If C(n+2, 8) : P(n−2, 4) = 57 : 16, find the integer value of n. (Use C(a, b) = a! / (b! (a − b)!) and P(a, b) = a! / (a − b)!)

Difficulty: Medium

Correct Answer: 19

Explanation:


Introduction / Context:
We must simplify a ratio mixing a combination and a permutation. Converting each to factorials and canceling common factors typically yields a manageable polynomial equation in n.


Given Data / Assumptions:

  • C(n+2, 8) / P(n−2, 4) = 57/16.
  • n is an integer large enough that all factorials are defined (n ≥ 19 from the solution).


Concept / Approach:
Write C(n+2, 8) = (n+2)! / (8! (n−6)!) and P(n−2, 4) = (n−2)! / (n−6)!. Then the ratio collapses to ((n+2)! / 8!) / (n−2)! = (n+2)(n+1)n(n−1) / 8!.


Step-by-Step Solution:

Let R = C(n+2, 8) / P(n−2, 4) = (n+2)(n+1)n(n−1) / 8!.8! = 40320. Set (n+2)(n+1)n(n−1) / 40320 = 57/16.Thus (n+2)(n+1)n(n−1) = 40320 * 57 / 16 = 2520 * 57 = 143640.Test n = 19: (21201918) = 420 * 342 = 143640 ⇒ works.


Verification / Alternative check:
Nearby n = 18 gives 116280; n = 20 gives 20212219 = larger than target, confirming n = 19 uniquely fits.


Why Other Options Are Wrong:
17, 18, 20 do not satisfy the exact product 143640.


Common Pitfalls:
Forgetting that the (n−6)! terms cancel; mis-evaluating 8! or arithmetic with large products.


Final Answer:
19

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