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

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Bunuel
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Melilla, Tasha, and Chester plan to fly on a jetliner that has seven [#permalink] New post   20 Jul 2017, 21:59
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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] New post   20 Jul 2017, 22:31
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_ _ 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] New post   29 Oct 2019, 19:51
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Bunuel wrote:
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


Let’s say the seat numbers are:

AB CDE FG

We see that there are 7C2 = (7 x 6)/2 = 21 ways Chester and Melilla can choose 2 of the 7 seats to sit. However, in order for them to sit next to each other without an aisle between them, they must choose seats AB, CD, DE, or FG. Therefore, the probability they sit next to each other is 4/21. (Note: Tasha is a non-factor here since she can sit at any seat that is not chosen by Chester or Melilla.)

Answer: C
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Re: Melilla, Tasha, and Chester plan to fly on a jetliner that has seven [#permalink] New post   20 Feb 2023, 10:55
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6/7 of the time, M is not in the exact middle, and then there is only one seat next to M, so there's a 1/6 chance C is next to M.
1/7 of the time, M is in the exact middle, and then there are two seats next to M, so a 2/6 chance C is next to M.

So the answer is (6/7)(1/6) + (1/7)(2/6) = 4/21
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Melilla, Tasha, and Chester plan to fly on a jetliner that has seven [#permalink] New post   09 Dec 2023, 21:41
pushpitkc wrote:
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

_ _ 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)



I completely understand the overall approach to this problem. However, for any option we have 2 choices - C M as well as M C . Should not the actual answer be 4*2/21 = 8/21 (I am aware that this option is not given as one of the answer choices)

Can someone please explain?
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Re: Melilla, Tasha, and Chester plan to fly on a jetliner that has seven [#permalink] New post   10 Dec 2023, 01:00
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BusinessAdmin wrote:
pushpitkc wrote:
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

_ _ 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)



I completely understand the overall approach to this problem. However, for any option we have 2 choices - C M as well as M C . Should not the actual answer be 4*2/21 = 8/21 (I am aware that this option is not given as one of the answer choices)

Can someone please explain?


The point is that 7C2 also gives the number of unordered pairs. So, if you distinguish between CM and MC in the numerator, you should also account for the same in the denominator and multiply that by 2, resulting in the same answer of 4/21.

Hope it's clear.
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Re: Melilla, Tasha, and Chester plan to fly on a jetliner that has seven [#permalink] New post   18 Apr 2024, 14:18
Bunuel wrote:
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

­_ _ aisle _ _ _ aisle _ _
1 2        3 4 5        6 7

Melilla and Chester can sit in any 2 seats out of 7 seats in 7C2=21 ways.
Since Melilla booked 3 tickets in the same row, and Melilla and Chester sit without any aisle between them, then 3 4 5 are the only options left.
The number of ways they can be seated:
1. M=1 C=2
2. M=2 C=1
3. M=2 C=3
4. M=3 C=2

4/21. Option (C) is correct.
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Re: Melilla, Tasha, and Chester plan to fly on a jetliner that has seven [#permalink]
18 Apr 2024, 14:18
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