In how many ways can 2 black, 3 white, and 4 red balls be arranged in a row so that a black ball does not end up next to a white ball?
First, we place the 2 black and 4 red balls in every possible order. The number of winning groups is $P_{6}^{2,4}=$ $\frac{6!}{2!4!}=15$. Among these, there are groups where the 2 black balls are next to each other. The number of such groups can be obtained by considering the two black balls as one: $P_{5}^{4}=\frac{5!}{4!}=5$. Therefore, in the other 10 groups, the black balls are not next to each other.

Let's first consider the 5 groups where the 2 black balls are adjacent. In each of these 6-element groups, out of the 7 possible positions, 3 positions (between the two black balls and to their right and left) are not available for the white balls. Therefore, the white balls can be placed in 4 positions (I, II, III, IV) in each case. The 3rd class repeated combinations of these I, II, III, IV elements, where one position can hold 2 or 3 balls, represent a possible arrangement for the 3 white balls, and vice versa, any valid arrangement of the 3 white balls can be characterized by one of these 3rd class repeated combinations. Thus, the number of all possible placements of the white balls in each group is $C_{4}^{i, 3}$, and from 5 groups, $5 \cdot C_{4}^{i, 3}=5 \cdot \frac{4 \cdot 5 \cdot 6}{1 \cdot 2 \cdot 3}=5 \cdot 20=100$ groups that meet our condition can be formed.

In the other 10 groups of 6 elements (where the 2 black balls are not adjacent), out of the 7 positions, 4 positions (to the right and left of each black ball) are not available. Therefore, 3 positions remain for the white balls: I, II, III. The number of 3rd class repeated combinations of these 3 elements is $C_{3}^{i, 3}$, and thus from these 10 groups, a total of $10 \cdot C^{i, 3}_{3}=10 \cdot \frac{3 \cdot 4 \cdot 5}{1 \cdot 2 \cdot 3}=10 \cdot 10=100$ groups that meet our condition can be formed.

Therefore, according to this, $100+100=200$ placements are possible that satisfy our condition.