52 (1138). In a box, there are multicolored balls: 5 white, 12 red, and 20 black. What is the smallest number of balls that need to be taken out of the box without looking inside, to ensure that among them there will be: a) at least one ball of each of the specified colors; b) 10 balls of one color?
Solution. a) In the worst case, 20 black and 12 red balls can be drawn, then the next ball drawn (white) will ensure the presence of balls of all the specified colors. Therefore, it is sufficient to draw 33 balls $(20+12+1=33)$ to ensure that there is at least one ball of each of the specified colors among them.

b) In the worst case, all white balls (5), 9 red, and 9 black balls can be drawn, then any ball drawn (red or black) will ensure the presence of 10 balls of one color (red or black). Therefore, it is sufficient to draw 24 balls to ensure that among the drawn balls there are 10 balls of one color.

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![](https://cdn.mathpix.com/cropped/2024_05_21_98932225530737b82071g-066.jpg?height=126&width=1205&top_left_y=248&top_left_x=448)

Fig. 16