1. **Problem statement:** We have a box with 5 red balls, 7 green balls, and 6 yellow balls. We want to find the number of ways to choose 6 balls such that exactly 2 balls of each color are chosen. 2. **Formula and rules:** The number of ways to choose $k$ items from $n$ items is given by the combination formula: $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ where $n!$ is the factorial of $n$. 3. **Applying the formula:** - Choose 2 red balls from 5: $\binom{5}{2}$ - Choose 2 green balls from 7: $\binom{7}{2}$ - Choose 2 yellow balls from 6: $\binom{6}{2}$ 4. **Calculate each combination:** $$\binom{5}{2} = \frac{5!}{2!3!} = \frac{120}{2 \times 6} = 10$$ $$\binom{7}{2} = \frac{7!}{2!5!} = \frac{5040}{2 \times 120} = 21$$ $$\binom{6}{2} = \frac{6!}{2!4!} = \frac{720}{2 \times 24} = 15$$ 5. **Total number of ways:** Since these choices are independent, multiply the results: $$10 \times 21 \times 15 = 3150$$ **Final answer:** There are $3150$ ways to choose 6 balls with exactly 2 balls of each color.