14. The \( 12^{\text {en }} \) grade class of 80 students is assembled in a large circle on the football fietd at halftime. Every student is connecled by a string to each of the other class members. a) How many pieces of string are necessary to connect all of the students to each othen b) Which rule did you use and why?

To solve the problem, we need to determine how many pieces of string are necessary to connect all 80 students in a circle, where each student is connected to every other student. ### a) How many pieces of string are necessary? To find the number of pieces of string needed to connect \( n \) students, we can use the formula for combinations, specifically \( \binom{n}{2} \), which represents the number of ways to choose 2 students from \( n \) students. This is because each piece of string connects a unique pair of students. For \( n = 80 \): \[ \text{Number of pieces of string} = \binom{80}{2} = \frac{80 \times 79}{2} = 3160 \] So, **3160 pieces of string** are necessary to connect all of the students to each other. ### b) Which rule did you use and why? The rule used here is the **combinatorial formula for combinations**, specifically \( \binom{n}{k} \), where \( n \) is the total number of items (students, in this case) and \( k \) is the number of items to choose (2, since each string connects 2 students). We used this rule because we are interested in counting the unique pairs of students that can be formed from the group of 80. Each pair corresponds to one piece of string, and since the order of selection does not matter (i.e., connecting student A to student B is the same as connecting student B to student A), combinations are the appropriate mathematical tool for this scenario.