Bunuel
There are 9 books; 2 by Sidney, 3 by Archer and the rest by John. These books are to be arranged linearly on a shelf such that:
i. no 2 books by John are adjacent.
ii. at least one book by Archer is at an end.
iii. except for John and Archer, the books by the same author are indistinguishable.
What is the possible number of arrangements?
(A) 8208
(B) 3450
(C) 3208
(D) 2595
(E) 1290
kapoor6295A fun one, but DEFinitely tougher than any P or C question you're going to see on the GMAT, so don't sweat it!!
Since an A must be on one end, there are three possible scenarios:
1. A on one end, J on the other.
2. A on one end, S on the other.
3. A on both ends.
Scenario 1:
There are 3 A's so we have 3 choices for which A is on the end and 2 choices for which end.
That's 6 ways to position the first A. Once we've done that, the J's can occupy one of four sets of slots starting from the A that's already placed:
AJxJxJxxJ
AJxJxxJxJ
AJxxJxJxJ
AxJxJxJxJ
The 4 J's can be arranged in 4*3*2*1 ways. So there are 4*4! ways to slot the J's.
That's 96 ways to position the J's.And once we've done that, there are 2 A's and 2 S's left to fill the four remaining slots. The 2 A's are distinguishable and the 2 S's are not. So, we have 4!/2 ways to slot the remaining four books.
That's 12 ways to position the remaining four books.6*96*12 = roughly 6*100*12 = 7200.
That's already enough to eliminate B, C, D, and E, so we don't even need to look at scenario 2 or scenario 3.
Answer choice A.
Edited to add: Wait, this is categorized as 600-700 level? Is there an easier way? Kudos if you show one!!