First post
anyways
I'm not sure if this is right but figured I'd share my thoughts
I would use some sort of sot method so if they people would be represented by
R D P J J J. With R=rita, D=daniela, P=Paul
now I'll use a slot method to see who can sit where
1st if paul sits in the first seat 3 ppl can sit in the next( the J J J's) then the next seat would be 4 ppl since it could be R or D or the remaining J's. Do this for all of Paul's seats with this methodology.
P 3 4 3 2 1
3 P 2 3 2 1
3 3 P 2 2 1
1 2 2 P 3 3
1 2 3 2 P 3
1 2 3 4 3 P
Now multiple the integers in each line to get the different combinations and then add them up
I got 288.
I'm not sure if I'm right. It would be nice to see if someone else could offer up a way to solve this.
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edvhou812
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I'm only practicing here, but is it 6!/2!*2!=180?
I don't know what to say, really. Three minutes to the biggest battle of our professional lives. You find out life's this game of inches, so is football. Because in either game - life or football - the margin for error is so small. I mean, one half a step too late or too early and you don't quite make it. One half second too slow, too fast and you don't quite catch it. I'll tell you this, in any fight it's the guy whose willing to die whose gonna win that inch. That's football guys, that's all it is. Now, what are you gonna do?
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shankar.ashwin
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I think the previous answer if wrong as Paul is common in both cases and so dividing it by (2!*2!) would be wrong.
My guess:
Total possible ways = 6! = 720.
Consider this, (Paul=A)
I have 6 positions for A, _ _ _ _ _ _.
Case 1: A takes the first seat, i.e : A _ _ _ _ _,
Rita=B, Daniele= C (So, ABC cant be together in any order): Here, B&C cannot take the seat to the right of A, so B could have any of the remaining 4 and C could take any of the remaining 3(ignoring the seat to the immediate right of A) A X _ _ _ _.
Now since A,B and C have occupied their places, the remaining 3 seats could be filled in 3! ways.
So, that gives 4*3*3! = 72 ways. The same could be derived when A sits in the right extreme as well, so thats 72 more cases.
Total possibilities = 144
Case2:
When A sits in 2,3,4 or 5.
Consider this _ A _ _ _ _.
Again, B & C cant take the seats adjacent to A, so B can take 3 and C can take 2. The remaining 3 can be filled in 3! ways.
So, 3*2*3! = 36 ways.
The same is valid for all 4 positions of A in this case,
So total possibilities = 36*4 = 144.
Hence, the ways in which the seating is possible is ((144+144=288)
My guess:
Total possible ways = 6! = 720.
Consider this, (Paul=A)
I have 6 positions for A, _ _ _ _ _ _.
Case 1: A takes the first seat, i.e : A _ _ _ _ _,
Rita=B, Daniele= C (So, ABC cant be together in any order): Here, B&C cannot take the seat to the right of A, so B could have any of the remaining 4 and C could take any of the remaining 3(ignoring the seat to the immediate right of A) A X _ _ _ _.
Now since A,B and C have occupied their places, the remaining 3 seats could be filled in 3! ways.
So, that gives 4*3*3! = 72 ways. The same could be derived when A sits in the right extreme as well, so thats 72 more cases.
Total possibilities = 144
Case2:
When A sits in 2,3,4 or 5.
Consider this _ A _ _ _ _.
Again, B & C cant take the seats adjacent to A, so B can take 3 and C can take 2. The remaining 3 can be filled in 3! ways.
So, 3*2*3! = 36 ways.
The same is valid for all 4 positions of A in this case,
So total possibilities = 36*4 = 144.
Hence, the ways in which the seating is possible is ((144+144=288)
Last edited by shankar.ashwin on Wed Aug 31, 2011 7:30 am, edited 1 time in total.
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GMATGuruNY
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Paul on either end:Fractal wrote:6 people are in the cinema. paul and rita are not allowed to sit next to each other. also paul and daniela are not allowed to sit next to each other.
number of possibilites how they can sit = ... ?
The number of options for Paul = 2.
Since Rita and Daniela cannot sit next to Paul, the number of remaining people who can sit next to Paul = 3.
The number of ways to arrange the 4 remaining people = 4! = 24.
To combine the options above, we multiply:
Number of arrangements = 2*3*24 = 144.
Paul in one of the 4 middle seats:
The number of options for Paul = 4.
Since Rita and Daniela cannot sit next to Paul, the number of remaining people who can sit to the left of Paul = 3.
Since Rita and Daniela cannot sit next to Paul, the number of remaining people who can sit to the right of Paul = 2.
The number of ways to arrange the 3 remaining people = 3! = 6.
To combine the options above, we multiply:
Number of arrangements = 4*3*2*6 = 144.
Total arrangements = 144+144 = 288.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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