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

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15 Aug 2008, 05:28

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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?

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

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$$.

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....

\(\frac{10!}{3!7!} = 10*9*8 / 3*2 = 10*3*4 = 120\) 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.

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!

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$$.

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!!!!

\(\frac{10!}{3!7!} = 10*9*8 / 3*2 = 10*3*4 = 120\) 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.

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." would you please explain? thanks