We can approach this problem by breaking it down into two parts: selecting two balls of the same color and selecting two balls of different colors.

Total Balls

The box contains:

• 2 red balls,
• 3 white balls,
• 4 green balls.

a) Two balls of the same color

To select two balls of the same color, we look at each color and count the number of ways we can choose 2 balls from the available ones.

• Red balls: There are 2 red balls, and the number of ways to select 2 out of 2 is: $\binom{2}{2} = 1$

• White balls: There are 3 white balls, and the number of ways to select 2 out of 3 is: $\binom{3}{2} = 3$

• Green balls: There are 4 green balls, and the number of ways to select 2 out of 4 is: $\binom{4}{2} = 6$

Now, add these possibilities: $1 + 3 + 6 = 10$ Thus, there are 10 ways to select two balls of the same color.

b) Two balls of different colors

To select two balls of different colors, we can choose one ball of one color and one ball of another color. Here’s how many ways we can do that:

• Red and White: There are 2 red balls and 3 white balls. The number of ways to select one red and one white is: $2 \times 3 = 6$

• Red and Green: There are 2 red balls and 4 green balls. The number of ways to select one red and one green is: $2 \times 4 = 8$

• White and Green: There are 3 white balls and 4 green balls. The number of ways to select one white and one green is: $3 \times 4 = 12$

Now, add these possibilities: $6 + 8 + 12 = 26$ Thus, there are 26 ways to select two balls of different colors.

Final Answer

• a) The number of ways to select two balls of the same color is 10.
• b) The number of ways to select two balls of different colors is 26.

Would you like any more details or have any questions?

Here are 5 related questions:

1. What is the total number of ways to select any two balls from the box?
2. How would the solution change if the box contained one additional red ball?
3. How many ways can three balls be selected, with at least one of each color?
4. If you select two balls at random, what is the probability that they are of the same color?
5. How would the solution differ if one more color of ball was added to the box?

Tip: When solving combination problems, it’s helpful to use binomial coefficients $\binom{n}{r}$ for counting selections without replacement.