Let's solve this step by step:

Problem Breakdown:

A box contains balls of different colors:

• 1 black ball
• 2 blue balls
• 3 green balls
• 4 yellow balls
• 5 red balls
• 6 white balls

The question involves two statements:

1. Statement 1: The smallest number $n$ such that any $n$ balls drawn randomly must contain at least one complete group of one color is 21.
2. Statement 2: The smallest number $m$ such that any $m$ balls drawn randomly must contain at least one ball of each color is 16.

Let's analyze each statement individually.

Statement 1 Analysis:

The smallest $n$ ensures that at least one full group (all balls of a single color) is drawn. To calculate this:

• The maximum number of balls that can be drawn without completing any color is the sum of the maximum possible incomplete draws:
  • Black: $0$ (can't avoid taking the 1 black ball, as it's the only one).
  • Blue: $1$ (can take at most 1 of the 2 blue balls without completing blue).
  • Green: $2$ (can take at most 2 of the 3 green balls).
  • Yellow: $3$ (can take at most 3 of the 4 yellow balls).
  • Red: $4$ (can take at most 4 of the 5 red balls).
  • White: $5$ (can take at most 5 of the 6 white balls).

Thus, the total is: $0 + 1 + 2 + 3 + 4 + 5 = 15$

To ensure one complete group, we must draw one more ball. Therefore: $n = 15 + 1 = 16$

Statement 1 is incorrect as it claims $n = 21$.

Statement 2 Analysis:

The smallest $m$ ensures that at least one ball of each color is drawn. To calculate this:

• To avoid having at least one ball of every color, we can maximize the number of balls drawn from all but one color:
  • Blue, green, yellow, red, and white: $2 + 3 + 4 + 5 + 6 = 20$.

So, if we draw 20 balls, it's possible to miss the black ball. To ensure at least one ball of every color, we need to draw one more ball. Therefore: $m = 20 + 1 = 21$

Statement 2 is incorrect as it claims $m = 16$.

Final Answer:

Neither statement 1 nor statement 2 is correct.

Correct option: (d) Neither 1 nor 2

Would you like a more detailed explanation or examples?

Related Questions:

1. How would the result change if the number of balls of each color were doubled?
2. Can this problem be solved using probability instead of combinatorics?
3. How does the pigeonhole principle apply to these types of problems?
4. What is the general formula for finding $n$ and $m$ in such problems?
5. What if the box also included additional colors or objects (e.g., shapes or sizes)?

Tip:

Always verify the logic by simulating scenarios with smaller numbers to check for boundary cases.