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# A group of 5 students bought movie tickets in one row next

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Joined: 10 Jun 2007
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A group of 5 students bought movie tickets in one row next  [#permalink]

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Updated on: 09 Sep 2013, 02:58
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18
00:00

Difficulty:

65% (hard)

Question Stats:

58% (01:13) correct 42% (01:33) wrong based on 807 sessions

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A group of 5 students bought movie tickets in one row next to each other. If Bob and Lisa are in this group, what is the probability that Bob and Lisa will each sit next to only one of the four other students from the group?

(A) 5%
(B) 10%
(C) 15%
(D) 20%
(E) 25%

Originally posted by bkk145 on 20 Jun 2007, 17:03.
Last edited by Bunuel on 09 Sep 2013, 02:58, edited 2 times in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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Joined: 02 Sep 2009
Posts: 49271
A group of 5 students bought movie tickets in one row next  [#permalink]

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09 Sep 2013, 02:59
7
6
A group of 5 students bought movie tickets in one row next to each other. If Bob and Lisa are in this group, what is the probability that Bob and Lisa will each sit next to only one of the four other students from the group?
(A) 5%
(B) 10%
(C) 15%
(D) 20%
(E) 25%

The question basically asks about the probability that Bob and Lisa sit at the ends.

The total # of sitting arrangements is 5!.

Desired arrangement is either BXYZL or LXYZB. Now, XYZ can be arranged in 3! ways, therefore total # of favorable arrangements is 2*3!.

P=(favorable)/(total)=(2*3!)/5!=1/10.

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Re: A group of 5 students bought movie tickets in one row next  [#permalink]

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20 Feb 2015, 13:41
1
Hi All,

There are several different 'math' approaches to this question. You have to be clear that there are 2 'acceptable' orientations of the 5 people though:

Bob, X, X, X, Lisa
Lisa, X, X, X, Bob

Since either Bob OR Lisa would have to be in the 'first spot', there's a 2/5 probability of that happening....

That would leave the other person to occupy the 'fifth spot'; since there are 4 people remaining, there's a 1/4 probability of that happening....

(2/5)(1/4) = 2/20 = 1/10 = 10%

GMAT assassins aren't born, they're made,
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Re: A group of 5 students bought movie tickets in one row next  [#permalink]

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21 Feb 2015, 23:26
"what is the probability that Bob and Lisa will each sit next to only one of the four other students from the group?"

Can someone please explain why the arrangement can't be " _ B _ L _ " or " _ L _ B _ "?
Does this not also satisfy that B & L also sit next to the other four students?
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Re: A group of 5 students bought movie tickets in one row next  [#permalink]

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21 Feb 2015, 23:41
1
Hi icetray,

The prompt tells us that the 5 seats are in a row. For Bob/Lisa to sit next to JUST 1 student, they would both have to sit on the ENDS of the row. In the examples that you listed, Bob and Lisa are sitting next to 2 students, which does not fit the "restrictions" of what we're asked to deal with.

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Re: A group of 5 students bought movie tickets in one row next  [#permalink]

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28 Feb 2016, 23:46
A group of 5 students bought movie tickets in one row next to each other. If Bob and Lisa are in this group, what is the probability that Bob and Lisa will each sit next to only one of the four other students from the group?

(A) 5%
(B) 10%
(C) 15%
(D) 20%
(E) 25%

The most simplified version:

, Lisa and Bob have 2 plots to choose from 5 seats i.e, extremities. other 3 have 3! ways to choose seats and Bob and Lisa have 2! ways. Multiplying we get 3!*2!=12 ways. To find Lisa's and Bob's position Divide. total ways 5 people= 5!/3!*2! = 120/12= 10%. Hope this made sense.
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Re: A group of 5 students bought movie tickets in one row next  [#permalink]

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07 Sep 2016, 16:18
bkk145 wrote:
A group of 5 students bought movie tickets in one row next to each other. If Bob and Lisa are in this group, what is the probability that Bob and Lisa will each sit next to only one of the four other students from the group?

(A) 5%
(B) 10%
(C) 15%
(D) 20%
(E) 25%

So bob and lisa have to sit next to one another and cannot occupy corner seats
Seating arrangement :
x - - - X

this can be done in 3! * 2! ways

so probability would be 3! * 2! / 5! = 12 /120 = 1/10 = 10%

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Re: A group of 5 students bought movie tickets in one row next  [#permalink]

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07 Sep 2016, 20:29
1
There are 2 different ways this can happen

LxxxB or BxxxL

Here is the probability of this happening (1/5)(3/4)(2/3)(1/2)(1/1)=2/20 or 1/10 10%
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Re: A group of 5 students bought movie tickets in one row next  [#permalink]

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17 Oct 2016, 02:04
Hi all,
I have a problem with the question.
what is the probability that Bob and Lisa will each sit next to only one of the four other students.
Is the question saying that Bob and Lisa will sit at the extremes. However, I solve it treating subgroup like (B _ L) X X X and then solve for possible arrangement and got the same answer. But the official answer says that Bob and Lisa sat at the extremes. Is my understanding wrong?
Help me.
TIA
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Re: A group of 5 students bought movie tickets in one row next  [#permalink]

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17 Oct 2016, 14:04
NaeemHasan wrote:
Hi all,
I have a problem with the question.
what is the probability that Bob and Lisa will each sit next to only one of the four other students.
Is the question saying that Bob and Lisa will sit at the extremes. However, I solve it treating subgroup like (B _ L) X X X and then solve for possible arrangement and got the same answer. But the official answer says that Bob and Lisa sat at the extremes. Is my understanding wrong?
Help me.
TIA

Hi NaeemHasan,

In the example that you listed, Lisa is sitting next to TWO people - and that is not in line with what the question states.

GMAT assassins aren't born, they're made,
Rich
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Re: A group of 5 students bought movie tickets in one row next  [#permalink]

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17 Oct 2016, 14:28
Probability that each will sit on the outside

1/5*3/4*2/3*1/2*1/1=1/20. This can happen two ways Bob is first Lisa is last, or Lisa is first and Bob is last. 1/20*2=2/20 or 10%
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Re: A group of 5 students bought movie tickets in one row next  [#permalink]

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19 Oct 2016, 02:23
EMPOWERgmatRichC wrote:
NaeemHasan wrote:
Hi all,
I have a problem with the question.
what is the probability that Bob and Lisa will each sit next to only one of the four other students.
Is the question saying that Bob and Lisa will sit at the extremes. However, I solve it treating subgroup like (B _ L) X X X and then solve for possible arrangement and got the same answer. But the official answer says that Bob and Lisa sat at the extremes. Is my understanding wrong?
Help me.
TIA

Hi NaeemHasan,

In the example that you listed, Lisa is sitting next to TWO people - and that is not in line with what the question states.

GMAT assassins aren't born, they're made,
Rich

Hi,
How? I allow only one person to sit between Bob and LIsa. Does this mean they sit next to two people? If so, Could you elaborate this, please.
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A group of 5 students bought movie tickets in one row next  [#permalink]

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19 Oct 2016, 20:57
2
Hi NaeemHasan,

In the example: B _ L _ _ _

Bob is sitting next to one person and Lisa is sitting next to two people.

The prompt asks for the probably that both Bob and Lisa will sit next to ONE person... NOT that Bob and Lisa will sit next to the SAME person. When sitting in a row of 6 people, the only way to sit next to just one person is to sit on the 'end' of the row. Thus, there are only two possible outcomes that meet this restriction:

B _ _ _ _ L
L _ _ _ _ B

GMAT assassins aren't born, they're made,
Rich
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Re: A group of 5 students bought movie tickets in one row next  [#permalink]

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11 Nov 2017, 00:08
IMO:

There are total 5 seats in a row.
Bob and Lisa requires to sit in such a way that each one of them have only 1 person next to them.
Hence they can be seated only at the end of the row.
Now there are total 2 ways to sit for Bob and Lisa -either Bob on leftmost corner and Lisa on rightmost corner or vice versa

Now, this question is about probability, so we need to find- (Favorable cases) and ( Total cases).

Favorable cases=
1) B/L_ _ _ L/B

In between 3 vacant seats can be filled by remaining 3x2x1=6 ways.
Edge seats can be filled in 2 ways-Lisa on left -Bob on right or Lisa on right and Bob on left.
Hence Favorable cases= 6x2=12 ways

Total cases=
_ _ _ _ _

5 seats can be filled in 5 ! ways.
Hence Total cases= 5!

Hence probability of Bob or Lisa at the edge seats is=

Favorable/ Total= 12/5! x 100 = 10%

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Re: A group of 5 students bought movie tickets in one row next  [#permalink]

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12 Nov 2017, 07:39
1
What I understand is that Bob and Lisa should have only person sitting next to them, this implies they take the edge seat
So total arrangement=5!=120
now the arrangement to meet the condition=3!2!( Arranging Bob and Lisa on edge seat=2!, Arranging the rest=3!)
So probability= 3!2!/5!=1/10=10%
Re: A group of 5 students bought movie tickets in one row next &nbs [#permalink] 12 Nov 2017, 07:39
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