Question
upstudy study bank question image url

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

Ask by Thornton Ramos. in Pakistan
Mar 18,2025

Upstudy AI Solution

Tutor-Verified Answer

Answer

To prove that the number of ways to pick a pair from \( n \) objects is \( \frac{n(n-1)}{2} \), use mathematical induction: 1. **Base Case (\( n = 2 \)):** - Ways to pick 2 objects: 1 - Calculation: \( \frac{2(2-1)}{2} = 1 \) - Base case holds. 2. **Inductive Step:** - Assume true for \( k \): \( \frac{k(k-1)}{2} \) ways. - For \( k + 1 \): - Ways from first \( k \) objects: \( \frac{k(k-1)}{2} \) - Ways including the new object: \( k \) - Total: \( \frac{k(k-1)}{2} + k = \frac{k(k + 1)}{2} \) - Thus, the formula holds for \( k + 1 \). By induction, the formula \( \frac{n(n-1)}{2} \) is valid for all \( n \geq 2 \).

Solution

To prove that the total number of ways of picking a pair of objects from a given set of \( n \) objects is \( \frac{n(n-1)}{2} \) using mathematical induction, we will follow these steps: ### Step 1: Base Case We start by checking the base case for \( n = 2 \). When \( n = 2 \), the number of ways to choose 2 objects from 2 is: \[ \binom{2}{2} = 1 \] Now, we calculate \( \frac{n(n-1)}{2} \): \[ \frac{2(2-1)}{2} = \frac{2 \cdot 1}{2} = 1 \] Thus, the base case holds true. ### Step 2: Inductive Step Assume that the statement is true for some integer \( k \), i.e., the number of ways to choose 2 objects from \( k \) objects is: \[ \frac{k(k-1)}{2} \] We need to show that it holds for \( k + 1 \). When we have \( k + 1 \) objects, we can choose 2 objects in two ways: 1. Choose 2 objects from the first \( k \) objects. 2. Choose 1 object from the first \( k \) objects and 1 object from the new \( (k + 1)^{th} \) object. The number of ways to choose 2 objects from the first \( k \) objects is: \[ \frac{k(k-1)}{2} \] The number of ways to choose 1 object from the first \( k \) objects is \( k \), and we can pair it with the new \( (k + 1)^{th} \) object. Thus, the number of ways to choose 1 object from the first \( k \) and 1 from the new object is: \[ k \] ### Total Ways for \( k + 1 \) Combining both cases, the total number of ways to choose 2 objects from \( k + 1 \) objects is: \[ \frac{k(k-1)}{2} + k \] Now, we need to simplify this expression: \[ \frac{k(k-1)}{2} + k = \frac{k(k-1)}{2} + \frac{2k}{2} = \frac{k(k-1) + 2k}{2} = \frac{k^2 - k + 2k}{2} = \frac{k^2 + k}{2} = \frac{k(k + 1)}{2} \] ### Conclusion Thus, we have shown that: \[ \text{Number of ways to choose 2 objects from } (k + 1) \text{ objects} = \frac{(k + 1)k}{2} \] This completes the inductive step. Since both the base case and the inductive step have been proven, by the principle of mathematical induction, the statement is true for all integers \( n \geq 2 \). Therefore, the total number of ways of picking a pair of objects from a given set of \( n \) objects is: \[ \frac{n(n-1)}{2} \]

Answered by UpStudy AI and reviewed by a Professional Tutor

error msg
Explain
Simplify this solution

Beyond the Answer

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 \( n \geq 2 \).

Try Premium now!
Try Premium and ask Thoth AI unlimited math questions now!
Maybe later Go Premium
Study can be a real struggle
Why not UpStudy it?
Select your plan below
Premium

You can enjoy

Start now
  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to Answer and Solution
  • Full Access to PDF Chat, UpStudy Chat, Browsing Chat
Basic

Totally free but limited

  • Limited Solution
Welcome to UpStudy!
Please sign in to continue the Thoth AI Chat journey
Continue with Email
Or continue with
By clicking “Sign in”, you agree to our Terms of Use & Privacy Policy