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

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AbeNeedsAnswers: Master | Next Rank: 500 Posts; Posts: 394; Joined: Sun Jul 02, 2017 10:59 amFollowed by: 5 members; Thanked: 1 times

by AbeNeedsAnswers » Tue May 19, 2020 7:11 pm

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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

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Source: Beat The GMAT — Problem Solving | Tue May 19, 2020 7:11 pm

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

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

by Brent@GMATPrepNow » Wed May 20, 2020 5:40 am

AbeNeedsAnswers wrote: ↑
Tue May 19, 2020 7:11 pm
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

Brent Hanneson - Creator of GMATPrepNow.com

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8152; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

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

by Scott@TargetTestPrep » Sun May 24, 2020 5:59 am

AbeNeedsAnswers wrote: ↑
Tue May 19, 2020 7:11 pm
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

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