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

To solve the given inequality \(^{n+1}C_{n-1} + ^{n}C_{n-1} \leq 50\), we need to simplify the expression and find the number of integers \( n \) satisfying the condition.

Recall the formula for combinations: \(^{n}C_{r} = \frac{n!}{r!(n-r)!}\).

Consider the terms individually:

• \(^{n+1}C_{n-1} = \frac{(n+1)!}{(n-1)!(2)!} = \frac{(n+1)n}{2}\)
• \(^{n}C_{n-1} = \frac{n!}{(n-1)!1!} = n\)

Substituting these into the inequality gives:

\(\frac{(n+1)n}{2} + n \leq 50\)

Combining terms:

\(\frac{n(n+1)}{2} + n = \frac{n^2 + n + 2n}{2} = \frac{n^2 + 3n}{2} \leq 50\)

Multiplying the entire inequality by 2 to clear the denominator:

\(n^2 + 3n \leq 100\)

Rearrange the inequality to standard quadratic form:

\(n^2 + 3n - 100 \leq 0\)

To solve for \( n \), factor the quadratic equation:

\((n + 13)(n - 10) \leq 0\)

The critical points from the factors are \( n = -13 \) and \( n = 10 \). By testing intervals around these points, we find the solution to be:

• \( n \in [-13, 10] \)

Since \( n \) must be a positive integer, we only consider \( n \) from 1 to 10.

Total positive integers satisfying this inequality are: 1, 2, 3, 4, 5, 6, 7, 8.

Thus, the number of positive integers \( n \) that satisfy the inequality is 8.