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UPSC Prelims 2023CSATQuantitative AptitudePermutations and Combinations - Pigeonhole Principle

Q6. A box contains 14 black balls, 20 blue balls, 26 green balls, 28 yellow balls, 38 red balls and 54 white balls. Consider the following statements:
1. The smallest number n such that any n balls drawn from the box randomly must contain one full group of at least one colour is 175.
2. The smallest number m such that any m balls drawn from the box randomly must contain at least one ball of each colour is 167.

Which of the above statements is/are correct?

A. 1 only
B. 2 only
C. Both 1 and 2✓ Correct
D. Neither 1 nor 2

Detailed Solution

✓ Correct Answer: Option C

For statement 1 (n): To guarantee one full group, we consider the worst-case scenario where we pick one less than the full count for each color. This sum is (14-1) + (20-1) + (26-1) + (28-1) + (38-1) + (54-1) = 13 + 19 + 25 + 27 + 37 + 53 = 174 balls.

The next ball (175th) *must* complete a full group of some color. So, n = 175. Statement 1 is correct. For statement 2 (m): To guarantee at least one ball of each color, we consider the worst-case scenario where we pick all balls of all but one color.

To maximize this, we pick all balls of the five largest groups: 20 (blue) + 26 (green) + 28 (yellow) + 38 (red) + 54 (white) = 166 balls. The next ball (167th) *must* be a black ball, ensuring at least one of each color. So, m = 167. Statement 2 is correct.

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