Hi M7MBA,

We're told that a bookseller has two display windows - and she plans to display 4 new fiction books in the left window and 3 new non-fiction books in the right window. Assuming she can put the four fiction books in any order, and separately, the three non-fiction books in any order, how many total configurations will there be for the two display windows. This is a mid-level Permutation question and requires just a bit of multiplication to solve.

To start, since we are putting the books in order (without any restrictions), we can use basic multiplication (re: factorials) to solve:

For the 4 fiction books, there are 4! = (4)(3)(2)(1) = 24 ways to arrange those books in the window
For the 3 non-fiction books, there are 3! = (3)(2)(1) = 6 ways to arrange those books in the window
In total, we have (24)(6) = 144 different ways to arrange the books.

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

M7MBA wrote:A bookseller has two display windows. She plans to display 4 new fiction books in the left window, and 3 new non-fiction books in the right window. Assuming she can put the four fiction books in any order, and separately, the three non-fiction books in any order, how many total configurations will there be for the two display windows?

A. 24
B. 72
C. 144
D. 336
E. 420

[spoiler]OA=C[/spoiler]

Source: Magoosh

The number of configurations is 4! x 3! = 24 x 6 = 144.

Answer: C