Rectifying my mistake - Combinatorics

Question

A basket has $50$ black balls, $50$ white balls and $50$ red balls. All the balls of the same color are identical to each other. In how many ways is it possible to pick $40$ balls out of the basket (without importance to order), s.t at least one red ball, one black ball and one white ball are picked?

My Answer is: $\dbinom{150}{40}-3\dbinom{100}{40}+\dbinom{3}{2}\dbinom{50}{40}$ Basically, the logic behind this answer is that I used the inclusion-exclusion theorem, where I excluded cases when no red ball or white ball or black ball is chosen and when no black and red balls are chosen and similarly the other pairs.

But the correct answer is: $\dbinom{39}{37}$. Can you please tell me where I was wrong?

Answer 1

You can use the Stars and Bars principle. If $x,y,z$ are the number of black,white and red balls you pick you need to find how many positive solutions the equation $x+y+z = 40$ has. It's exactly

$$\binom{40-1}{3-1} = \binom{39}{2} = \binom{39}{37}$$

The reason you answer is wrong is that by using the exclusion-inclusion formula we're considering the balls as distinguishable, while the question clearly states they are identical.

Answer 2

$\binom{150}{40}-\binom 3 2\binom{100}{40}+\binom 3 1\binom{50}{40}$ counts the distinct ways to select at least one ball from each category when selecting 40 balls (in total) from three categories each of 50 distinctive balls.

$\binom{39}{37}$ counts the distinct ways to select 37 otherwise indistinguishable balls from three categories, plus one ball of each category to ensure there is at least one from each selected.

The problem states the balls of each colour are identical, hence the second answer is the appropriate one.