Question

If \(P(2n+1,n-1):P(2n-1,n) = 3:5\), find \(n\).

• A. 2
• B. 4
• C. 6
• D. 8
• E. 10

Hint:

In permutation ratios, simplify factorials step by step before solving quadratic equations.

Answer 1

The Correct Option is B

Step 1: Use permutation formula.
\[ P(n,r) = \frac{n!}{(n-r)!} \]
Step 2: Expand ratio.
\[ \frac{P(2n+1,n-1)}{P(2n-1,n)} = \frac{3}{5} \] \[ \frac{(2n+1)!}{(n+2)!} \cdot \frac{(n-1)!}{(2n-1)!} = \frac{3}{5} \]
Step 3: Simplify.
\[ = \frac{(2n+1)(2n)}{(n+2)(n+1)n} = \frac{3}{5} \]
Step 4: Cross-multiply.
\[ 5(2n+1)(2n) = 3(n+2)(n+1)n \]
Step 5: Expand.
LHS = \(20n^2 + 10n\). RHS = \(3n^3 + 9n^2 + 6n\).
So: \[ 3n^3 - 11n^2 - 4n = 0 \]
Step 6: Factorize.
\[ n(3n^2 - 11n - 4) = 0 \] Quadratic: \(3n^2 - 11n - 4 = 0\).
Discriminant = \(121 + 48 = 169\), root = 13.
\[ n = \frac{11 \pm 13}{6} \] \[ n = 4 \quad \text{or} \quad n = -\tfrac{1}{3} \] Reject negative.
Final Answer:
\[ \boxed{4} \]