8. Using induction prove that the total number of ways of picking a pair of objects from a given set of \( n \) objects is \( \frac{n(n-1)}{2} \).

To prove by induction that the total number of ways to pick a pair of objects from a set of \( n \) objects is \( \frac{n(n-1)}{2} \), we first establish the base case when \( n = 2 \). Base Case (\( n = 2 \)): For a set with 2 objects, the number of ways to choose 2 objects is clearly 1, and the formula gives \( \frac{2(2-1)}{2} = 1 \). Therefore, the base case holds. Inductive Step: Assume the formula holds for \( n = k \), that is, the number of ways to pick 2 objects from a set of \( k \) objects is \( \frac{k(k-1)}{2} \). We need to show it holds for \( n = k + 1 \). Consider a set with \( k + 1 \) objects. We can form a pair in two ways: 1. Pairs that include the \( (k + 1)^{th} \) object. There are \( k \) objects remaining, hence \( k \) pairs are formed with the \( (k + 1)^{th} \) object. 2. Pairs that do not include the \( (k + 1)^{th} \) object. By the inductive hypothesis, the number of ways to pick 2 objects from those \( k \) objects is \( \frac{k(k-1)}{2} \). So, the total number of ways to choose 2 objects from \( k + 1 \) objects is: \[ k + \frac{k(k-1)}{2} \] To simplify this expression, we convert \( k \) into a fraction equivalent to have a common denominator: \[ k = \frac{2k}{2} \] Thus, \[ k + \frac{k(k-1)}{2} = \frac{2k}{2} + \frac{k(k-1)}{2} = \frac{2k + k(k - 1)}{2} = \frac{k^2 + k}{2} = \frac{(k + 1)k}{2} \] This confirms that the formula holds for \( n = k + 1 \). By induction, we conclude that the total number of ways to pick a pair of objects from a set of \( n \) objects is indeed \( \frac{n(n-1)}{2} \) for all \( n \geq 2 \).