HSC 2013-14 MX1 Marathon (archive) (5 Viewers)

[VBN2470]

Re: HSC 2013 3U Marathon Thread

Disclaimer: I can't do this

Is the answer 2128 ways?

 

[HeroicPandas]

Re: HSC 2013 3U Marathon Thread

is it 4944?

 

[JJ345]

Re: HSC 2013 3U Marathon Thread

my first method was totally wrong i realised after 2 hours
total number of ways they can be arranged is 7! =5040
number of ways all couples sit together = 3!2!2!2!2! =96
number of ways 2 couples sit together but other 2 random = 5!2!2! =480
number of ways only 1 couple sit together = 6!2!=1440
therefore number of ways none sit together = 5040-96-480-1440 =3024
once again i dont know for sure

I think you've double counted some cases, like the bolded case-->What if in that 5! arrangement two couples who aren't mean to be together are 'accidently' put together, then there's 3 couples together--> a latter case which you account for separately. Also, I think the couple that is being put together must be chosen like 4C1 or something before every case because they're all distinct...but i'm guessing the numbers will be way to large.

 

[Sy123]

Re: HSC 2013 3U Marathon Thread

my first method was totally wrong i realised after 2 hours
total number of ways they can be arranged is 7! =5040
number of ways all couples sit together = 3!2!2!2!2! =96
number of ways 2 couples sit together but other 2 random = 5!2!2! =480
number of ways only 1 couple sit together = 6!2!=1440
therefore number of ways none sit together = 5040-96-480-1440 =3024
once again i dont know for sure

Is the answer 2128 ways?

is it 4944?

The answer is 12 apparently lol

 

[JJ345]

Re: HSC 2013 3U Marathon Thread

The answer is 12 apparently lol

If this was meant to be *that* question, then I think you copied it down wrong, this was the original:
4 couples are to be seated around a circular table, with men and women alternating. The couples are not allowed to sit next to their partners. Find the number of ways the party can be seated if each person must have a neighbour on each side.

You missed the men and women alternating part-->But I know where 12 comes from now.

 

[Sy123]

Re: HSC 2013 3U Marathon Thread

If this was meant to be *that* question, then I think you copied it down wrong, this was the original:
4 couples are to be seated around a circular table, with men and women alternating. The couples are not allowed to sit next to their partners. Find the number of ways the party can be seated if each person must have a neighbour on each side.

You missed the men and women alternating part-->But I know where 12 comes from now.

Oh ok, my bad

 

[Sy123]

Re: HSC 2013 3U Marathon Thread

 

[integral95]

Re: HSC 2013 3U Marathon Thread

isn't it ln(x)/(1+x^2)?

Cause I can't seem to get 0 with your integral lol

 

[Sy123]

Re: HSC 2013 3U Marathon Thread

isn't it ln(x)/(1+x^2)?

Cause I can't seem to get 0 with your integral lol

Yep, once again my bad haha

 

[RealiseNothing]

Re: HSC 2013 3U Marathon Thread

Disclaimer: I can't do this

If the women and men alternate like JJ said this is a very easy question tbh. You can arrange the men in 3! ways (it's a circle). Now the women can only be arranged in 2 ways, hence 3! x 2 = 12.

A bit of explanation for why the women can only be arranged in 2 ways: There are 4 seats for each women to sit at. 2 of these seats will be next to their partner, and so there are only 2 available seats for each women to sit at. The first woman chooses her seat in 2 different ways, and then the rest of the women in fact only have 1 possible seat to choose. Consider the following:

Woman 1 can sit in A or B
Woman 2 can sit in B or C
Woman 3 can sit in C or D
Woman 4 can sit in D or A

If woman 1 choose seat B, it forces woman 2 to choose seat C, which forces woman 3 to choose seat D, which forces woman 4 to choose seat A. So after woman 1 has chosen either A or B, it sets off a "chain reaction" where the other 3 are forced into only one possible seat.

Hence the women can be seated in 2x1x1x1 = 2 ways.

 

[Sy123]

Re: HSC 2013 3U Marathon Thread

If the women and men alternate like JJ said this is a very easy question tbh. You can arrange the men in 3! ways (it's a circle). Now the women can only be arranged in 2 ways, hence 3! x 2 = 12.

A bit of explanation for why the women can only be arranged in 2 ways: There are 4 seats for each women to sit at. 2 of these seats will be next to their partner, and so there are only 2 available seats for each women to sit at. The first woman chooses her seat in 2 different ways, and then the rest of the women in fact only have 1 possible seat to choose. Consider the following:

Woman 1 can sit in A or B
Woman 2 can sit in B or C
Woman 3 can sit in C or D
Woman 4 can sit in D or A

If woman 1 choose seat B, it forces woman 2 to choose seat C, which forces woman 3 to choose seat D, which forces woman 4 to choose seat A. So after woman 1 has chosen either A or B, it sets off a "chain reaction" where the other 3 are forced into only one possible seat.

Hence the women can be seated in 2x1x1x1 = 2 ways.

Yep I already had this explained to me,

Would a generalization to n couples be hard to do?

 

[Carrotsticks]

Re: HSC 2013 3U Marathon Thread

Yep I already had this explained to me,

Would a generations to n couples be hard to do?

=)

 

[Sy123]

Re: HSC 2013 3U Marathon Thread

=)

Yea I'm not sure why I wrote a different word to what I was thinking of

Dat English in 6 days

 

[RealiseNothing]

Re: HSC 2013 3U Marathon Thread

=)

please don't make him generalise

 

[asianese]

Re: HSC 2013 3U Marathon Thread

carret and his generalising.

 

[superSAIyan2]

Re: HSC 2013 3U Marathon Thread

do you use the technique in 4U where Integrate f(u) du = Integration f(x) dx?

 

[Sy123]

Re: HSC 2013 3U Marathon Thread

Disclaimer: I can't do this

Can someone do this version anyway?

What I got was:

If we take the number of arrangements of n couples around a round table so that no couple is together as

Then we need to find:



To find , I considered an already set up arrangements of couples, then there are 6 spots for the 2 couples to sit and therefore to sit the two couple apart it is 6C2

Therefore



Similarly




And by manual computation we find



??

Is this correct? Can someone affirm or deny it?

 

[Sy123]

Re: HSC 2013 3U Marathon Thread

do you use the technique in 4U where Integrate f(u) du = Integration f(x) dx?

(if its a definite integral)

Well yes you do use it but I wouldn't call it a 4U technique..... (or is it? I don't know what is said in the 4U syllabus)

 

[superSAIyan2]

Re: HSC 2013 3U Marathon Thread

^ Alright thanks. I'm quite sure it isn't taught in 3U so i wasn't sure if we're allowed to use it

 

[Carrotsticks]

Re: HSC 2013 3U Marathon Thread

Can someone do this version anyway?

What I got was:

If we take the number of arrangements of n couples around a round table so that no couple is together as

Then we need to find:



To find , I considered an already set up arrangements of couples, then there are 6 spots for the 2 couples to sit and therefore to sit the two couple apart it is 6C2

Therefore



Similarly




And by manual computation we find



??

Is this correct? Can someone affirm or deny it?

I don't understand the bolded part. Can you clarify?

The problem is very difficult in its generalised form. I can imagine that it would use some form of the inclusion/exclusion principle.