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

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gmatt1476
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If k and n are positive integers such that n > k, then k! + (n-k)*(k-1 [#permalink] New post   20 Apr 2020, 13:04
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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
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E
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BrentGMATPrepNow
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If k and n are positive integers such that n > k, then k! + (n-k)*(k-1 [#permalink] New post   Updated on: 10 Apr 2022, 15:20
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gmatt1476 wrote:
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

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

Originally posted by BrentGMATPrepNow on 30 Apr 2020, 12:28.
Last edited by BrentGMATPrepNow on 10 Apr 2022, 15:20, edited 1 time in total.
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Re: If k and n are positive integers such that n > k, then k! + (n-k)*(k-1 [#permalink] New post   20 Apr 2020, 13:17
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Answer: E

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

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If k and n are positive integers such that n > k, then k! + (n-k)*(k-1 [#permalink] New post   20 Apr 2020, 14:19
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gmatt1476 wrote:
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



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

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
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Re: If k and n are positive integers such that n > k, then k! + (n-k)*(k-1 [#permalink] New post   22 Apr 2020, 21:30
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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
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Re: If k and n are positive integers such that n > k, then k! + (n-k)*(k-1 [#permalink] New post   02 May 2020, 17:08
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gmatt1476 wrote:
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


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

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

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

(k - 1)! * n

Answer: E
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Re: If k and n are positive integers such that n > k, then k! + (n-k)*(k-1 [#permalink] New post   25 Aug 2020, 00:21
Why is k(k-1)! = k! ?
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Re: If k and n are positive integers such that n > k, then k! + (n-k)*(k-1 [#permalink] New post   10 Sep 2020, 20:25
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TheyHaveMySon wrote:
Why is k(k-1)! = k! ?


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.
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If k and n are positive integers such that n > k, then k! + (n-k)*(k-1 [#permalink] New post   21 Mar 2021, 00:27
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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 :thumbsup:
Devmitra Sen (GMAT Quant Expert)
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Re: If k and n are positive integers such that n > k, then k! + (n-k)*(k-1 [#permalink] New post   27 May 2021, 08:30
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gmatt1476 wrote:
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


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Answer: Option E

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Re: If k and n are positive integers such that n > k, then k! + (n-k)*(k-1 [#permalink] New post   27 Dec 2023, 07:00
This does not work with n=2 and k=1 btw
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Re: If k and n are positive integers such that n > k, then k! + (n-k)*(k-1 [#permalink] New post   27 Dec 2023, 07:11
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emmanuel1990 wrote:
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) \(n*(k-1)!\)

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


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.
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Re: If k and n are positive integers such that n > k, then k! + (n-k)*(k-1 [#permalink]
27 Dec 2023, 07:11
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