If ((n+3)! − (n+2)!)/(n+1)! = 49, what is the value of n?

Correct answer: 5

The governing concept is factorial notation and cancellation. Use (n+3)! = (n+3)(n+2)! to factor the numerator: (n+3)! − (n+2)! = (n+2)![(n+3) − 1] = (n+2)!(n+2). Dividing by (n+1)! gives (n+2)!/(n+1)! × (n+2) = (n+2)(n+2) = (n+2)^2. Hence (n+2)^2 = 49. In the usual school definition of factorial, the relevant factorial arguments must be non-negative integers, so n+2 is positive. Therefore n+2 = 7 and n = 5, making option D correct. The algebraic root n = −9 is not admissible in this factorial setting. Options A, B, and C produce values different from 49.