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Melilla, Tasha, and Chester plan to fly on a jetliner that has seven

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V
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Posts: 54371
Melilla, Tasha, and Chester plan to fly on a jetliner that has seven  [#permalink]

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New post 20 Jul 2017, 21:59
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A
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C
D
E

Difficulty:

  65% (hard)

Question Stats:

48% (02:49) correct 52% (03:12) wrong based on 38 sessions

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Melilla, Tasha, and Chester plan to fly on a jetliner that has seven seats per row. The middle row consists of three seats, which are separated on each side by an aisle and two seats. The website allows users to choose specific seats, but a computer glitch only assigns rows, not the specific seat within that row. Assuming Melilla plans to book three tickets in the same row, what is the probability that Chester will sit next to Melilla, without any aisle between them?

A. 3/35
B. 1/5
C. 4/21
D. 6/35
E. 1/7

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Melilla, Tasha, and Chester plan to fly on a jetliner that has seven  [#permalink]

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New post 20 Jul 2017, 22:31
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1
_ _ Aisle1_ _ _Aisle 2_ _ is the seating arrangement.

The total ways in which we could pick 2 seats out of 7, is 7c2 = \(\frac{7*6}{2} = 21\)
because there are 7 seats in the same row, and the computer could assign any of the seats.

If they have to sit without any aisle between then,
there are four combinations in which they could be seated without any aisle between them.
They are as follows :

C M Aisle1 _ _ _ Aisle 2 _ _
_ _ Aisle 1 C M _ Aisle 2 _ _
_ _ Aisle 1 _ C M Aisle 2 _ _
_ _ Aisle1 _ _ _ Aisle 2 C M

The probability is \(\frac{4}{21}\)(Option C)
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Re: Melilla, Tasha, and Chester plan to fly on a jetliner that has seven  [#permalink]

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New post 21 Jul 2017, 22:05
pushpitkc wrote:
_ _ Aisle1_ _ _Aisle 2_ _ is the seating arrangement.

The total combinations in which Chester will sit next to Melila, is 7c2 = \(\frac{7*6}{2} = 21\)
because there are 7 seats in the same row, and the computer could assign any of the seats.
If they have to sit without any aisle between then,

There are four combinations in which they could be seated without any aisle between them.
They are as follows :

C M Aisle1 _ _ _ Aisle 2 _ _
_ _ Aisle 1 C M _ Aisle 2 _ _
_ _ Aisle 1 _ C M Aisle 2 _ _
_ _ Aisle1 _ _ _ Aisle 2 C M

The probability is \(\frac{4}{21}\)(Option C)


pushpitkc
7c2 is the number of ways to pick 2 seats from 7 seats,,,neednt necessarily be that chester and melila are next to each other,,
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Re: Melilla, Tasha, and Chester plan to fly on a jetliner that has seven  [#permalink]

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New post 22 Jul 2017, 02:40
mohshu wrote:
pushpitkc wrote:
_ _ Aisle1_ _ _Aisle 2_ _ is the seating arrangement.

The total combinations in which Chester will sit next to Melila, is 7c2 = \(\frac{7*6}{2} = 21\)
because there are 7 seats in the same row, and the computer could assign any of the seats.
If they have to sit without any aisle between then,

There are four combinations in which they could be seated without any aisle between them.
They are as follows :

C M Aisle1 _ _ _ Aisle 2 _ _
_ _ Aisle 1 C M _ Aisle 2 _ _
_ _ Aisle 1 _ C M Aisle 2 _ _
_ _ Aisle1 _ _ _ Aisle 2 C M

The probability is \(\frac{4}{21}\)(Option C)


pushpitkc
7c2 is the number of ways to pick 2 seats from 7 seats,,,neednt necessarily be that chester and melila are next to each other,,



Hi Mohshu,

The question is asking to select 2 seats[for Melila and Chester to sit nxt to each other] from 7.

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Re: Melilla, Tasha, and Chester plan to fly on a jetliner that has seven   [#permalink] 22 Jul 2017, 02:40
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Melilla, Tasha, and Chester plan to fly on a jetliner that has seven

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