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

Final Answer:

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Re: A group of 5 students bought movie tickets in one row next to each oth [#permalink]
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"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 to each oth [#permalink]
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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 to each oth [#permalink]
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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 to each oth [#permalink]
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 to each oth [#permalink]
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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.

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Re: A group of 5 students bought movie tickets in one row next to each oth [#permalink]
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 to each oth [#permalink]
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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 5 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,
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Originally posted by EMPOWERgmatRichC on 19 Oct 2016, 20:57.
Last edited by EMPOWERgmatRichC on 13 Apr 2020, 16:09, edited 1 time in total.
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Re: A group of 5 students bought movie tickets in one row next to each oth [#permalink]
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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%
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Re: A group of 5 students bought movie tickets in one row next to each oth [#permalink]
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%


Can Anyone explain me. It is said that Bob and Lisa must seat next to each other. Why It is B_ _ _ L or L _ _ _ B?
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A group of 5 students bought movie tickets in one row next to each oth [#permalink]
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User_user wrote:
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%


Can Anyone explain me. It is said that Bob and Lisa must seat next to each other. Why It is B_ _ _ L or L _ _ _ B?


Hi User_user,

The prompt asks for the probably that both Bob and Lisa will EACH sit next to JUST ONE person... When sitting in a row of 5 people, the only way to sit next to just one person is to sit on the 'end' of the row. There is no additional information (re: "restrictions") about where anyone is supposed to sit, so there are two possible outcomes that meet the given information:

B _ _ _ L
L _ _ _ B

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

Originally posted by EMPOWERgmatRichC on 13 Apr 2020, 13:14.
Last edited by EMPOWERgmatRichC on 13 Apr 2020, 16:10, edited 1 time in total.
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Re: A group of 5 students bought movie tickets in one row next to each oth [#permalink]
How long are people taking to solve this one? I know that the average time is ~2 min. but just trying to figure out pacing.
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Re: A group of 5 students bought movie tickets in one row next to each oth [#permalink]
jjigae1985 wrote:
How long are people taking to solve this one? I know that the average time is ~2 min. but just trying to figure out pacing.


This question has 2 aspects that can determine your pacing -

1) Understanding exactly what the question is asking
2) Knowledge on combinatorics and probability

I took exactly 1:00 min to answer this one, most of which went into understanding what the qs really asked i.e. B _ _ _ L or L _ _ _ B. With good knowledge on combinatorics, calculations would seem pretty easy to do so can be done in less than a min if you understood what its asking
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Re: A group of 5 students bought movie tickets in one row next to each oth [#permalink]
Bunuel 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%

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.

Answer: B.


Why is it that the indistinguishable objects formula is not used here, when XYZ are indistinguishable, and B and L are indistinguishable from each other?
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Re: A group of 5 students bought movie tickets in one row next to each oth [#permalink]
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onlydode wrote:
Bunuel 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%

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.

Answer: B.


Why is it that the indistinguishable objects formula is not used here, when XYZ are indistinguishable, and B and L are indistinguishable from each other?


Because the objects are people, which are actually distinguishable. Bob and Lisa are distinguishable from each other, and each of the other three students is also distinguishable from one another, making for instance BXYZL sitting arrangement different from LXYZB or from LZYXB sitting arrangementS.
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Re: A group of 5 students bought movie tickets in one row next to each oth [#permalink]
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