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

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Math Expert
Joined: 02 Sep 2009
Posts: 62285

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06 Jun 2017, 10:37
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Question Stats:

89% (00:51) correct 11% (01:40) wrong based on 146 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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Updated on: 06 Jun 2017, 12:12
(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

Originally posted by quantumliner on 06 Jun 2017, 11:42.
Last edited by quantumliner on 06 Jun 2017, 12:12, edited 2 times in total.
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06 Jun 2017, 12:03
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

hi quantumliner

you have got the right answer but the option mentioned by you is wrong. Suggest you rectify it.
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06 Jun 2017, 17:45
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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18 Aug 2017, 02:41
Bunuel,
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18 Aug 2017, 03:20
NaeemHasan wrote:
Bunuel,

The correct answer is D. Edited. Thank you.
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18 Aug 2017, 05:44
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

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18 Aug 2017, 09:40
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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18 Aug 2017, 13:50
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 $$(n + 3)^5$$

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31 Aug 2017, 17:01
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
Re: [(n+3)!]^5/[(n+2)!]^5 =   [#permalink] 31 Aug 2017, 17:01
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