Question

The number of positive integers satisfying the inequality \(^{n+1}C_{n-1} + ^{n}C_{n-1} \leq 50\) is

• A. 9
• B. 8
• C. 7
• D. 6

Hint:

\(^nC_r = ^nC_{n-r}\). Simplify using this property.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Simplify combinations using \(^nC_r = \frac{n!}{r!(n-r)!}\).

Step 2: Detailed Explanation:
\(^{n+1}C_{n-1} = ^{n+1}C_2 = \frac{(n+1)n}{2}\).
\(^nC_{n-1} = ^nC_1 = n\).
Inequality: \(\frac{n(n+1)}{2} + n \leq 50\)
\(\frac{n^2 + n + 2n}{2} \leq 50 \implies \frac{n^2 + 3n}{2} \leq 50\)
\(n^2 + 3n - 100 \leq 0\)
Roots: \(n = \frac{-3 \pm \sqrt{9 + 400}}{2} = \frac{-3 \pm \sqrt{409}}{2}\). \(\sqrt{409} \approx 20.22\), so \(n \approx 8.61\).
\(n\) positive integer, \(n \leq 8\). Also \(n \geq 1\) for combinations.
Also \(n+1 \geq n-1\) always true. So \(n = 1, 2, \ldots, 8\). Total 8 values.

Step 3: Final Answer:
Option (B) 8.