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[(n+3)!]^5/[(n+2)!]^5 =

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Bunuel
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[(n+3)!]^5/[(n+2)!]^5 = [#permalink] New post   06 Jun 2017, 11:37
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A
B
C
D
E

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5% (low)

Question Stats:

90% (00:51) correct 10% (01:29) wrong based on 88 sessions
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\(\frac{[(n+3)!]^5}{[(n+2)!]^5} =\)

A. 2
B. (n+1)^5
C. (n+2)^5
D. (n+3)^5
E. ((n+3/(n+2))^5

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D
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quantumliner
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Re: [(n+3)!]^5/[(n+2)!]^5 = [#permalink] New post   Updated on: 06 Jun 2017, 13:12
(n+3)! can be written as (n+3)(n+2)!

now \([(n+3)!]^5\) can be written as \([(n+3)(n+2)!]^5\) = \((n+3)^5\) * \([(n+2)!]^5\)

therefore

\([(n+3)!]^5\)/\([(n+2)!]^5\) = [\((n+3)^5\) *\([(n+2)!]^5\)]/\([(n+2)!]^5\) = (n+3)^5


Answer is D

Originally posted by quantumliner on 06 Jun 2017, 12:42.
Last edited by quantumliner on 06 Jun 2017, 13:12, edited 2 times in total.
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Re: [(n+3)!]^5/[(n+2)!]^5 = [#permalink] New post   06 Jun 2017, 13:03
quantumliner wrote:
(n+3)! can be written as (n+3)(n+2)!

now \([(n+3)!]^5\) can be written as \([(n+3)(n+2)!]^5\) = \((n+3)^5\) * \([(n+2)!]^5\)

therefore

\([(n+3)!]^5\)/\([(n+2)!]^5\) = [\((n+3)^5\) *\([(n+2)!]^5\)]/\([(n+2)!]^5\) = (n+3)^5


Answer is E


hi quantumliner

you have got the right answer but the option mentioned by you is wrong. Suggest you rectify it.
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Re: [(n+3)!]^5/[(n+2)!]^5 = [#permalink] New post   06 Jun 2017, 18:45
The answer should be D. We can solve this question under 30 seconds if we plug in a number for n. For instance, lets say n=2. According to the equation, we will get [(2+3)!^5]/[(2+2)!^5] which will equal to 5^5 as your final answer. Now, we just have to plug in n as 2 and see which of the multiple choices will give us 5^5 as our final answer. Only D will.

Kindly give me a kudos if you liked my approach! Thanks!!
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Re: [(n+3)!]^5/[(n+2)!]^5 = [#permalink] New post   18 Aug 2017, 03:41
Bunuel,
Is the answer really A?
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Re: [(n+3)!]^5/[(n+2)!]^5 = [#permalink] New post   18 Aug 2017, 04:20
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NaeemHasan wrote:
Bunuel,
Is the answer really A?


The correct answer is D. Edited. Thank you.
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Re: [(n+3)!]^5/[(n+2)!]^5 = [#permalink] New post   18 Aug 2017, 06:44
Bunuel wrote:
\(\frac{[(n+3)!]^5}{[(n+2)!]^5} =\)

A. 2
B. (n+1)^5
C. (n+2)^5
D. (n+3)^5
E. ((n+3/(n+2))^5


\([\frac{(n+3)!}{(n+2)!}]^5=\)
(n+3)! = (n+3)(n+2)!
(n+3)!/(n+2)! = n+3

\([\frac{(n+3)!}{(n+2)!}]^5=\) = (n+3)^5

Answer D
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Re: [(n+3)!]^5/[(n+2)!]^5 = [#permalink] New post   18 Aug 2017, 10:40
We need to find the value of the expression \(\frac{[(n+3)!]^5}{[(n+2)!]^5}\)

Plug n=0,
The expression \(\frac{[(n+3)!]^5}{[(n+2)!]^5} = \frac{[(3)!]^5}{[(2)!]^5} = \frac{[3*2!]^5}{[(2)!]^5} = \frac{[(3)^5*(2!)^5]}{[(2)!]^5} = (3)^5\)

Only Option D\((n+3)^5\) gives us the same value, and is our correct answer.
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Re: [(n+3)!]^5/[(n+2)!]^5 = [#permalink] New post   18 Aug 2017, 14:50
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Bunuel wrote:
\(\frac{[(n+3)!]^5}{[(n+2)!]^5} =\)

A. 2
B. (n+1)^5
C. (n+2)^5
D. (n+3)^5
E. ((n+3/(n+2))^5

I plugged in n = 0 and n = 1, then used law of exponents:
\(\frac{(a)^n}{(b)^n}\) = \((\frac{a}{b})^n\)

If n = 0:

\(\frac{(3*2*1)^5}{(2*1)^5}\) = \(\frac{(6)^5}{(2)^5}\) = \((\frac{6}{2})^5\) = \(3^5\)

n = 0. Result is \((n + 3)^5\)

To ascertain, I checked n = 1:

\(\frac{(4*3*2)^5}{(3*2)^5}\) = \(\frac{(24)^5}{(6)^5}\) = \((\frac{24}{6})^5\) = \(4^5\)

n = 1, result is \((n + 3)^5\)

Answer D
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Re: [(n+3)!]^5/[(n+2)!]^5 = [#permalink] New post   31 Aug 2017, 18:01
Bunuel wrote:
\(\frac{[(n+3)!]^5}{[(n+2)!]^5} =\)

A. 2
B. (n+1)^5
C. (n+2)^5
D. (n+3)^5
E. ((n+3/(n+2))^5


Pick a simple value- 1


a^x/b^x =(a/b)^x

(4!)^5/ (3!)^5 =

4^5

D

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Re: [(n+3)!]^5/[(n+2)!]^5 = [#permalink]
31 Aug 2017, 18:01
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