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# Ten nominees for three business awards are randomly seated

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SVP
Joined: 21 Jul 2006
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15 Aug 2008, 04:28
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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

Thanks
Director
Joined: 27 May 2008
Posts: 540

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15 Aug 2008, 05:49
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

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
SVP
Joined: 30 Apr 2008
Posts: 1863
Location: Oklahoma City
Schools: Hard Knocks

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15 Aug 2008, 06:36
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

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 Director Joined: 27 May 2008 Posts: 540 Re: PS: Probability [#permalink] ### Show Tags 15 Aug 2008, 06:51 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.... Director Joined: 27 May 2008 Posts: 540 Re: PS: Probability [#permalink] ### Show Tags 15 Aug 2008, 07:23 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, 08:38, edited 1 time in total. SVP Joined: 30 Apr 2008 Posts: 1863 Location: Oklahoma City Schools: Hard Knocks Re: PS: Probability [#permalink] ### Show Tags 15 Aug 2008, 08:01 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

Director
Joined: 27 May 2008
Posts: 540

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15 Aug 2008, 08:42

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 ....
SVP
Joined: 30 Apr 2008
Posts: 1863
Location: Oklahoma City
Schools: Hard Knocks

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15 Aug 2008, 08:43
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]

_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Director Joined: 27 May 2008 Posts: 540 Re: PS: Probability [#permalink] ### Show Tags 15 Aug 2008, 09:10 J allen, hate to dissappoint you ... but it seems there are some double countings in you sheet. Line 1 and 16 Line 10 and 25 But i like the method ... do we have MS Excel in GMAT test center jallenmorris wrote: When all else fails...head to excel! BRB Ok, I think I've found the answer. It's $$\frac{7}{12}$$ durgesh79 wrote: :x this question is driving me crazy.... first 1/3 then 7/12 now i'm with 5/12 OA please or i cant sleep tonight .... SVP Joined: 30 Apr 2008 Posts: 1863 Location: Oklahoma City Schools: Hard Knocks Re: PS: Probability [#permalink] ### Show Tags 15 Aug 2008, 09:15 1 This post received KUDOS Yes there are lines that are identical, but if you notice, there are 80 lines and I'm only counting 70 of them...so 70/120 = 7/12. EDIT: We're right with 7/12, but 7/12 represents the opposite of what the question asks for. We are asked to find the probability that NO empty seat is adjacent to another empty seat. This would be 1 - 7/12 for 5/12. that is an answer!!!! durgesh79 wrote: J allen, hate to dissappoint you ... but it seems there are some double countings in you sheet. Line 1 and 16 Line 10 and 25 But i like the method ... do we have MS Excel in GMAT test center jallenmorris wrote: When all else fails...head to excel! BRB Ok, I think I've found the answer. It's $$\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

Director
Joined: 27 May 2008
Posts: 540

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15 Aug 2008, 09:19
2
KUDOS
thanks J allen, it was Fun
now lets give kudos to eachother

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!!!!
Manager
Joined: 14 Jul 2008
Posts: 96
Schools: HBS, Stanford.

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15 Aug 2008, 09:27
hi what is source for this question? is it OG?

seems pretty tough for GMAT.
_________________

I did it all for the Starwood points.

Manager
Joined: 14 Jul 2008
Posts: 96
Schools: HBS, Stanford.

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15 Aug 2008, 09:28
jallenmorris wrote:

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

I did it all for the Starwood points.

SVP
Joined: 30 Apr 2008
Posts: 1863
Location: Oklahoma City
Schools: Hard Knocks

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15 Aug 2008, 09:29
Manhattan GMAT is not GMAT Prep.

PatrickBateman wrote:
jallenmorris wrote:

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$$. GMAT Club Premium Membership - big benefits and savings Manager Joined: 14 Jul 2008 Posts: 96 Schools: HBS, Stanford. Re: PS: Probability [#permalink] ### Show Tags 15 Aug 2008, 09:33 jallenmorris wrote: Manhattan GMAT is not GMAT Prep. PatrickBateman wrote: jallenmorris wrote: EDIT: Did a google search for the question and found: http://www.manhattanreview.com/forum/showthread.php?p=1217 just saw this....apparently source is gmatprep the title of the thread is "GMAT Prep Combinatronics Ten nominees" _________________ I did it all for the Starwood points. SVP Joined: 21 Jul 2006 Posts: 1508 Re: PS: Probability [#permalink] ### Show Tags 15 Aug 2008, 09:35 the OA to this is C, but the source of this question is Gmatprep. Thanks a lot guys for your great contributions! SVP Joined: 30 Apr 2008 Posts: 1863 Location: Oklahoma City Schools: Hard Knocks Re: PS: Probability [#permalink] ### Show Tags 15 Aug 2008, 09:51 I just saw the link, i never bothered to look that closely at it. _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

GMAT Club Premium Membership - big benefits and savings

Manager
Joined: 14 Jul 2008
Posts: 96
Schools: HBS, Stanford.

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15 Aug 2008, 09:58
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.
_________________

I did it all for the Starwood points.

Senior Manager
Joined: 31 Jul 2008
Posts: 289

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15 Aug 2008, 09:58
good one
SVP
Joined: 21 Jul 2006
Posts: 1508

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15 Aug 2008, 10:03
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!

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."
thanks
Re: PS: Probability   [#permalink] 15 Aug 2008, 10:03

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