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?

Correct Option: Both 1 and 2

Let us analyze the two statements using the Pigeonhole Principle and combinatorics. The box contains: Black (B) = 14 Blue (Bl) = 20 Green (G) = 26 Yellow (Y) = 28 Red (R) = 38 White (W) = 54 Total balls = 14 + 20 + 26 + 28 + 38 + 54 = 180 balls. Statement 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.' To *avoid* getting a full group of *any* colour, we can draw the maximum possible number of balls for each colour without completing that colour's total count. That means we can draw: 13 black balls (out of 14), 19 blue balls (out of 20), 25 green balls (out of 26), 27 yellow balls (out of 28), 37 red balls (out of 38), and 53 white balls (out of 54). Total worst-case balls without completing any single group = 13 + 19 + 25 + 27 + 37 + 53 = 174 balls. If we draw 1 more ball (i.e., 174 + 1 = 175 balls), by the Pigeonhole Principle, at least one colour group must be completely exhausted (fully drawn). Thus, n = 175 is correct. Statement 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.' To *avoid* getting at least one ball of *each* colour, the worst-case scenario is drawing all balls of the *largest* categories except leaving the smallest category untouched. The largest category is White (54 balls). If we draw all balls except White, i.e., all Black (14) + Blue (20) + Green (26) + Yellow (28) + Red (38) = 126 balls, we still have zero white balls. To guarantee at least one white ball (and thus at least one ball of every colour), we must draw 126 + 1 = 127 balls. Wait! Let us re-read statement 2: 'must contain at least one ball of each colour is 167.' Wait, to avoid getting *all* colours, the maximum number of balls we can draw without getting at least one ball of *every* colour is by completely omitting the smallest group (Black, which has 14 balls). If we draw all balls of the other 5 groups: 20 + 26 + 28 + 38 + 54 = 166 balls, we have zero black balls. Drawing one more ball (166 + 1 = 167 balls) guarantees that we will get at least one black ball, thereby ensuring we have at least one ball of *each* of the 6 colours! Thus, m = 167 is correct. Therefore, both Statement 1 and Statement 2 are correct. Option (c) is correct.