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If n > 4, what is the value of the integer n ?

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If n > 4, what is the value of the integer n ?  [#permalink]

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New post 21 Sep 2019, 16:10
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If n > 4, what is the value of the integer n ?


(1) \(\frac{n!}{(n - 3)!} = \frac{3!n!}{4!(n - 4)!}\)

(2) \(\frac{n!}{3!(n - 3)!} + \frac{n!}{4!(n - 4)!} = \frac{(n + 1)!}{4!(n - 3)!}\)




DS47661.01

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Re: If n > 4, what is the value of the integer n ?  [#permalink]

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New post 21 Sep 2019, 16:22
for statement (1), it can be simplified to:
\(n(n-1)(n-2) = \frac{n(n-1)(n-2)(n-3)}{4}\)
\(4 = n-3\)
\(n = 7\) --> sufficient

for statement (2), upon simplification it will end with \(n = n\), which means no definite value --> insufficient

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If n > 4, what is the value of the integer n ?  [#permalink]

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New post Updated on: 19 Oct 2019, 01:20
Statement 1-
nC3=nC4
n=7

sufficient

Statement 2-

nC3+nC4= (n+1)C4
This is true for all n(>4).

{Pascal Rule- nC(k-1) + nCk= (n+1)Ck

Insufficient


gmatt1476 wrote:
If n > 4, what is the value of the integer n ?


(1) \(\frac{n!}{(n - 3)!} = \frac{3!n!}{4!(n - 4)!}\)

(2) \(\frac{n!}{3!(n - 3)!} + \frac{n!}{4!(n - 4)!} = \frac{(n + 1)!}{4!(n - 3)!}\)




DS47661.01

Originally posted by nick1816 on 18 Oct 2019, 16:30.
Last edited by nick1816 on 19 Oct 2019, 01:20, edited 1 time in total.
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Re: If n > 4, what is the value of the integer n ?  [#permalink]

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New post 19 Oct 2019, 00:41
nick - Can you explain why statement B is always true ?

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Re: If n > 4, what is the value of the integer n ?  [#permalink]

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New post 19 Oct 2019, 00:55
You can select k items from n+1 elements in 2 ways.

Either directly select k items from n+1 elements

or

[you can select k items from n elements] + [select k-1 items from those n element and add (n+1)th item]


Dug50 wrote:
nick - Can you explain why statement B is always true ?

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If n > 4, what is the value of the integer n ?  [#permalink]

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New post 19 Oct 2019, 00:59
gmatt1476 wrote:
If n > 4, what is the value of the integer n ?


(1) \(\frac{n!}{(n - 3)!} = \frac{3!n!}{4!(n - 4)!}\)

(2) \(\frac{n!}{3!(n - 3)!} + \frac{n!}{4!(n - 4)!} = \frac{(n + 1)!}{4!(n - 3)!}\)




DS47661.01


If n > 4, what is the value of the integer n ?


(1) \(\frac{n!}{(n - 3)!} = \frac{3!n!}{4!(n - 4)!}\)
1/(n-3) = 3!/4! = 1/4
n-3 =4
n=7
SUFFICIENT

(2) \(\frac{n!}{3!(n - 3)!} + \frac{n!}{4!(n - 4)!} = \frac{(n + 1)!}{4!(n - 3)!}\)
n!/4!(n-3)! [4 + n-3] = n!/4!(n-3)! (n+1) = (n+1)!/4!(n-3)!
Always true for all values of n except n=3
NOT SUFFICIENT

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If n > 4, what is the value of the integer n ?  [#permalink]

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New post 19 Oct 2019, 01:05
If n > 4, what is the value of the integer n ?


(1) \(\frac{n!}{(n - 3)!} = \frac{3!n!}{4!(n - 4)!}\)
Two ways..
(a) Combination formula
\(\frac{n!}{(n - 3)!} = \frac{3!n!}{4!(n - 4)!}\).....\(\frac{n!}{3!(n - 3)!} = \frac{n!}{4!(n - 4)!}.....nC3=nC4\)
Thus \(n=3+4=7\)..SUFF
(b) Arithmetic way
\(\frac{n!}{(n - 3)!} = \frac{3!n!}{4!(n - 4)!}\).......\(\frac{n(n-1)(n-2)(n-3)!}{(n - 3)!} = \frac{3!n(n-1)(n-2)(n-3)(n-4)!}{4!(n - 4)!}\)....\(4!*n(n-1)(n-2)=3!*(n)(n-1)(n-2)(n-3).....4=n-3...n=7..SUFF\)



(2) \(\frac{n!}{3!(n - 3)!} + \frac{n!}{4!(n - 4)!} = \frac{(n + 1)!}{4!(n - 3)!}\)
Two ways..
(a) Combination formula

\(\frac{n!}{3!(n - 3)!} + \frac{n!}{4!(n - 4)!} = \frac{(n + 1)!}{4!(n - 3)!}\).....\(.....nC3+nC4=(n+1)C4\)
Thus n could be anything as it is true for all..SUFF
(b) Arithmetic way
\(\frac{n!}{3!(n - 3)!} + \frac{n!}{4!(n - 4)!} = \frac{(n + 1)!}{4!(n - 3)!}\) ......
Take out n!/3!(n-3)!
\(\frac{n!}{3!(n-3)!}(1+\frac{n-3}{4} )=\frac{n!}{3!(n-3)!}(\frac{n+1}{4} ).....\frac{n+1}{4}=\frac{n+1}{4}\)....Always true..
This also tells us why nC3+nC4=(n+1)C4

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If n > 4, what is the value of the integer n ?   [#permalink] 19 Oct 2019, 01:05
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