If k and n are positive integers such that n k, then k! + (n-k)*(k-1

Question

If k and 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)!

PS70371.02

BrentGMATPrepNow · Accepted Answer

STRATEGY : Upon reading any GMAT Problem Solving question, we should always ask,   Can I use the answer choices to my advantage?   
In this case, we can easily test the answer choices for equivalency. 
Now let's give ourselves up to 20 seconds to identify a faster approach. 
In this case, we can also try to simplify the expression so that it matches one of the answer choices.
I think testing for equivalency will be faster, so I'll go with that...

Key concept: If two expressions are equivalent, they must evaluate to the same value for every possible value of x. 
For example, since the expression 2x + 3x is equivalent to the expression 5x, the two expressions will evaluate to the same number for every value of x. 
So, if x = 7, the expression 2x + 3x = 2(7) + 3(7) = 14 + 21 =  35 , and the expression 5x = 5(7) =  35

Let's test   n = 3   and   k = 2  . 
Plug   n = 3   and   k = 2   into the given expression to get:  k! + (n - k)(k -1)! =  2 ! + ( 3  -  2 )( 2  -1)! = 2! + (1)(1!) = 2 + 1 =  3  :

Now we'll plug   n = 3   and   k = 2   into the five answer choices see which one(s) evaluate(s) to   3  :
(A)  (k)(n!) = ( 2 )( 3 !) = (2)(6) =  12  . No good. Eliminate. 
(B)  (k!)(n) = ( 2 !)( 3 ) = (2)(3) =  6  . No good. Eliminate. 
(C)  (n - k)! = ( 3  -  2 )! = 1! =  1  . No good. Eliminate. 
(D)  (n)(k + 1)! = ( 3 )( 2  + 1)! = (3)(3!) = (3)(6) =  18  . No good. Eliminate. 
(E)  (n)(k - 1)! = ( 3 )( 2  - 1)! = (3)(1!) = (3)(1) =  3  . Perfect!!

Answer: E

----------------------------------------------------------

ALTERNATE APPROACH   : Apply some algebra

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

abannore · Answer

Answer: E

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

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sujoykrdatta · Answer

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

Alternatively (if you prefer working with numbers instead): 
Let k = n = 2 (note: It works even if n > k is not satisfied; else, we may take n=3, k=2; etc.)
=>  k!+(n-k)*(k-1)!  =  2! + 0*1!  = 2

Working with the options, only Option E satisfies

Answer E

takozzz · Answer

Because k and n are positive integers such that n > k
Let's make n=3,k=2, so k!+(n−k)∗(k−1)!=2!+(3-2)!=3
Then we put the number n=3,k=2 to the answer choice to see which one will get the same number
choice E , n∗(k−1)!=3*(2-1)!=3

ScottTargetTestPrep · Answer

We can factor out (k - 1)! from k! and thus we have:

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

(k - 1)! *

(k - 1)! * n

Answer: E

TheyHaveMySon · Answer

Why is  k(k-1)! = k!  ?

bschoolprep13 · Answer

Sometimes it's helpful to just substitute values to test a formula. In this case, k(k-1)! = k! can be proved as:

k = 5

k! = 5! = 5*4*3*2*1 = 120
(k-1)! = (5-1)! = 4! = 4*3*2*1 = 24

So, k(k-1)! = 5(24) = 120 = 5! = k!

Hope this helps.

CrackverbalGMAT · Answer

Solution: k! + (n-k)(k-1)! = k (k-1)! + (n-k)(k-1)! =(k-1)! (k + n - k ) = (k-1)! n (option e) Hope this helps

Devmitra Sen (GMAT Quant Expert)

GMATinsight · Answer

Answer: Option E

Video solution by  GMATinsight

https://www.youtube.com/watch?v=m5xfEyOplMY

emmanuel1990 · Answer

This does not work with n=2 and k=1 btw

Bunuel · Answer

Recall that 0! = 1, so if n = 2 and k = 1, then:

k! + (n-k)*(k-1)! = 1! + (2 - 1)*(1 - 1)! = 1 + 1*0!=1+1=2

and

n*(k-1)! = 2*(1-1)!=2*0!=2*1=2

Hope it helps.

DanTheGMATMan · Answer

Factor out the lowest value from each term and lit it rip:

https://www.youtube.com/watch?v=AJEoPV_4WvU

ravi1522 · Answer

k!+(n-k)*(k-1)!
k*(k-1)!+(n-k)(k-1)!
(k-1)!(k+n-k)
(k-1)!*n. 
Hence Answer E.

mra01 · Answer

It does work. 0 factorial is 1.

sujoykrdatta · Answer

k! + (n-k)  ×  (k-1)! 
= (k-1)!  ×  k + (n-k)  ×  (k-1)! 
= (k-1)!  ×  
= (k-1)! × n
 
Ans E

HammerToFall · Answer

It does not work. A, B and E all equal the same in that case.

SatyP12345 · Answer

Hi Bunel - a bit confused here because if you have n = 2 and k = 1 then options A and B also work, right? What would you do in this case? Thanks for all the help!

If k and n are positive integers such that n > k, then k! + (n-k)*(k-1

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