5. A box contain 6 distinct red balls and 4 distinct blue balls. In how many ways can 3 balls be drawn randomly from the box if: a. the color is not considered? b. b. 2 balls is red and one is blue? c. c. all three balls are red?

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Let's break down the problem step by step.

### Known Conditions:
- There are 6 distinct red balls.
- There are 4 distinct blue balls.
- We need to find the number of ways to draw 3 balls under different conditions.

### a. The color is not considered
When the color is not considered, we can simply choose any 3 balls from the total of 10 balls (6 red + 4 blue).

The number of ways to choose 3 balls from 10 is given by the combination formula:

$$
C(n, r) = \frac{n!}{r!(n-r)!}
$$

where $$ n $$ is the total number of items to choose from, and $$ r $$ is the number of items to choose.

Here, $$ n = 10 $$ and $$ r = 3 $$.

Let's calculate this:

$$
C(10, 3) = \frac{10!}{3!(10-3)!} = \frac{10!}{3! \cdot 7!}
$$

Calculating $$ C(10, 3) $$:

$$
C(10, 3) = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120
$$

### b. 2 balls are red and 1 is blue
For this case, we need to choose 2 red balls from the 6 available and 1 blue ball from the 4 available.

The number of ways to choose 2 red balls is:

$$
C(6, 2) = \frac{6!}{2!(6-2)!} = \frac{6!}{2! \cdot 4!}
$$

Calculating $$ C(6, 2) $$:

$$
C(6, 2) = \frac{6 \times 5}{2 \times 1} = \frac{30}{2} = 15
$$

The number of ways to choose 1 blue ball is:

$$
C(4, 1) = \frac{4!}{1!(4-1)!} = \frac{4!}{1! \cdot 3!} = 4
$$

Now, we multiply the two results to find the total number of ways to choose 2 red and 1 blue:

$$
\text{Total ways} = C(6, 2) \times C(4, 1) = 15 \times 4 = 60
$$

### c. All three balls are red
In this case, we need to choose all 3 balls from the 6 red balls.

The number of ways to choose 3 red balls is:

$$
C(6, 3) = \frac{6!}{3!(6-3)!} = \frac{6!}{3! \cdot 3!}
$$

Calculating $$ C(6, 3) $$:

$$
C(6, 3) = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = \frac{120}{6} = 20
$$

### Summary of Results:
- a. The number of ways to draw 3 balls (color not considered): **120**
- b. The number of ways to draw 2 red and 1 blue: **60**
- c. The number of ways to draw 3 red balls: **20**
- a. 120 ways to draw 3 balls without considering color. - b. 60 ways to draw 2 red and 1 blue ball. - c. 20 ways to draw 3 red balls.

5. A box contain 6 distinct red balls and 4 distinct blue balls. In how many ways can 3 balls be drawn randomly from the box if: a. the color is not considered? b. b. 2 balls is red and one is blue? c. c. all three balls are red?

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