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

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
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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

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

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

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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]

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

I feel the answer should be D. As on solving 2nd condition,we get certain value of x as 5.

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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.
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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!

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.

can someone please explain scenario 2 using the combinations formula? i'm not sure i'm understanding why n could be anything

how is (N-3)! = (N-3).(N-4)! ?

(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)!