2. There are 8 boys and 12 girls in a class, and 5 of them are to be chosen for a trip. If 2 particular girls are to be included, the number of ways to make the selection is \( \begin{array}{llll}\text { A. } 812 & \text { B. } 816 & \text { C. } 818 & \text { D. } 820\end{array} \)

To solve the problem, we need to select 5 students from a class of 8 boys and 12 girls, with the condition that 2 particular girls must be included in the selection. Since 2 particular girls are already included, we need to choose 3 more students from the remaining students. After including the 2 girls, we have: - Total girls left: \(12 - 2 = 10\) - Total boys: 8 So, the total number of students left to choose from is \(10 + 8 = 18\). Now, we need to choose 3 students from these 18 remaining students. The number of ways to choose 3 students from 18 can be calculated using the combination formula: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] where \(n\) is the total number of items to choose from, and \(r\) is the number of items to choose. In our case, \(n = 18\) and \(r = 3\): \[ \binom{18}{3} = \frac{18!}{3!(18-3)!} = \frac{18!}{3! \cdot 15!} = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} \] Calculating this step-by-step: 1. Calculate the numerator: \[ 18 \times 17 = 306 \] \[ 306 \times 16 = 4896 \] 2. Calculate the denominator: \[ 3 \times 2 \times 1 = 6 \] 3. Now divide the numerator by the denominator: \[ \frac{4896}{6} = 816 \] Thus, the number of ways to select the 5 students, including the 2 particular girls, is \(816\). The answer is \( \text{B. } 816 \).