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\(7\) people, \(A, B, C, D, E, F,\) and \(G,\) go to a movie and sit next to each other in \(7\) adjacent seats in the

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by VJesus12 » Thu Oct 07, 2021 2:17 am

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\(7\) people, \(A, B, C, D, E, F,\) and \(G,\) go to a movie and sit next to each other in \(7\) adjacent seats in the front row of the theater. How many different arrangements are possible? If \(A\) will not sit to the left of \(F\) and \(F\) will not sit to the left of \(E.\) How many different arrangements are possible.

A) \(\dfrac{7!}2\)

B) \(\dfrac{7!}3\)

C) \(\dfrac{7!}4\)

D) \(\dfrac{7!}5\)

E) \(\dfrac{7!}6\)

Answer: E

Source: Veritas Prep
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Source: — Problem Solving |

The total number of arrangements without limitations is 7!.

The limitations stated indicate that
EFA is the relative seating order of those three with the remaining 4 distributed without limitations, including among the 3.

Recognize that 7! allows for E,F,and A to be distributed in any order.

E,F,and A can be distributed in 3! ways =6.

However, only 1 of the 6 is permissible,and so needs to be divided out.

Answer [spoiler]7!/6, E[/spoiler]
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