Question

The number of ways of forming the ordered pairs (p, q) such that p>q by choosing p and q from the first 50 natural numbers is:

• A. \(1275 \)
• B. \(1250 \)
• C. \(1225 \)
• D. \(1200 \)

Hint:

When forming ordered pairs or selections with a strict inequality (like \(p>q\) or \(p<q\)), the problem often simplifies to a combination problem. Once the elements are chosen, their order is uniquely determined by the inequality.

Answer 1

The Correct Option is C

Step 1: Understand the problem statement.
We need to form ordered pairs \((p, q)\) from the first 50 natural numbers (i.e., from the set \(\{1, 2, \ldots, 50\}\)) such that \(p > q\).
Step 2: Relate the problem to combinations.
If we select any two distinct numbers from the set of 50 natural numbers, there is only one way to assign them to \(p\) and \(q\) such that \(p > q\). Specifically, the larger of the two chosen numbers will be \(p\), and the smaller will be \(q\).
For example, if we choose the numbers 10 and 25, then \(p\) must be 25 and \(q\) must be 10 to satisfy \(p > q\).
Therefore, the problem reduces to finding the number of ways to choose 2 distinct numbers from a set of 50 numbers. This is a combination problem.
Step 3: Calculate the number of combinations.
The number of ways to choose \(k\) items from a set of \(n\) distinct items is given by the combination formula:
\[ C(n, k) = \binom{n}{k} = \frac{n!}{k!(n-k)!} \] In this case, \(n = 50\) (total natural numbers) and \(k = 2\) (numbers to choose for the pair).
\[ \text{Number of ways} = C(50, 2) = \binom{50}{2} = \frac{50!}{2!(50-2)!} = \frac{50!}{2!48!} \] \[ = \frac{50 \times 49 \times 48!}{2 \times 1 \times 48!} \] \[ = \frac{50 \times 49}{2} \] \[ = 25 \times 49 \] \[ = 1225 \] So, there are 1225 ways to form such ordered pairs.
The final answer is \(\boxed{1225}\).