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

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

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21 Sep 2019, 16:10
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Difficulty:

95% (hard)

Question Stats:

48% (02:59) correct 52% (02:49) wrong based on 108 sessions

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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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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

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

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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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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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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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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

IMO A
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Posts: 8281
If n > 4, what is the value of the integer n ?  [#permalink]

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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

A
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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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