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Probability

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by ritula » Mon Aug 25, 2008 3:00 am
Jake, Lena, Fred, John and Inna need to drive home from a party in a car that can seat 7 people. If only Inna or Jake can drive, how many seat allocations are possible?

1. 30
2. 42
3. 60
4. 120
5. 125
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Re: Probability

by sudhir3127 » Mon Aug 25, 2008 3:25 am
ritula wrote:Jake, Lena, Fred, John and Inna need to drive home from a party in a car that can seat 7 people. If only Inna or Jake can drive, how many seat allocations are possible?

1. 30
2. 42
3. 60
4. 120
5. 125
IMO C.60
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Re: Probability

by ritula » Mon Aug 25, 2008 3:39 am
Kindly explain also.
sudhir3127 wrote:
ritula wrote:Jake, Lena, Fred, John and Inna need to drive home from a party in a car that can seat 7 people. If only Inna or Jake can drive, how many seat allocations are possible?

1. 30
2. 42
3. 60
4. 120
5. 125
IMO C.60
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by pepeprepa » Mon Aug 25, 2008 3:59 am
Sudhir, could you explain how you see the question and what you do because I have much more possibilities than the answers?
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by mals24 » Mon Aug 25, 2008 5:01 am
I got 720 as my answer which is nowhere close to the options but i can't seem to find whats wrong with my working:

There are 7 seats and 5 people.
If Inna drives, there are 6 seats and 4 people remaining. Out of the remaining people the first person can sit in any of the 6 seats, the 2nd in any of the remainng 5, 3 in the remainig 4 and last person in the remaining 3 seats.
Hence the 4 people can be arranged in 6*5*4*3 = 360 ways.

Same if Jake drives, the remaining 4 people can be arranged in 6*5*4*3 = 360 ways.

Therefore the total number of arrangements possible: 2*360 = 720.
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by 4meonly » Mon Aug 25, 2008 7:20 am
Hmm. I've got 720, too

2*(6P4)=720

Where is my mistake?
Last edited by 4meonly on Mon Aug 25, 2008 8:12 am, edited 1 time in total.
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by Ankdes » Mon Aug 25, 2008 8:06 am
I was thinking it would be 2*(6P4) getting you to 30. Not sure how Sudhir got to 60.

7 seats
one driving means 6 seats left and 4 people to sit in them (6P4)
Then you know either Jake or Lena can lean back in the drivers seat... so (*2)

Are Jake and Lean dating? But regardless...someone smarter than I please chime in with a different explanation

Gracias
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by Ian Stewart » Mon Aug 25, 2008 9:12 am
Could you please let us know the source of the question? This is a seating arrangement question, so order should matter, but it's clear from the answer choices that the question designer does not want order to matter. Unless some information is missing, it's not a well-written question, and the answer choices don't make any sense to me. I don't see anything wrong with the approach taken by mals24 and 4meonly above.
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by rishi235 » Mon Aug 25, 2008 9:26 am
Wow....
I'm getting the answer as 1) 30....hehehe

Heres my working...
2C1 * 6C4 = 30...
Can't get it...

Thanks.... :)
Last edited by rishi235 on Mon Aug 25, 2008 9:47 am, edited 1 time in total.
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by kris610 » Mon Aug 25, 2008 9:38 am
Ian Stewart wrote:Could you please let us know the source of the question? This is a seating arrangement question, so order should matter, but it's clear from the answer choices that the question designer does not want order to matter. Unless some information is missing, it's not a well-written question, and the answer choices don't make any sense to me. I don't see anything wrong with the approach taken by mals24 and 4meonly above.
I've to agree with Ian here:

With each of the two drivers, you can have 6C4*4! ways if order matters. If order does not matter, it could be 30.
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by 4meonly » Mon Aug 25, 2008 9:47 am
rishi235 wrote:Wow....
I'm getting the answer as 1) 30....hehehe

Heres my working...
2C1 * 6C4 = 30...
Someone kindly figure out my mistake...

Thanks.... :)
You use 6C4 - this is used when order is not important
But here order is important, so we should use 6P4
Try it. You'll get 720
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by rishi235 » Mon Aug 25, 2008 9:52 am
oh i get my mistake ...Thanks 4meonly...

I solved the problem using mals24's method...got 720 too ... :roll:
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by kris610 » Mon Aug 25, 2008 10:19 am
4meonly wrote:
rishi235 wrote:Wow....
I'm getting the answer as 1) 30....hehehe

Heres my working...
2C1 * 6C4 = 30...
Someone kindly figure out my mistake...

Thanks.... :)
You use 6C4 - this is used when order is not important
But here order is important, so we should use 6P4
Try it. You'll get 720
I don't think 6P4 is appropriate here even if the order is important. In that case, you first find the number of ways you can *select* 4 seats from the available 6 -- 6C4. Once you find that, you multiply that by 4! i.e. each of the selections can be arranged in 24 different ways -- 6C4*4!
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by Ian Stewart » Mon Aug 25, 2008 11:48 am
I'd simply add that this question is not worth devoting much attention to. If this were on the GMAT, the correct answer would be 720, and the simple fact that each person in the car is mentioned by name suggests that the order in which they sit should be important. I'm still curious where the question is from, but it is not a realistic GMAT question, at least not with the five answer choices that are provided.
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by ritula » Mon Aug 25, 2008 9:53 pm
The sourc of this question is a test in www.TCYonline.com
Even i got the answer as 720 but cldnt find it in the options. so thought mayb i was wrong sum where.
Ian Stewart wrote:Could you please let us know the source of the question? This is a seating arrangement question, so order should matter, but it's clear from the answer choices that the question designer does not want order to matter. Unless some information is missing, it's not a well-written question, and the answer choices don't make any sense to me. I don't see anything wrong with the approach taken by mals24 and 4meonly above.
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