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555-605 (Medium)|   Algebra|      
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arjunnath
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If k and n are positive integers such that n > k, then k! + (n - k)*( 06 Nov 2021, 07:09
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25% (medium)

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77% (01:32) correct 23% (02:03) wrong based on 235 sessions

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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)!
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E
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DashingNightmare2
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If k and n are positive integers such that n > k, then k! + (n - k)*( 09 Nov 2021, 23:59
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arjunnath
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)!
Show more


k! + (n - k)*(k - 1)!
= k*(k-1)! + (n - k)*(k - 1)!
Taking (k - 1)! common before and after +

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

Hence E.
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frappuchino
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If k and n are positive integers such that n > k, then k! + (n - k)*( 09 Aug 2023, 05:39
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DashingNightmare2
arjunnath
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)!


k! + (n - k)*(k - 1)!
= k*(k-1)! + (n - k)*(k - 1)!
Taking (k - 1)! common before and after +

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

Hence E.
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Hi. I'm still not sure I understand this. How is k! + (n - k)*(k - 1)! = k*(k-1)! + (n - k)*(k - 1)!
Could someone please explain. Thanks
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If k and n are positive integers such that n > k, then k! + (n - k)*( 10 Aug 2023, 04:44
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anureetjuneja

Hi. I'm still not sure I understand this. How is k! + (n - k)*(k - 1)! = k*(k-1)! + (n - k)*(k - 1)!
Could someone please explain. Thanks
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We know the definition of 5! i.e. 5!= 5*4*3*2*1

k! = k * (k-1) * (k-2) * (k-3) *........... *1 --(1)
k-1! = (k-1) * (k-2) * (k-3) *...........* 1 --(2)
We can substitute (2) into (1)
k! = k * k-1!
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If k and n are positive integers such that n > k, then k! + (n - k)*( 13 Oct 2023, 12:17
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What however if we assume that n=2 and k=1? By plugging in the values into the given formula we receive 1. When you plug the values into the Answer E formula you receive 0. How can this be?
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If k and n are positive integers such that n > k, then k! + (n - k)*( 13 Oct 2023, 20:02
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flyinghirsch
What however if we assume that n=2 and k=1? By plugging in the values into the given formula we receive 1. When you plug the values into the Answer E formula you receive 0. How can this be?
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Note that 0! equals 1 (not 0).
Also note that factorials only exist for postive integers. So we can't assume k=0 as then k-1! will be undefined.
You can read all about factorials here : https://gmatclub.com/forum/math-number-theory-88376.html
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If k and n are positive integers such that n > k, then k! + (n - k)*( 14 Oct 2023, 13:18
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arjunnath
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)!
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Bunuel, this question is mistakenly classified as a Data Sufficiency Question !
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If k and n are positive integers such that n > k, then k! + (n - k)*( 14 Oct 2023, 13:23
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LucienH
arjunnath
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)!

Bunuel, this question is mistakenly classified as a Data Sufficiency Question !
Show more
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Moved to the PS forum. Thank you!
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If k and n are positive integers such that n > k, then k! + (n - k)*( 10 Dec 2024, 06:25
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