Factorial DS

D

1) n! / (3! (N-3)!) = N! / (4! (N-4)!)

1 / (3! (N-3)) = 1 / (4!)
1 / (N-3) = 1 / 4

2) n! / (3! (N-3)!) + N! / (4! (N-4)!) = (N +1)! / (4! (N-4)!)
n! / (3! (N-3)!) = (N +1)! / (4! (N-4)!) - N! / (4! (N-4)!)
n! / (3! (N-3)!) = N! * N / (4! (N-4)!)
1 / (3! (N-3)) = N / (4!)
1 / (N-3) = N / (4)
N^2 - 3N -4 = 0
(N+1) ( N-4) = 0
N = -1 or 4, N = 4 because -1! does not exist
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are "n" and "N" different?

No. They are the same. It is a typo

I would go for D.

1) n! / (3! (n-3)!) = n! / (4! (n-4)!)
=> n(n-1)(n-2)(n-3)! / 3! (n-3)! = n(n-1)(n-2)(n-3)(n-4)! / 4! (n-4)!
=> n(n-1)(n-2) / 3! = n(n-1)(n-2)(n-3) / 4!
=> n(n-1)(n-2) / n(n-1)(n-2)(n-3) = 3! / 4!
=> 1 / (n-3) = 3! / 4!
=> 1 / (n-3) = 1 / 4
=> n-3 = 4
=> n = 7

Solving in the similar way, n will come out to be 7 in the second case.

IMO A

Cos A gives 7 and B gives 4 and its given N>4

anand0408 wrote:IMO A
Cos A gives 7 and B gives 4 and its given N>4

ditto anand0408.

Also, can we really use what Pussyin Boots said about -1! not existent? I mean I agree that it doesn't, but does it really mean that N cannot be -1? (assume there was no condition that N>4)

Another reason it can't be D is because GMAT never gives conflicting conditions in 1 and 2 that are both correct. N cannot be both 7 and 4.

OA Please

Non-existance of answer is still an answer. I stand by D. As far as I know, Data Sufficiency asks if (A) or (B) is enough to come up with the answer, even if they are difference in each case.

Here is a question:

What's x equals to?
(A) x-3 = 3
(B) x*5 = 10
Clearly the answer is D