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manhattan gmat ds DEC 26 CHALLENGE

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mand-y
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manhattan gmat ds DEC 26 CHALLENGE [#permalink] New post   26 Jan 2006, 09:36
00:00
A
B
C
D
E

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(N/A)

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0% (00:00) correct 0% (00:00) wrong based on 3 sessions
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Prime Seats
There are x people and y chairs in a room where x and y are positive prime numbers. How many ways can the x people be seated in the y chairs (assuming that each chair can seat exactly one person)?

(1) x + y = 12
(2) There are more chairs than people.
plz explain no OA
THANKS



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Re: manhattan gmat ds DEC 26 CHALLENGE [#permalink] New post   26 Jan 2006, 09:56
well I'm trying to rub the rust off my brain after some months of inactivity
should be (C)
b/c
1) are there 7 seats for 5 people or viceversa?
2) there are 7 chairs and 5 people!

where's the trick

:roll: :roll:
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lhotseface
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Re: manhattan gmat ds DEC 26 CHALLENGE [#permalink] New post   26 Jan 2006, 14:25
Ok. I see the trick now...

For I,

chairs = 5, people = 7. No. of arrangements = 7p5 = 7*6*5*4*3 ways...two people will be left out and will levitate in the air.

people =5, chairs = 7. 1st person can sit in 7 ways, 2nd person can sit in 6 ways...7*6*5*4*3...two chairs will be empty.


A is SUFF.
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ywilfred
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Re: manhattan gmat ds DEC 26 CHALLENGE [#permalink] New post   26 Jan 2006, 23:36
(1) (x,y) = (5,7) (7,5)

For the first case, (5,7), number of ways to the number of ways of assigning 7 chairs to 5 people. The number of ways of doing this is 7P5 (think of the number of permutations for picking 5 chairs out of 7).
For the second case, (7,5), number of ways is to assign 7 people to 5 chairs. 2 people will be left out. The way to do this is to permute 5 people from 7.

A is suff since both cases arrive at the same solution.

(2) There are more chairs than people. Insufficinet since we do not know how many people and how many chairs.

Ans A
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Re: manhattan gmat ds DEC 26 CHALLENGE [#permalink] New post   27 Jan 2006, 10:22
Was on C untill wilfred developed A. Cool explanation.



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Re: manhattan gmat ds DEC 26 CHALLENGE [#permalink]
27 Jan 2006, 10:22
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