It is currently 23 Nov 2017, 20:49

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Current Student
Joined: 21 Feb 2015
Posts: 10

Kudos [?]: 3 [0], given: 11

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

### Show Tags

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?

Kudos [?]: 3 [0], given: 11

EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 10178

Kudos [?]: 3536 [1], given: 173

Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: 340 Q170 V170
Re: A group of 5 students bought movie tickets in one row next [#permalink]

### Show Tags

21 Feb 2015, 23:41
1
KUDOS
Expert's post
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.

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

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

# Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save \$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************

Kudos [?]: 3536 [1], given: 173

Intern
Joined: 13 Sep 2015
Posts: 7

Kudos [?]: 3 [0], given: 25

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

### Show Tags

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.

Kudos [?]: 3 [0], given: 25

Current Student
Joined: 03 May 2015
Posts: 262

Kudos [?]: 88 [0], given: 23

Location: South Africa
GPA: 3.49
WE: Web Development (Insurance)
Re: A group of 5 students bought movie tickets in one row next [#permalink]

### Show Tags

19 Jul 2016, 23:40
mavery wrote:
I may be oversimplifying this, but isn't it just asking "what is the probability that Bob and Lisa are sitting on the outside of the group?"

1/5 * 1/4 = 1/20 = 5%

Disclaimer: I really suck at these problems, so I'm probably wrong.

Your approach is correct. But you forgot that there are 2 cases It could be B XXX L or L XXX B. So answer is your answer X 2
_________________

Kudos if I helped

Kudos [?]: 88 [0], given: 23

Manager
Joined: 18 Jun 2016
Posts: 105

Kudos [?]: 21 [0], given: 76

Location: India
Concentration: Technology, Entrepreneurship
GMAT 1: 700 Q49 V36
Re: A group of 5 students bought movie tickets in one row next [#permalink]

### Show Tags

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%

_________________

If my post was helpful, feel free to give kudos!

Kudos [?]: 21 [0], given: 76

Manager
Joined: 26 Jan 2016
Posts: 115

Kudos [?]: 24 [1], given: 55

Location: United States
GPA: 3.37
Re: A group of 5 students bought movie tickets in one row next [#permalink]

### Show Tags

07 Sep 2016, 20:29
1
KUDOS
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%

Kudos [?]: 24 [1], given: 55

Intern
Joined: 15 Aug 2012
Posts: 19

Kudos [?]: 4 [0], given: 44

Schools: AGSM '19
Re: A group of 5 students bought movie tickets in one row next [#permalink]

### Show Tags

16 Sep 2016, 18:11
joannaecohen wrote:
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%

This is what I did too and was gonna reply until I saw this post. I'm glad my reasoning was right or we're both wrong

To explain again. Based on what the question asked, possibilities are BXXXL or LXXXB.

Probability of BXXXL happening:

B (1/5)*X(3/4)*X(2/3)*X(1/2)*L(1/1) = 1/20

Probability of LXXXB happening is also the same as above: 1/20

So probability of BXXXL OR LXXXB happening is 1/20+1/20=2/20=1/10=10%

Kudos [?]: 4 [0], given: 44

Manager
Joined: 06 Oct 2015
Posts: 89

Kudos [?]: 16 [0], given: 49

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

### Show Tags

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

Kudos [?]: 16 [0], given: 49

Math Expert
Joined: 02 Sep 2009
Posts: 42338

Kudos [?]: 133161 [0], given: 12415

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

### Show Tags

17 Oct 2016, 03:14
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

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

Check here: a-group-of-5-students-bought-movie-tickets-in-one-row-next-47450.html#p1265145
_________________

Kudos [?]: 133161 [0], given: 12415

EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 10178

Kudos [?]: 3536 [0], given: 173

Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: 340 Q170 V170
Re: A group of 5 students bought movie tickets in one row next [#permalink]

### Show Tags

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
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

# Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save \$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************

Kudos [?]: 3536 [0], given: 173

Manager
Joined: 26 Jan 2016
Posts: 115

Kudos [?]: 24 [0], given: 55

Location: United States
GPA: 3.37
Re: A group of 5 students bought movie tickets in one row next [#permalink]

### Show Tags

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%

Kudos [?]: 24 [0], given: 55

Manager
Joined: 06 Oct 2015
Posts: 89

Kudos [?]: 16 [0], given: 49

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

### Show Tags

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.

Kudos [?]: 16 [0], given: 49

EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 10178

Kudos [?]: 3536 [2], given: 173

Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: 340 Q170 V170
A group of 5 students bought movie tickets in one row next [#permalink]

### Show Tags

19 Oct 2016, 20:57
2
KUDOS
Expert's post
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
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

# Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save \$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************

Kudos [?]: 3536 [2], given: 173

Intern
Joined: 12 Nov 2009
Posts: 22

Kudos [?]: [0], given: 17

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

### Show Tags

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%

Kudos [?]: [0], given: 17

Intern
Joined: 25 Mar 2017
Posts: 1

Kudos [?]: 0 [0], given: 1

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

### Show Tags

12 Nov 2017, 07:39
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%

Kudos [?]: 0 [0], given: 1

Re: A group of 5 students bought movie tickets in one row next   [#permalink] 12 Nov 2017, 07:39

Go to page   Previous    1   2   [ 35 posts ]

Display posts from previous: Sort by