1. **Problem Statement:** We need to find the number of different ways to select a panel of 12 jurors and 2 alternate jurors from a group of 27 potential jurors. 2. **Understanding the problem:** We are choosing 14 jurors in total, but they are divided into two groups: 12 jurors and 2 alternates. The order within each group does not matter, but the groups are distinct. 3. **Formula used:** The number of ways to choose $k$ items from $n$ items is given by the combination formula: $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ 4. **Step 1: Choose 12 jurors out of 27:** $$\binom{27}{12} = \frac{27!}{12! \times 15!}$$ 5. **Step 2: Choose 2 alternate jurors out of the remaining 15:** $$\binom{15}{2} = \frac{15!}{2! \times 13!}$$ 6. **Step 3: Multiply the two results because these are independent choices:** $$\binom{27}{12} \times \binom{15}{2}$$ 7. **Calculating the values:** $$\binom{27}{12} = 17383860$$ $$\binom{15}{2} = 105$$ 8. **Final number of ways:** $$17383860 \times 105 = 1825305300$$ **Answer:** There are $1825305300$ different ways to choose the panel and alternates.