Let us break this problem into steps and solve each part systematically:

Given Information:

• 6 black balls, 2 white balls, and 4 red balls.
• Total balls = $6 + 2 + 4 = 12$.

We are drawing 3 balls simultaneously.

a) How many total draws are possible?

The total number of ways to choose 3 balls from 12 is given by the combination formula: $\binom{n}{k} = \frac{n!}{k!(n-k)!}$

Here $n = 12$ and $k = 3$: $\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$

Thus, 220 total draws are possible.

b) How many draws consist of balls of three different colors?

To have balls of three different colors, we must select:

• 1 black ball,
• 1 white ball,
• 1 red ball.

The number of ways to select 1 ball of each color is: $\binom{6}{1} \times \binom{2}{1} \times \binom{4}{1}$

Calculating each term: $\binom{6}{1} = 6, \quad \binom{2}{1} = 2, \quad \binom{4}{1} = 4$

So: $6 \times 2 \times 4 = 48$

Thus, 48 draws consist of balls of three different colors.

c) How many draws consist of balls of the same color?

To have all 3 balls of the same color, they must all be:

1. Black balls, or
2. White balls, or
3. Red balls.

The number of ways to choose 3 balls of the same color is:

• For black balls: $\binom{6}{3}$,
• For white balls: $\binom{2}{3}$ (not possible as there are only 2 white balls),
• For red balls: $\binom{4}{3}$.

Calculating each term: $\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20, \quad \binom{2}{3} = 0, \quad \binom{4}{3} = \frac{4 \times 3 \times 2}{3 \times 2 \times 1} = 4$

Adding these: $20 + 0 + 4 = 24$

Thus, 24 draws consist of balls of the same color.

Final Answers:

a) Total draws possible: 220
b) Draws with three different colors: 48
c) Draws with balls of the same color: 24

Would you like a detailed explanation of any step or further assistance?

5 Related Questions:

1. How would the results change if we drew 4 balls instead of 3?
2. What is the probability of getting exactly two red balls in a draw of 3?
3. How many ways can you draw 2 balls of the same color and 1 of a different color?
4. What is the probability of drawing no black balls in a single draw of 3?
5. How many different distributions are possible if 4 balls are drawn at once?

Tip:

When solving combinatorics problems, always verify whether repetition is allowed (permutations vs. combinations) to avoid calculation errors.