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Manhattan 700 + Combination .. Experts' help needed..

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by rohu27 » Sun Jan 30, 2011 6:32 am
this has already been discussed

https://www.beatthegmat.com/combinatoric ... 73855.html
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by Stuart@KaplanGMAT » Sun Jan 30, 2011 7:12 am
btgyes wrote:Acc to me ,

it should be

!5 + !4 +!3 +!2 +!1 =153 is ans.

Please see where am doing mistake...

Thanks
Hi,

it's very difficult to find your mistake without understanding how you arrived at your answer.

It looks like you've put Joey in line and then counted how many spots are left behind him for Frankie. If that's the case, then at least one big problem is you're ignoring where all the other people can stand.

For example, your 1! (not that it matters on the GMAT, since you don't get any marks for scratchwork, but when expressing a factorial the exclamation mark comes after the number, not before it) represents how many places you can put Frankie when Joey is 5th in line. However, you're ignoring that there are 4! arrangements for the other 4 people in line.

As it notes in the other discussion on this thread, the solution is much simpler; there are only two possibilities for Frankie and Joey - Frankie in front of Joey or vice-versa. Since we're randomly arranging people, each of those possibilities will occur 50% of the time. So, the correct answer to the question is simply 6! (the total number of ways to arrange 6 people) divided by 2, i.e.:

6!/2 = 6*5*4*3*2/2 = 6*5*4*3 = 30*12 = 360
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by btgyes » Thu Feb 03, 2011 6:22 am
Stuart Kovinsky wrote:
btgyes wrote:Acc to me ,

it should be

!5 + !4 +!3 +!2 +!1 =153 is ans.

Please see where am doing mistake...

Thanks
Hi,

it's very difficult to find your mistake without understanding how you arrived at your answer.

It looks like you've put Joey in line and then counted how many spots are left behind him for Frankie. If that's the case, then at least one big problem is you're ignoring where all the other people can stand.

For example, your 1! (not that it matters on the GMAT, since you don't get any marks for scratchwork, but when expressing a factorial the exclamation mark comes after the number, not before it) represents how many places you can put Frankie when Joey is 5th in line. However, you're ignoring that there are 4! arrangements for the other 4 people in line.

As it notes in the other discussion on this thread, the solution is much simpler; there are only two possibilities for Frankie and Joey - Frankie in front of Joey or vice-versa. Since we're randomly arranging people, each of those possibilities will occur 50% of the time. So, the correct answer to the question is simply 6! (the total number of ways to arrange 6 people) divided by 2, i.e.:

6!/2 = 6*5*4*3*2/2 = 6*5*4*3 = 30*12 = 360
Sir,

Acc to me, we solve this sum in this way.



lets consider these sitting arrangements.

A B C D E F

NOW lets say, Joey is sitting at A position

so now Frankie can sit 5! ways ie 120 ways.

in similar ways , Joey sits at B then Frankie can sits 4! ways ie 24 ways...

in this way.....

5! +4! +3! +2! + 1! = 153 ways.

i understand yr reasoning but which scenario am not taking into account...>

Plz comment

Thanks
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by GMATGuruNY » Thu Feb 03, 2011 6:46 am
btgyes wrote:
Stuart Kovinsky wrote:
btgyes wrote:Acc to me ,

it should be

!5 + !4 +!3 +!2 +!1 =153 is ans.

Please see where am doing mistake...

Thanks
Hi,

it's very difficult to find your mistake without understanding how you arrived at your answer.

It looks like you've put Joey in line and then counted how many spots are left behind him for Frankie. If that's the case, then at least one big problem is you're ignoring where all the other people can stand.

For example, your 1! (not that it matters on the GMAT, since you don't get any marks for scratchwork, but when expressing a factorial the exclamation mark comes after the number, not before it) represents how many places you can put Frankie when Joey is 5th in line. However, you're ignoring that there are 4! arrangements for the other 4 people in line.

As it notes in the other discussion on this thread, the solution is much simpler; there are only two possibilities for Frankie and Joey - Frankie in front of Joey or vice-versa. Since we're randomly arranging people, each of those possibilities will occur 50% of the time. So, the correct answer to the question is simply 6! (the total number of ways to arrange 6 people) divided by 2, i.e.:

6!/2 = 6*5*4*3*2/2 = 6*5*4*3 = 30*12 = 360
Sir,

Acc to me, we solve this sum in this way.



lets consider these sitting arrangements.

A B C D E F

NOW lets say, Joey is sitting at A position

so now Frankie can sit 5! ways ie 120 ways.

in similar ways , Joey sits at B then Frankie can sits 4! ways ie 24 ways...

in this way.....

5! +4! +3! +2! + 1! = 153 ways.

i understand yr reasoning but which scenario am not taking into account...>

Plz comment

Thanks
You're not accounting correctly for how the arrangements are restricted.

5! is correct. If Joey is in the A position, Frankie can sit in any of the other positions. The number of ways to arrange the other 5 mobsters in positions B-F = 5! = 120.

But if Joey is in the B position, the problem becomes more restricted, because Frankie cannot be in the A position:
Number of choices for A = 4 (anyone but Frankie)
Number of arrangements for positions C-F = number of ways to arrange the remaining 4 mobsters = 4!
Total possible arrangements = 4*4! = 96.

Similar logic should be applied when Joey is in positions C-F.

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by btgyes » Thu Feb 03, 2011 7:24 am
Thanks a lot EXPERTS.........


i got what i was missing....

i was not considering that position before JOEY can be occupied by other four people (as Frankie is the only one who can't sit ahead of joey)



Thanks a lot ....

Probability is my favorite topic.. my probability of giving wrong ans to probability ques is less than 1/1000.

I think this is that one question of previous 1000 questions.... :)

anyways .. Thanks.....
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