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Difficulty: 505-555 Level,   Algebra,            
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If k and n are positive integers such that n > k, then k! + (n-k)*(k-1 [#permalink]
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gmatt1476
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]
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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]
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gmatt1476
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]
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]
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TheyHaveMySon
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]
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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:
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Re: If k and n are positive integers such that n > k, then k! + (n-k)*(k-1 [#permalink]
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gmatt1476
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]
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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]
Expert Reply
emmanuel1990
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]
Expert Reply
­Factor out the lowest value from each term and lit it rip:

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