|
Author |
Message |
|
TAGS:
|
|
|
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4441
Followers: 10
Kudos [?]:
82
[0], given: 0
|
In how many ways can 4 arts students and 4 science students [#permalink]
 22 Feb 2004, 09:30
In how many ways can 4 arts students and 4 science students be arranged alternately round the table.
_________________
Best Regards,
Paul
|
|
|
|
|
|
|
Director
Joined: 03 Jul 2003
Posts: 716
Followers: 2
Kudos [?]:
4
[0], given: 0
|
Assuming that the students are distingushable,
then it is 7!.
|
|
|
|
|
|
Director
Joined: 13 Nov 2003
Posts: 811
Location: BULGARIA
Followers: 1
Kudos [?]:
6
[0], given: 0
|
XAXAXAXAX- A can sit in 3! and S can occupy 5 places or 4! ways for S students seating, so i would go with 3!x4!=144.
|
|
|
|
|
|
Director
Joined: 03 Jul 2003
Posts: 716
Followers: 2
Kudos [?]:
4
[0], given: 0
|
kpadma wrote: Assuming that the students are distingushable,
then it is 7!.
It is either 4!*3! or 7*4!*3! . But I feel it is 7*4!*3!
Paul: what is the answer?
|
|
|
|
|
|
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4441
Followers: 10
Kudos [?]:
82
[0], given: 0
|
 Answer is 144.
Explanation:
Assuming A_B_C_D_ for art students positions and _E_F_G_H for science students. There are (4-1)! ways of sitting art students around the table and (4-1)! ways of sitting science students around the table. Also, there are 4C1 way for any given group's student to sit between any member of the other group. For instance, E could be between A/B or between B/C or between C/D or right after D. Therefore, answer is 3! * 3! * 4C1 = 144. Good job guys!
_________________
Best Regards,
Paul
|
|
|
|
|
|
GMAT Instructor
Joined: 07 Jul 2003
Posts: 773
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 5
Kudos [?]:
9
[0], given: 0
|
Paul wrote: :yes Answer is 144.
Explanation:
Assuming A_B_C_D_ for art students positions and _E_F_G_H for science students. There are (4-1)! ways of sitting art students around the table and (4-1)! ways of sitting science students around the table. Also, there are 4C1 way for any given group's student to sit between any member of the other group. For instance, E could be between A/B or between B/C or between C/D or right after D. Therefore, answer is 3! * 3! * 4C1 = 144. Good job guys!
I agree with your answer but the explanation does not make sense.
Let's put student A as our point of reference and anchor her to the 1st position on the table. There are now 3 available positions at the table specifically for art sutdetns 4 avaibable position for the science student. Hence, there are 3! x 4! difference distinct arrangements of the "kichen sind athors)
_________________
Best,
AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993
|
|
|
|
|
|
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4441
Followers: 10
Kudos [?]:
82
[0], given: 0
|
AkamaiBrah wrote: I agree with your answer but the explanation does not make sense.
Let's put student A as our point of reference and anchor her to the 1st position on the table. There are now 3 available positions at the table specifically for art sutdetns 4 avaibable position for the science student. Hence, there are 3! x 4! difference distinct arrangements of the "kichen sind athors)
Everyone please ignore my explanation. I must be  Please follow Akamai's explanation.
_________________
Best Regards,
Paul
|
|
|
|
|
|
Manager
Joined: 16 Jan 2004
Posts: 69
Location: NJ
Followers: 1
Kudos [?]:
0
[0], given: 0
|
4 Arts and 4 Science
3!*2! * 3!*2!
Ways to sit 4 Arts students is (4-1) = 3!. Now you can have this in 2 ways
so total is 3!2! = 12
same for science = 12
total = 12*12=144
I have multiplied 2! in both cases because both set of students have two seperate arrangements.
One which starts with arts student: A S A S A S A S
and One which starts with science student: S A S A S A S A
so both groups of arts and science students can be seated in 2! ways respectively
|
|
|
|
|
|
GMAT Instructor
Joined: 07 Jul 2003
Posts: 773
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 5
Kudos [?]:
9
[0], given: 0
|
pakoo wrote: 4 Arts and 4 Science
3!*2! * 3!*2!
Ways to sit 4 Arts students is (4-1) = 3!. Now you can have this in 2 ways
so total is 3!2! = 12
same for science = 12
total = 12*12=144
I have multiplied 2! in both cases because both set of students have two seperate arrangements.
One which starts with arts student: A S A S A S A S
and One which starts with science student: S A S A S A S A
so both groups of arts and science students can be seated in 2! ways respectively
I'm sorry, but your explanation does not make sense to me.
_________________
Best,
AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993
|
|
|
|
|
|
Intern
Joined: 05 Feb 2004
Posts: 3
Location: Germany
Followers: 0
Kudos [?]:
0
[0], given: 0
|
Thanks for the solution AkamaiBrah. I understand what you mean.
Consider just the arts students first. They can be seated in a circle in (4-1)! = 3!
Now there are 4 empty places in between each Arts students and each of these empty places needs to be filled with a science student. 4 places and 4 science students. Thus, 4! ways
Total no. of ways = 3! x 4! = 144 (Answer)
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
close
