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A certain gallery is hanging a series of 7 paintings

Expert replies
by BTGmoderatorDC » Wed Feb 21, 2018 4:10 am
A certain gallery is hanging a series of 7 paintings. All the paintings will be exhibited in a row along a single wall. Exactly 2 of the paintings are on panel, the remainder are on canvas. In how many ways can the paintings be exhibited if the works on panel must be the second and sixth in the row?

A. 240
B. 200
C. 122
D. 80
E. 16

Can some experts give me the best solution in this problem?

OA A
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Source: — Problem Solving |

by GMATGuruNY » Wed Feb 21, 2018 4:29 am
lheiannie07 wrote:A certain gallery is hanging a series of 7 paintings. All the paintings will be exhibited in a row along a single wall. Exactly 2 of the paintings are on panel, the remainder are on canvas. In how many ways can the paintings be exhibited if the works on panel must be the second and sixth in the row?

A. 240
B. 200
C. 122
D. 80
E. 16
Number of options for the 2nd painting = 2. (Either of the 2 panel paintings.)
Number of options for the 6th painting = 1. (Must be the 1 remaining panel painting.)
Number of ways to arrange the remaining 5 paintings = 5! = 120.
To combine the options above, we multiply:
2*1*120 = 240.

The correct answer is A.
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by Scott@TargetTestPrep » Mon Jun 17, 2019 4:34 pm
BTGmoderatorDC wrote:A certain gallery is hanging a series of 7 paintings. All the paintings will be exhibited in a row along a single wall. Exactly 2 of the paintings are on panel, the remainder are on canvas. In how many ways can the paintings be exhibited if the works on panel must be the second and sixth in the row?

A. 240
B. 200
C. 122
D. 80
E. 16

Can some experts give me the best solution in this problem?

OA A
There are 2! ways to arrange the 2 paintings on panel and 5! ways to arrange the 5 paintings on canvas. Therefore, the total number of ways to arrange the 7 paintings is 2! x 5! = 2 x 120 = 240.

Answer: A

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