Question

If \[ \frac{n + 2C8}{n - 2P4} = \frac{57}{16}, { then the value of } n { is:} \]

• A. 20
• B. 19
• C. 18
• D. 17

Hint:

When solving equations with factorials, simplify step by step and ensure to handle combinations and permutations properly.

Answer 1

The Correct Option is B

Step 1: Recall that combinations and permutations are given by the formulas: \[ nCk = \frac{n!}{k!(n-k)!} \quad {and} \quad nPk = \frac{n!}{(n-k)!}. \] Thus, \[ n + 2C8 = \frac{(n+2)!}{8!(n-6)!}, \quad n - 2P4 = \frac{(n-2)!}{(n-6)!}. \] Step 2: Substitute these into the equation: \[ \frac{\frac{(n+2)!}{8!(n-6)!}}{\frac{(n-2)!}{(n-6)!}} = \frac{57}{16}. \] Simplify: \[ \frac{(n+2)!}{8!(n-2)!} = \frac{57}{16}. \] Step 3: Further simplifying the factorials: \[ \frac{(n+2)(n+1)}{8!} = \frac{57}{16}. \] This gives: \[ (n+2)(n+1) = 57 \quad \Rightarrow \quad n^2 + 3n + 2 = 57. \] Step 4: Solve for \( n \): \[ n^2 + 3n - 55 = 0. \] Using the quadratic formula: \[ n = \frac{-3 \pm \sqrt{3^2 - 4(1)(-55)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 220}}{2} = \frac{-3 \pm \sqrt{229}}{2}. \] The positive root gives \( n = 19 \).