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# A bookseller has two display windows. She plans to display 4 new fict

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Math Expert
Joined: 02 Sep 2009
Posts: 60605
A bookseller has two display windows. She plans to display 4 new fict  [#permalink]

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24 Feb 2015, 07:47
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87% (01:09) correct 13% (01:32) wrong based on 87 sessions

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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

Kudos for a correct solution.

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Posts: 8341
Re: A bookseller has two display windows. She plans to display 4 new fict  [#permalink]

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24 Feb 2015, 09:19
Bunuel 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

Kudos for a correct solution.

4 on left can be arranged in 4!.. and 3 on right in 3!..
total=4!*3!=144
ans C
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Re: A bookseller has two display windows. She plans to display 4 new fict  [#permalink]

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24 Feb 2015, 09:37
Two displays books could be arranged n! # of total arrangements are combinations multiplied together,
3!*4! =6*24=144
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Re: A bookseller has two display windows. She plans to display 4 new fict  [#permalink]

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24 Feb 2015, 11:18
Answer is combination of left window:

4! = 4 * 3 * 2

AND (means multiply)

3! = 3*2

together 4 * 3 * 2 * 3 * 2 = 144

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Posts: 1868
Re: A bookseller has two display windows. She plans to display 4 new fict  [#permalink]

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28 Feb 2015, 08:40
Bunuel 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

Kudos for a correct solution.

Left hand side books can be ordered in 4! ways = 24 ways
Right hand side books can be ordered in 3! ways = 6 ways

Total number of ways = 24 * 6
= 144
Hence option C.

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Posts: 60605
Re: A bookseller has two display windows. She plans to display 4 new fict  [#permalink]

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02 Mar 2015, 06:37
Bunuel 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

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

The left window will have permutations of the 4 fiction books, so the number of possibilities for that window is
permutations = 4! = (4)(3)(2)(1) = 24

The right window will have permutations of the 3 non-fiction books, so the number of possibilities for that window is
permutations = 3! = (3)(2)(1) = 6

Any of the 24 displays of the left window could be combined with any of the 6 displays of the right window, so the total number of configurations is 24*6 = 144

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Re: A bookseller has two display windows. She plans to display 4 new fict  [#permalink]

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14 Dec 2019, 19:47
Hi All,

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.

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Re: A bookseller has two display windows. She plans to display 4 new fict  [#permalink]

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23 Dec 2019, 18:20
Bunuel 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

Kudos for a correct solution.

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

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Re: A bookseller has two display windows. She plans to display 4 new fict   [#permalink] 23 Dec 2019, 18:20
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