It is currently 21 Sep 2017, 05:30

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Ten nominees for three business awards are randomly seated

  post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
SVP
SVP
avatar
Joined: 21 Jul 2006
Posts: 1513

Kudos [?]: 978 [0], given: 1

Ten nominees for three business awards are randomly seated [#permalink]

Show Tags

New post 15 Aug 2008, 05:28
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Ten nominees for three business awards are randomly seated at the 10 places at a around table. Three of these nominees are then called up together to receive the three awards. What is the probability that no two adjacent seats will be left empty?

a) 1/3

b) 2/5

c) 5/12

d) 7/16

e) 1/2



Please explain.

Thanks

Kudos [?]: 978 [0], given: 1

Director
Director
avatar
Joined: 27 May 2008
Posts: 541

Kudos [?]: 354 [0], given: 0

Re: PS: Probability [#permalink]

Show Tags

New post 15 Aug 2008, 06:49
tarek99 wrote:
Ten nominees for three business awards are randomly seated at the 10 places at a around table. Three of these nominees are then called up together to receive the three awards. What is the probability that no two adjacent seats will be left empty?

a) 1/3

b) 2/5

c) 5/12

d) 7/16

e) 1/2



Please explain.

Thanks


number of ways 10 people can be seated on a round table = 9!

lets says 3 poople who got award are 1,2,3
number of ways 12 are always together = 2 * 8!
number of ways 23 are always together = 2 * 8!
number of ways 31 are always together = 2 * 8!

(consider two people as a unit and they can switch places in two ways)

Total number of ways when 2 of them are together = 3 * 2 * 8!
P = 3 * 2 * 8! / 9! = 6/9 = 2/3

desired probability = 1-2/3 = 1/3 option A

Kudos [?]: 354 [0], given: 0

SVP
SVP
User avatar
Joined: 30 Apr 2008
Posts: 1870

Kudos [?]: 607 [0], given: 32

Location: Oklahoma City
Schools: Hard Knocks
Re: PS: Probability [#permalink]

Show Tags

New post 15 Aug 2008, 07:36
I was approaching the question a different way but couldn't figure it out my way. Can you help?

10 people, the first 1 called up doesn't matter.

Second one must not be adjacent to any others, so there are 2 of the 9 remaining forbidden seats. The seats on either side of Winner #1.

Now, here is where it gets difficult for me. If you have more than 1 seat between the 2 empty seats, you have 4 forbidden seats (one of each side of the 2 empty seats). But, what if there is only 1 seat between the two empty seats, then there are only 3 forbidden seats, and not 4. How do you account for these 2 different scenarios in this approach?

tarek99 wrote:
Ten nominees for three business awards are randomly seated at the 10 places at a around table. Three of these nominees are then called up together to receive the three awards. What is the probability that no two adjacent seats will be left empty?

a) 1/3

b) 2/5

c) 5/12

d) 7/16

e) 1/2



Please explain.

Thanks

_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

GMAT Club Premium Membership - big benefits and savings

Kudos [?]: 607 [0], given: 32

Director
Director
avatar
Joined: 27 May 2008
Posts: 541

Kudos [?]: 354 [0], given: 0

Re: PS: Probability [#permalink]

Show Tags

New post 15 Aug 2008, 07:51
jallenmorris wrote:
I was approaching the question a different way but couldn't figure it out my way. Can you help?

10 people, the first 1 called up doesn't matter.

Second one must not be adjacent to any others, so there are 2 of the 9 remaining forbidden seats. The seats on either side of Winner #1.

Now, here is where it gets difficult for me. If you have more than 1 seat between the 2 empty seats, you have 4 forbidden seats (one of each side of the 2 empty seats). But, what if there is only 1 seat between the two empty seats, then there are only 3 forbidden seats, and not 4. How do you account for these 2 different scenarios in this approach?



i started the question in exact same way and lost it ... :?
and i'm still doubtful about my above answer....

Kudos [?]: 354 [0], given: 0

Director
Director
avatar
Joined: 27 May 2008
Posts: 541

Kudos [?]: 354 [0], given: 0

Re: PS: Probability [#permalink]

Show Tags

New post 15 Aug 2008, 08:23
guys i dont recommmend doing this, but after spending close to 15 minutes on one question this is what happenes ....

below is the list of all possible scenarios, with x marked on fav scenarios.

1 2 3
1 2 4
1 2 5
1 2 6
1 2 7
1 2 8
1 2 9
1 2 10
1 3 4
1 3 5 x
1 3 6 x
1 3 7 x
1 3 8 x
1 3 9 x
1 3 10 x
1 4 5
1 4 6 x
1 4 7 x
1 4 8 x
1 4 9 x
1 4 10 x
1 5 6
1 5 7 x
1 5 8 x
1 5 9 x
1 5 10 x
1 6 7
1 6 8 x
1 6 9 x
1 6 10 x
1 7 8
1 7 9 x
1 7 10 x
1 8 9
1 8 10 x
1 9 10

count x = 21
count total = 36

P = 21/36 = 7/12 .... which is not an option .....

as per this logic, my earlier answer is wrong ... or not... may be i'm still confused :?

OA and OE please....

EDIT : another mistake, i should not count the highlighted x
count x = 15
count total = 36

P = 15/36 = 5/12

Last edited by durgesh79 on 15 Aug 2008, 09:38, edited 1 time in total.

Kudos [?]: 354 [0], given: 0

SVP
SVP
User avatar
Joined: 30 Apr 2008
Posts: 1870

Kudos [?]: 607 [0], given: 32

Location: Oklahoma City
Schools: Hard Knocks
Re: PS: Probability [#permalink]

Show Tags

New post 15 Aug 2008, 09:01
Wouldn't there be 120 total options? 3C10

\(\frac{10!}{3!7!} = 10*9*8 / 3*2 = 10*3*4 = 120\) and then we must remove out any possible scenarios with seats together.

For each seat, there will be 2 possible ways it can have an adjacent empty set, or 2C3 = 3 ways per seat. But 3 is only for the first one. Lets say you start with Seat #1 at a round table. On either side of Seat #1 is Seat #10 and Seat #2. 2C3 filles the 2 empty seats scenario and is what we must count. But once you have this, if you move to Seat #2, you cannot count #1 next to it as that situation has already been counted when analyzing Seat #1. So, to account for this, when you use #1, you must move to seat # 3 or you risk double counting.

So Seat #1 = 3; #3 = 3; #5=3; #7=3; #9 = 3...that's 15 ways that 2 adjacent seats can be next to each other. Now figure out when all three seats will be next to each other.

This would be (3C5 or 3P5 because order matters?) because the order for Seat #1 could be
I find it less confusing to just list out the number of ways all 3 will be together.

1-2-3
2-3-4
3-4-5
4-5-6
5-6-7
6-7-8
7-8-9
8-9-10
9-10-1
10-1-2

Now about 2 seats together - where x could be any seat except for the seats surrounding the 2 chosen. We need to consider 1-2-4 different than 1-2-5 so we have 6 seats left that can be selected and will not satisfy the 3 seats in a row we've already counted.
1-2-x (*6) [running total]
2-3-x (*6) [12]
3-4-x (*6) [18]
4-5-x (*6) [24]
5-6-x (*6) [30]
6-7-x (*6) [36]
7-8-x (*6) [42]
8-9-x (*6) [48]
9-10-x (*6) [54]
10-1-x (*6) [60]

so we have 70/120...for 7/12.

HA! We may be wrong, but at least we're not alone!

EDIT: Did a google search for the question and found: http://www.manhattanreview.com/forum/showthread.php?p=1217
_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

GMAT Club Premium Membership - big benefits and savings

Kudos [?]: 607 [0], given: 32

Director
Director
avatar
Joined: 27 May 2008
Posts: 541

Kudos [?]: 354 [0], given: 0

Re: PS: Probability [#permalink]

Show Tags

New post 15 Aug 2008, 09:42
:x

this question is driving me crazy....

first 1/3
then 7/12
now i'm with 5/12

OA please or i cant sleep tonight ....

Kudos [?]: 354 [0], given: 0

SVP
SVP
User avatar
Joined: 30 Apr 2008
Posts: 1870

Kudos [?]: 607 [0], given: 32

Location: Oklahoma City
Schools: Hard Knocks
Re: PS: Probability [#permalink]

Show Tags

New post 15 Aug 2008, 09:43
When all else fails...head to excel! BRB

Ok, I think I've found the answer. It's \(\frac{7}{12}\)

durgesh79 wrote:
:x

this question is driving me crazy....

first 1/3
then 7/12
now i'm with 5/12

OA please or i cant sleep tonight ....

Attachments

Probability.xls [30.5 KiB]
Downloaded 89 times

To download please login or register as a user


_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

GMAT Club Premium Membership - big benefits and savings

Kudos [?]: 607 [0], given: 32

Director
Director
avatar
Joined: 27 May 2008
Posts: 541

Kudos [?]: 354 [0], given: 0

Re: PS: Probability [#permalink]

Show Tags

New post 15 Aug 2008, 10:10
J allen, hate to dissappoint you ... but it seems there are some double countings in you sheet.

Line 1 and 16
Line 10 and 25

But i like the method ... do we have MS Excel in GMAT test center :lol: :twisted:

jallenmorris wrote:
When all else fails...head to excel! BRB

Ok, I think I've found the answer. It's \(\frac{7}{12}\)

durgesh79 wrote:
:x

this question is driving me crazy....

first 1/3
then 7/12
now i'm with 5/12

OA please or i cant sleep tonight ....

Kudos [?]: 354 [0], given: 0

1 KUDOS received
SVP
SVP
User avatar
Joined: 30 Apr 2008
Posts: 1870

Kudos [?]: 607 [1], given: 32

Location: Oklahoma City
Schools: Hard Knocks
Re: PS: Probability [#permalink]

Show Tags

New post 15 Aug 2008, 10:15
1
This post received
KUDOS
Yes there are lines that are identical, but if you notice, there are 80 lines and I'm only counting 70 of them...so 70/120 = 7/12.

EDIT: We're right with 7/12, but 7/12 represents the opposite of what the question asks for. We are asked to find the probability that NO empty seat is adjacent to another empty seat. This would be 1 - 7/12 for 5/12. that is an answer!!!!

durgesh79 wrote:
J allen, hate to dissappoint you ... but it seems there are some double countings in you sheet.

Line 1 and 16
Line 10 and 25

But i like the method ... do we have MS Excel in GMAT test center :lol: :twisted:

jallenmorris wrote:
When all else fails...head to excel! BRB

Ok, I think I've found the answer. It's \(\frac{7}{12}\)

durgesh79 wrote:
:x

this question is driving me crazy....

first 1/3
then 7/12
now i'm with 5/12

OA please or i cant sleep tonight ....

_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

GMAT Club Premium Membership - big benefits and savings

Kudos [?]: 607 [1], given: 32

2 KUDOS received
Director
Director
avatar
Joined: 27 May 2008
Posts: 541

Kudos [?]: 354 [2], given: 0

Re: PS: Probability [#permalink]

Show Tags

New post 15 Aug 2008, 10:19
2
This post received
KUDOS
thanks J allen, it was Fun :wink:
now lets give kudos to eachother :lol:


jallenmorris wrote:
Yes there are lines that are identical, but if you notice, there are 80 lines and I'm only counting 70 of them...so 70/120 = 7/12.

EDIT: We're right with 7/12, but 7/12 represents the opposite of what the question asks for. We are asked to find the probability that NO empty seat is adjacent to another empty seat. This would be 1 - 7/12 for 5/12. that is an answer!!!!

Kudos [?]: 354 [2], given: 0

Manager
Manager
User avatar
Joined: 14 Jul 2008
Posts: 96

Kudos [?]: 6 [0], given: 0

Schools: HBS, Stanford.
Re: PS: Probability [#permalink]

Show Tags

New post 15 Aug 2008, 10:27
hi what is source for this question? is it OG?

seems pretty tough for GMAT.
_________________

I did it all for the Starwood points.

Kudos [?]: 6 [0], given: 0

Manager
Manager
User avatar
Joined: 14 Jul 2008
Posts: 96

Kudos [?]: 6 [0], given: 0

Schools: HBS, Stanford.
Re: PS: Probability [#permalink]

Show Tags

New post 15 Aug 2008, 10:28
jallenmorris wrote:
EDIT: Did a google search for the question and found: http://www.manhattanreview.com/forum/showthread.php?p=1217



just saw this....apparently source is gmatprep :(
_________________

I did it all for the Starwood points.

Kudos [?]: 6 [0], given: 0

SVP
SVP
User avatar
Joined: 30 Apr 2008
Posts: 1870

Kudos [?]: 607 [0], given: 32

Location: Oklahoma City
Schools: Hard Knocks
Re: PS: Probability [#permalink]

Show Tags

New post 15 Aug 2008, 10:29
Manhattan GMAT is not GMAT Prep.

PatrickBateman wrote:
jallenmorris wrote:
EDIT: Did a google search for the question and found: http://www.manhattanreview.com/forum/showthread.php?p=1217



just saw this....apparently source is gmatprep :(

_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

GMAT Club Premium Membership - big benefits and savings

Kudos [?]: 607 [0], given: 32

Manager
Manager
User avatar
Joined: 14 Jul 2008
Posts: 96

Kudos [?]: 6 [0], given: 0

Schools: HBS, Stanford.
Re: PS: Probability [#permalink]

Show Tags

New post 15 Aug 2008, 10:33
jallenmorris wrote:
Manhattan GMAT is not GMAT Prep.

PatrickBateman wrote:
jallenmorris wrote:
EDIT: Did a google search for the question and found: http://www.manhattanreview.com/forum/showthread.php?p=1217



just saw this....apparently source is gmatprep :(



the title of the thread is "GMAT Prep Combinatronics Ten nominees"
_________________

I did it all for the Starwood points.

Kudos [?]: 6 [0], given: 0

SVP
SVP
avatar
Joined: 21 Jul 2006
Posts: 1513

Kudos [?]: 978 [0], given: 1

Re: PS: Probability [#permalink]

Show Tags

New post 15 Aug 2008, 10:35
the OA to this is C, but the source of this question is Gmatprep.

Thanks a lot guys for your great contributions!

Kudos [?]: 978 [0], given: 1

SVP
SVP
User avatar
Joined: 30 Apr 2008
Posts: 1870

Kudos [?]: 607 [0], given: 32

Location: Oklahoma City
Schools: Hard Knocks
Re: PS: Probability [#permalink]

Show Tags

New post 15 Aug 2008, 10:51
I just saw the link, i never bothered to look that closely at it.
_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

GMAT Club Premium Membership - big benefits and savings

Kudos [?]: 607 [0], given: 32

Manager
Manager
User avatar
Joined: 14 Jul 2008
Posts: 96

Kudos [?]: 6 [0], given: 0

Schools: HBS, Stanford.
Re: PS: Probability [#permalink]

Show Tags

New post 15 Aug 2008, 10:58
so....uhhh....what is the way to do this <2 min?

Don't think they'll like us busting out Excel on our mobiles at the test center. :P
_________________

I did it all for the Starwood points.

Kudos [?]: 6 [0], given: 0

Senior Manager
Senior Manager
avatar
Joined: 31 Jul 2008
Posts: 294

Kudos [?]: 54 [0], given: 0

Re: PS: Probability [#permalink]

Show Tags

New post 15 Aug 2008, 10:58
good one :shock:

Kudos [?]: 54 [0], given: 0

SVP
SVP
avatar
Joined: 21 Jul 2006
Posts: 1513

Kudos [?]: 978 [0], given: 1

Re: PS: Probability [#permalink]

Show Tags

New post 15 Aug 2008, 11:03
jallenmorris wrote:
Wouldn't there be 120 total options? 3C10

\(\frac{10!}{3!7!} = 10*9*8 / 3*2 = 10*3*4 = 120\) and then we must remove out any possible scenarios with seats together.

For each seat, there will be 2 possible ways it can have an adjacent empty set, or 2C3 = 3 ways per seat. But 3 is only for the first one. Lets say you start with Seat #1 at a round table. On either side of Seat #1 is Seat #10 and Seat #2. 2C3 filles the 2 empty seats scenario and is what we must count. But once you have this, if you move to Seat #2, you cannot count #1 next to it as that situation has already been counted when analyzing Seat #1. So, to account for this, when you use #1, you must move to seat # 3 or you risk double counting.

So Seat #1 = 3; #3 = 3; #5=3; #7=3; #9 = 3...that's 15 ways that 2 adjacent seats can be next to each other. Now figure out when all three seats will be next to each other.

This would be (3C5 or 3P5 because order matters?) because the order for Seat #1 could be
I find it less confusing to just list out the number of ways all 3 will be together.

1-2-3
2-3-4
3-4-5
4-5-6
5-6-7
6-7-8
7-8-9
8-9-10
9-10-1
10-1-2

Now about 2 seats together - where x could be any seat except for the seats surrounding the 2 chosen. We need to consider 1-2-4 different than 1-2-5 so we have 6 seats left that can be selected and will not satisfy the 3 seats in a row we've already counted.
1-2-x (*6) [running total]
2-3-x (*6) [12]
3-4-x (*6) [18]
4-5-x (*6) [24]
5-6-x (*6) [30]
6-7-x (*6) [36]
7-8-x (*6) [42]
8-9-x (*6) [48]
9-10-x (*6) [54]
10-1-x (*6) [60]

so we have 70/120...for 7/12.

HA! We may be wrong, but at least we're not alone!

EDIT: Did a google search for the question and found: http://www.manhattanreview.com/forum/showthread.php?p=1217



thanks buddy for your input here, but i'm having a hard time understanding how we can have 3 different possibilities for each said. This is what you said "For each seat, there will be 2 possible ways it can have an adjacent empty set, or 2C3 = 3 ways per seat. But 3 is only for the first one."
would you please explain?
thanks

Kudos [?]: 978 [0], given: 1

Re: PS: Probability   [#permalink] 15 Aug 2008, 11:03

Go to page    1   2    Next  [ 30 posts ] 

Display posts from previous: Sort by

Ten nominees for three business awards are randomly seated

  post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.