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Difficulty:
75%
(hard)
Question Stats:
62% (02:49) correct
38%
(02:38)
wrong
based on 2019
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Date
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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
(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
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
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
\(n(n-1)(n-2) = \frac{n(n-1)(n-2)(n-3)}{4}\)
\(4 = n-3\)
\(n = 7\) --> sufficient
for statement (2), upon simplification it will end with \(n = n\), which means no definite value --> insufficient
A
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
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
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]
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
Kinshook
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gmatt1476
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
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.
(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
_________________
(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.
(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
_________________
Posts: 1
I feel the answer should be D. As on solving 2nd condition,we get certain value of x as 5.
Posted from my mobile device
Posted from my mobile device
Posts: 1,015
Even during the exam if the combination formula does not strike, then also its fine. Most of the stuff cancels out easily. Just be aware of the trap for statement 2.
Can some one please explain the simplification proces of the statements step by step? Could you please do not skip any mini step in the middle as well.
Thanks in advance!
Thanks in advance!
Posts: 49
Don’t get flummoxed by complicated equations and factorials. Try to break down factorials to cancel a few of them from both sides.
Check the attachment. Answer is A.
Hope this helps.
Check the attachment. Answer is A.
Hope this helps.
Attachments
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can someone please explain scenario 2 using the combinations formula? i'm not sure i'm understanding why n could be anything
chetan2u
how is (N-3)! = (N-3).(N-4)! ?
Schools: Oxford Said'25 (A)
Posts: 10
DV221
(N-3)! can be written as (N-3) (N-4) (N-5)... until you reach 1, whatever the value of N is.
Here, you will notice if you ignore (N-3) for a minute, the rest of the expansion denotes, in fact, the expansion for (N-4)!.
And hence, (N-3)! = (N-3) (N-4)!
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