ${3 \choose 1}$ options to choose the color of the forth ball, and ${3 \choose 1}$ and to choose the color of the fifth ball. it means ${3 \choose 1} ^ 2$
Hence, I thought to write that for every $0≤k≤49$ we have the same pattern of options, thus we I would do a sigma of this pattern. But it is not correct because there are limitations on the number of balls from a certain color (for example, imagine that we choose $15$ white and 1 from each other color, the number of options for $18$ balls rely on only $2$ colors. so, I would like to have a direction of how to approach to this question correctly.
By the way, firstly I thought to use Inclusion–exclusion principle but I saw that it became complicated because if we want to use the complement principle, we need to separate to three different cases. (because the complement of at one least red, at least one black and at least one white by De-Morgan is no black balls or no white balls or no red balls.)
I know that there is a similar question - How many ways can you select $5$ marbles if you select at least one from each color? but here the upper bounds don't exceed to their sum (50), so I am not sure this approach will help to solve the exercise.
Take out one black ball, one white ball and one red ball.
Now, there are $19$ black balls, $14$ white balls and $17$ red balls left.
As order doesn't matter for this particular problem and balls of the same color are assumed identical, the number we look for is completely determined by the number of black, white and red balls we take out. We have $20$ options for black balls: $0,1,\ldots,19$ (recall we already took one out and we may not take any more), and similarly $15$ for white balls and $18$ for red balls, so there are $$20\cdot 15\cdot 18=5400$$ ways of picking at least a black ball, a white ball and a red ball.