Arrangement in a circle.... (1 Viewer)

[kpq_sniper017]

I came across this question in Cambridge 3U Maths Yr 12 - it was the first question of the "Extension" section of one of the exercises.

Q: A group of n men and n+1 women sit around a circular table. How many arrangements are possible if no two men are to sit next to one another?
A: n!(n+1)!/(2n)! (given in the Answer section)

edit: got rid of my working (probably turning people off by the multitude of words).

But, it'd be great if somebody could work through it.

 

[raymes]

what uve got to think about is though (and this is what makes it complicated), two women can be sitting next to each other anywhere around the table i.e. not only the final two positions

e.g.
you could have: (if u fix a man where n=3)
MWMWMWW or
MWWMWMW

 

[kpq_sniper017]

hhhmm.....
still confused. *why does the answer give a no. of arrangements <1?* it doesn't make sense.

so how do u take into consideration that the two women sitting next to each other can be sitting in any position?
is my working OK apart from that fact?

 

[raymes]

to tell you the truth i dont really know even though i had a bit of a look at the question

im reconsidering whether what i said earlier was true cause even if two women sit next to each other anywhere around the table, its only a rotation of the table really, therefore not a different combination...i think

its a toughy

 

[~*HSC 4 life*~]

ppppft i hate perms

 

[Collin]

Oh I swear perms and combs were quite possibly the worst topic in the entire 2,3 and 4 unit syllabuses. Even worse than statistics and probability (and they were pretty bad)

 

[kpq_sniper017]

Originally posted by raymes
to tell you the truth i dont really know even though i had a bit of a look at the question

im reconsidering whether what i said earlier was true cause even if two women sit next to each other anywhere around the table, its only a rotation of the table really, therefore not a different combination...i think

its a toughy

that was what i was thinking as well.....
so i just left the last two spaces for the women.

 

[kpq_sniper017]

where are cm_tutor and affinity when u need them???

1. pet hate is perms & combs. lol - however, i kind if like the challenge and getting the answer out in the end.

i just hope somebody can solve this one, coz it's pretty interesting and i'd really like to know the answer.

*i'm still yet to figure out why plugging n=3 into the answer given in the book gives 0.2.....

 

[Collin]

Mate not trying to be rude, but I can't be bothered trying to work it out. The point I'm here for is though, alot of textbooks around have many many mistakes throughout them, some of the textbooks we used were loaded with stupid mistakes. So maybe you were right for the questions you got stuck on, perhaps.

 

[CM_Tutor]

There is definitely something wrong with the answer to this question, as it is not possible for a number of arrangements to be anything other than an integer.

 

[kpq_sniper017]

Originally posted by CM_Tutor
There is definitely something wrong with the answer to this question, as it is not possible for a number of arrangements to be anything other than an integer.

that's what i thought as well. when i subbed in other values for "n" i got irrational numbers - probably can't even be expressed as surds (what do you call those, btw? - teacher said the word but i've forgotten - i think)

the answer i got was: 2n(n+1)[(n+1)!]² + (n!)².

subbing in values for "n", i get integer numbers, but i don't know if my answer is right. it only took me about 5 minutes to get my answer (which led me to believe i was wrong).....considering it was in the "extension" section.

anyone have any idea of how to figure this one out?

 

[CM_Tutor]

Originally posted by pcx_demolition017
that's what i thought as well. when i subbed in other values for "n" i got irrational numbers - probably can't even be expressed as surds (what do you call those, btw? - teacher said the word but i've forgotten - i think)

How do you get irrational answers, let alone transcendental ones? n!(n+1)!/(2n)! is a fraction of integers, so must be rational.

 

[kpq_sniper017]

Originally posted by CM_Tutor


How do you get irrational answers, let alone transcendental ones? n!(n+1)!/(2n)! is a fraction of integers, so must be rational.

ok....ignore that one

but i still get an answer <1 for each time i sub n in.
thanks cm...transcendental...that's the one

can anyone veryify that my answer of 2n(n+1)[(n+1)!]² + (n!)² is correct/incorrect?

 

[nike33]

my answer...

consider the circle...the only way the situation can occur is if two women sit next to each other then all the rest are in alternating

ie M F F M F M F M F ...etc

for now consider all the males fixed in their position. the number of females have (n+1)! ways of moving around still containing to the restriction.. now 'fix' all the females and there are n males moving around..hence they have n! possibilities.. therefore the total combinations is n!(n+1)!..well its consistant with n =1,2,3 ...........i think ..off to do afinities qn now hehe

 

[Heinz]

I have a feeling that the question is asking for the probability that no two men are to sit next to one another...

 

[nike33]

yes..if my answer of n!(n+1)! is correct (i think it is) then there will be n+1+n people ie 2n+1...but with a circle you minus the one due to its symmetry hence it becomes (2n)! is the total amount of arangements and as a probability it becomes
n!(n+1)! / (2n)!

 

[KanyeEast69420]

sample space: (n + n + 1 -1)! = (2n)!

fix one woman in a fixed position, so (n + 1 -1)! = (n)! ways to arrange the rest of the women
since its a circle, there are (n+1)! gaps remaining between the women, but there are only n men, so there are (n+1)P(n) or (n+1)! ways to arrange the men

therefore probability that no two men sit next to each other is (n)!(n+1)!/(2n)!

 

[Socialism]

just 20 years late kanye