OG: If k an n are positive integers such that n > k, then k! + (n – k)(k – 1)!

[AbeNeedsAnswers]

If k an n are positive integers such that n > k, then k! + (n – k)(k – 1)! is equivalent to which of the following?

A. (k)(n!)
B. (k!)(n)
C. (n – k)!
D. (n)(k + 1)!
E. (n)(k – 1)!

E

[Brent@GMATPrepNow]

If k an n are positive integers such that n > k, then k! + (n – k)(k – 1)! is equivalent to which of the following?

A. (k)(n!)
B. (k!)(n)
C. (n – k)!
D. (n)(k + 1)!
E. (n)(k – 1)!

E

Key Concept #1: k! = (k)(k-1)(k-2)(k-3)....(3)(2)(1)
Key Concept #2: (k-1)! = (k-1)(k-2)(k-3)....(3)(2)(1)
Key Concept #3: (x-y)(z) = xz - yz


So....
Take: k! + (n-k)(k-1)!
Apply concept #2 to get: k! + (n - k)(k-1)(k-2)(k-3)....(3)(2)(1)
Apply concept #3 to get: k! + (n)(k-1)(k-2)(k-3)....(3)(2)(1) - (k)(k-1)(k-2)(k-3)....(3)(2)(1)

Notice that (k)(k-1)(k-2)(k-3)....(3)(2)(1) = k!
And (k-1)(k-2)(k-3)....(3)(2)(1) = (k - 1)!

Substitute to get: k! - (n)(k - 1)! - k!
Simplify to get: (n)(k - 1)!

Answer: E

Cheers,
Brent

[Scott@TargetTestPrep]

If k an n are positive integers such that n > k, then k! + (n – k)(k – 1)! is equivalent to which of the following?

A. (k)(n!)
B. (k!)(n)
C. (n – k)!
D. (n)(k + 1)!
E. (n)(k – 1)!

E

We can re-express k! as k(k - 1)!. We can then factor out (k - 1)! from both terms of the expression and obtain:

k(k - 1)! + (n – k)(k – 1)!

(k - 1)!(k + n - k)

(k - 1)!(n)

Answer: E