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# A rectangular table seats 4 people on each of two sides, with every pe

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Manager
Joined: 10 Feb 2008
Posts: 76
A rectangular table seats 4 people on each of two sides, with every  [#permalink]

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24 Mar 2010, 01:59
1
13
00:00

Difficulty:

75% (hard)

Question Stats:

53% (01:48) correct 47% (02:19) wrong based on 151 sessions

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A rectangular table seats 4 people on each of two sides, with every person directly facing another person across the table. If eight people choose their seats at random, what is probability that any two of them directly face other?

(A) 1/56
(B) 1/8
(C) 1/7
(D) 15/56
(E) 4/7

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A rectangular table seats 4 people on each of two sides, with every  [#permalink]

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24 Mar 2010, 12:41
3
anaik100 wrote:
A rectangular table seats 4 people on each of two sides, with every person directly facing another person across the table. If eight people choose their seats at random, what is probability that any two of them directly face other?

(A) 1/56
(B) 1/8
(C) 1/7
(D) 15/56
(E) 4/7

Person A can take any place, probability that person B will take the opposite seat is 1/7 (as there are 7 seats left).

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A rectangular table seats 4 people on each of two sides, with every pe  [#permalink]

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24 Jul 2016, 09:04
2
A B C D
------------
| TABLE |
------------
E F G H

A to H are 8 people.

Prob to select any 1 person = 1

Prob to select the person opposite to the chosen person = 1/7

For ex. If we select A as the person than prob of choosing E is 1/7.

Hence, answer will be C.

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Joined: 31 Aug 2014
Posts: 4
Re: A rectangular table seats 4 people on each of two sides, with every pe  [#permalink]

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28 Jul 2016, 10:27
2
We are given :

+ + + +
TABLE
+ + + +

Probability = # Favorable Outcomes / # Total Outcomes

Total Outcomes = 8 * 7 (Since a spot on the table for 2 people can be selected in 8 * 7 ways) = 56

Favorable Outcomes = 8 * 1 (One person can pick a seat in 8 ways, but to sit opposite to that person, there is only 1 option, thus 8*1) = 8

Probability = 8/56 = 1/7, thus the Answer is C.

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Re: A rectangular table seats 4 people on each of two sides, with every pe  [#permalink]

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28 Jul 2016, 21:38
3
Bunuel wrote:
A rectangular table seats 4 people on each of two sides, with every person directly facing another person across the table. If eight people choose their seats at random, what is probability that any two of them directly face other?

(A) 1/56
(B) 1/8
(C) 1/7
(D) 15/56
(E) 4/7

The question asks for the probability of any two people (say A and B) facing each other.

Method 1:
A can take any seat. After that B should take the seat facing A which is 1 of the 7 remaining seats. So probability of A and B facing each other = 1/7

Method 2:
In how many ways can 8 people can sit on a rectangular table with 4 chairs on 2 opposite sides? The first person can sit in 4 ways (there are 4 distinct seats with respect to the table. Think circular arrangement). Now there are 7 distinct seats and 7 people so they can sit in 7! ways.
Total number of ways = 4*7!

In how many ways can 8 people sit such that A and B face each other? A goes and sits in 4 ways (as before). B sits directly opposite him in 1 way only. Now we have 6 distinct seats and 6 people. They can sit in 6! ways.

Probability = 4*6!/4*7! = 1/7
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Re: A rectangular table seats 4 people on each of two sides, with every pe  [#permalink]

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31 Jul 2017, 01:28
1
1
anaik100 wrote:
A rectangular table seats 4 people on each of two sides, with every person directly facing another person across the table. If eight people choose their seats at random, what is probability that any two of them directly face other?

(A) 1/56
(B) 1/8
(C) 1/7
(D) 15/56
(E) 4/7

Consider those 2 persson A and B so -

Fav condition:
A can sit in any of seat around the table no of ways - 8
B has to take seat opposite to A no of ways - 1
other member can take seat other 6 seats no of ways - 6!

All condition:
all possible combination without restriction no of ways - 8!

$$prob = fav/all => 8*1*6!/8!$$
$$prob = 1/7$$
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Re: A rectangular table seats 4 people on each of two sides, with every pe  [#permalink]

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09 Aug 2017, 07:08
VeritasPrepKarishma wrote:
Bunuel wrote:
A rectangular table seats 4 people on each of two sides, with every person directly facing another person across the table. If eight people choose their seats at random, what is probability that any two of them directly face other?

(A) 1/56
(B) 1/8
(C) 1/7
(D) 15/56
(E) 4/7

The question asks for the probability of any two people (say A and B) facing each other.

Method 1:
A can take any seat. After that B should take the seat facing A which is 1 of the 7 remaining seats. So probability of A and B facing each other = 1/7

Method 2:
In how many ways can 8 people can sit on a rectangular table with 4 chairs on 2 opposite sides? The first person can sit in 4 ways (there are 4 distinct seats with respect to the table. Think circular arrangement). Now there are 7 distinct seats and 7 people so they can sit in 7! ways.
Total number of ways = 4*7!

In how many ways can 8 people sit such that A and B face each other? A goes and sits in 4 ways (as before). B sits directly opposite him in 1 way only. Now we have 6 distinct seats and 6 people. They can sit in 6! ways.

Probability = 4*6!/4*7! = 1/7

To solve this question we can do it by 2 methods....
Let the 2 persons facing each other be A & B
Method 1 : Place A at a position... Now there are 7 spaces for B to seat but in only one case it is opposite to B. So probability of B facing A = 1/7.

Method 2 : Counting total no. of arrangements
So, Total no. of ways of seating all the 8 people = 8!
(VeritasPrepKarishma I disagree with you here for total no. of ways. Lets A is seated at corner seat then left and right seating of A is very different in nature. Hence it may not be considered as circular arrangement.)

Total no. of desired ways in which A & B faces each other = 4 (pair of opposite seats) * 2 (arranging A and B on the pair of opposite seat) * 6! (arranging other 6 people on the 6 seats) = 4*2*6!

So required probability = 4*2*6!/8! = 1/7

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A rectangular table seats 4 people on each of two sides, with every pe  [#permalink]

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28 Sep 2018, 16:00
Hello. Can someone tell me if my approach is valid?

Total possible ways are 8C2 = 28

Favorable outcome (choosing 1 person from 1 side) = 4c1= 4

4C1/ 8C2 = 4/28 = 1/7
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Re: A rectangular table seats 4 people on each of two sides, with every pe  [#permalink]

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28 Sep 2018, 18:59
shashankism wrote:
VeritasPrepKarishma wrote:
Bunuel wrote:
A rectangular table seats 4 people on each of two sides, with every person directly facing another person across the table. If eight people choose their seats at random, what is probability that any two of them directly face other?

(A) 1/56
(B) 1/8
(C) 1/7
(D) 15/56
(E) 4/7

The question asks for the probability of any two people (say A and B) facing each other.

Method 1:
A can take any seat. After that B should take the seat facing A which is 1 of the 7 remaining seats. So probability of A and B facing each other = 1/7

Method 2:
In how many ways can 8 people can sit on a rectangular table with 4 chairs on 2 opposite sides? The first person can sit in 4 ways (there are 4 distinct seats with respect to the table. Think circular arrangement). Now there are 7 distinct seats and 7 people so they can sit in 7! ways.
Total number of ways = 4*7!

In how many ways can 8 people sit such that A and B face each other? A goes and sits in 4 ways (as before). B sits directly opposite him in 1 way only. Now we have 6 distinct seats and 6 people. They can sit in 6! ways.

Probability = 4*6!/4*7! = 1/7

To solve this question we can do it by 2 methods....
Let the 2 persons facing each other be A & B
Method 1 : Place A at a position... Now there are 7 spaces for B to seat but in only one case it is opposite to B. So probability of B facing A = 1/7.

Method 2 : Counting total no. of arrangements
So, Total no. of ways of seating all the 8 people = 8!
(VeritasPrepKarishma I disagree with you here for total no. of ways. Lets A is seated at corner seat then left and right seating of A is very different in nature. Hence it may not be considered as circular arrangement.)

Total no. of desired ways in which A & B faces each other = 4 (pair of opposite seats) * 2 (arranging A and B on the pair of opposite seat) * 6! (arranging other 6 people on the 6 seats) = 4*2*6!

So required probability = 4*2*6!/8! = 1/7

Look at this:

Attachment:

Fig1.jpeg [ 22.95 KiB | Viewed 661 times ]

If the first person comes and sits at A, it is exactly the same as sitting at D (table corner on his left and a person on his right, the shorter side of the rectangle on the left). Similarly, positions B and C are exactly the same.
Using same logic for the shorter sides of the rectangle too, we will get 4 unique places to sit for the first person, not 8 (with respect to the table)
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Posts: 379
Re: A rectangular table seats 4 people on each of two sides, with every pe  [#permalink]

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28 Sep 2018, 21:14
hdavies wrote:
Hello. Can someone tell me if my approach is valid?

Total possible ways are 8C2 = 28

Favorable outcome (choosing 1 person from 1 side) = 4c1= 4

4C1/ 8C2 = 4/28 = 1/7

Can you please comment on the above approach whether it is right?
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Re: A rectangular table seats 4 people on each of two sides, with every pe &nbs [#permalink] 28 Sep 2018, 21:14
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# A rectangular table seats 4 people on each of two sides, with every pe

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