• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

probability question help (1 Viewer)

Benmc

Member
Joined
Feb 21, 2005
Messages
118
Location
Bathurst
Gender
Male
HSC
2006
how do you do this question.
7 people are to be seated at a round table. two people, kev and jill, refuse to sit next to each other. how many seating arrangements are then possible?
 

rama_v

Active Member
Joined
Oct 22, 2004
Messages
1,151
Location
Western Sydney
Gender
Male
HSC
2005
No restrictions = 6!
But two cant sit together
so its 6! - the times they do sit together
= 6! - (5!x2!)
= 6! - (240)
= 480

I think, this is my worst topic :(
 
Last edited:

word.

Member
Joined
Sep 20, 2005
Messages
174
Gender
Male
HSC
2005
find the number arrangements of them together
you do this by treating them as one group
so there are '6' people around the table
number of arrangements around the table = 5!;
within the 2 people = 2!,
5! * 2!

minus the total arrangements
6! - 5!*2!
 

Condog

New Member
Joined
Mar 9, 2005
Messages
12
Location
Wollongong
Gender
Male
HSC
2005
i believe this is how its done:
there are 6! ways the people can sit around the table
there are 5! ways in which two people can be seated together but they can be either side of each other so its 2(5!) (u take the 2 people as one person)
tthen the number of arrangements which the two people cannot be together is
6!-2(5!)
 

Benmc

Member
Joined
Feb 21, 2005
Messages
118
Location
Bathurst
Gender
Male
HSC
2006
rama_v said:
No restrictions = 6!
But two cant sit together
so its 6! - the times they do sit together
= 6! - (5!x2! / 5)
= 6! - (48)
= 672

I think, this is my worst topic :(

nah the answer is 480
 

rama_v

Active Member
Joined
Oct 22, 2004
Messages
1,151
Location
Western Sydney
Gender
Male
HSC
2005
Yeah, I edited it just then lol,
I divided by 5 for no apparent reason
shows how much i know about permutations :p
 

Dumsum

has a large Member;
Joined
Aug 16, 2004
Messages
1,552
Location
Maroubra South
Gender
Male
HSC
2005
jake2.0 said:
well actually its a combination
I'm pretty sure this is a permutation.



Who else thinks that Kev and Jill will end up hooking up in the future? Lol. These math problems are always all about people hating each other :(
 

Stefano

Sexiest Member
Joined
Sep 27, 2004
Messages
296
Location
Sydney, Australia
Gender
Male
HSC
2005
Condog said:
i believe this is how its done:
there are 6! ways the people can sit around the table
there are 5! ways in which two people can be seated together but they can be either side of each other so its 2(5!) (u take the 2 people as one person)
tthen the number of arrangements which the two people cannot be together is
6!-2(5!)
Exactly correct.


Benmc - just remember that the amount of ways the two peope CANNOT sit together is the total way of seating everyone minus the amount of ways those two people can sit together.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top