Let's break down each part of the problem.

Total number of marbles:

• Red: 5
• White: 4
• Blue: 6

Total = $5 + 4 + 6 = 15$ marbles.

Now, let's solve each question step by step.

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Part (a): How many ways can three marbles be selected so that at least one is white?

First, calculate the total number of ways to select any 3 marbles from the 15 total marbles. This is a combination problem because the order of selection does not matter.

$\text{Total ways} = \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455$

Next, we calculate how many ways there are to select 3 marbles with no white ones (i.e., selecting only from the red and blue marbles). There are $5 + 6 = 11$ red and blue marbles.

$\text{Ways with no white marbles} = \binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165$

Now, subtract the number of ways with no white marbles from the total number of ways to find the number of ways that include at least one white marble.

$\text{Ways with at least one white marble} = 455 - 165 = 290$

So, the answer for part (a) is 290 ways.

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Part (b): How many ways can three marbles be selected so that all three marbles are the same color?

Here, you must consider each color separately. The number of ways to select all three marbles as the same color depends on whether there are at least 3 marbles of that color.

• Red marbles: There are 5 red marbles, so the number of ways to select 3 red marbles is:

$\binom{5}{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$

• White marbles: There are 4 white marbles, so the number of ways to select 3 white marbles is:

$\binom{4}{3} = \frac{4 \times 3 \times 2}{3 \times 2 \times 1} = 4$

• Blue marbles: There are 6 blue marbles, so the number of ways to select 3 blue marbles is:

$\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$

Now, sum the possibilities:

$\text{Total ways for part (b)} = 10 + 4 + 20 = 34$

So, the answer for part (b) is 34 ways.

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Part (c): How many ways can three marbles be selected so that there is one of each color?

To select one marble of each color (red, white, blue), we simply select 1 marble from each color group. The number of ways to do this is:

• Ways to select 1 red: $\binom{5}{1} = 5$
• Ways to select 1 white: $\binom{4}{1} = 4$
• Ways to select 1 blue: $\binom{6}{1} = 6$

Now, multiply these possibilities:

$\text{Total ways for part (c)} = 5 \times 4 \times 6 = 120$

So, the answer for part (c) is 120 ways.

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Final Answers:

• (a): 290 ways
• (b): 34 ways
• (c): 120 ways

Would you like any more details or clarifications?

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Here are 5 related questions for you to explore further:

1. What if there were 7 white marbles instead of 4? How would that change the answer for part (a)?
2. How many ways can you select 4 marbles so that at least one is white?
3. What if you could select only 2 marbles and both had to be the same color? How many ways would there be?
4. How would the problem change if the order of selecting the marbles mattered?
5. How many ways are there to select 3 marbles such that at least two are red?

Tip:

When solving combination problems, always verify whether the order of selection matters. If it doesn't, use combinations (denoted as $\binom{n}{r}$), and if it does, use permutations.