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SigmundFraud
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Joined: June 16th, 2012, 11:41 am

Theatre Row Problem

July 2nd, 2014, 8:36 am

The Theater Row: Eight eligible bachelors and seven beautiful models happen randomly to have purchased single seats in the same 15-seat row of a theatre. On the average, how many pairs of adjacent seats are ticketed for marriageable couples?
 
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Traden4Alpha
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Joined: September 20th, 2002, 8:30 pm

Theatre Row Problem

July 2nd, 2014, 9:59 am

1. In today's era of gay marriage, the answer is 14. 2. The lower bound is 14*(1 - (7/15)^2 - (8/15)^2 ) = 6.96888...3. ... more later
 
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EBal
Posts: 431
Joined: May 20th, 2005, 1:30 pm

Theatre Row Problem

July 2nd, 2014, 11:51 am

Let [$]a(w, m)[$] is number of such pairs formed with w women and m men if a woman is in the leftmost seat, and [$]b(w, m)[$] - number of pairs if a man is in the leftmost seat. Then we have [$]a(w,m)=\frac{(w+m-2)!}{(w-1)!(m-1)!}+b(w-1,m)+a(w-1,m)\,,\quad b(w,m)=\frac{(w+m-2)!}{(w-1)!(m-1)!}+a(w,m-1)+b(w,m-1)[$]. Introduce the transform[$]A(x,y)=\sum_{w=1}^\infty\sum_{m=1}^\infty a(w,m)x^wy^m[$] and same for [$]B(x,y)[$]. Then [$]A, B[$] satisfy[$]A=\frac{xy}{(1-x - y)}+xA+xB\,,\quad B=\frac{xy}{(1-x-y)}+yA+yB[$]. This can be solved, for the quantity of interest [$]A+B=\frac{2xy}{(1-x-y)^2}[$]. All that remains to do is to find the coefficient in front of [$]x^7y^8[$].
Last edited by EBal on July 1st, 2014, 10:00 pm, edited 1 time in total.
 
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EBal
Posts: 431
Joined: May 20th, 2005, 1:30 pm

Theatre Row Problem

July 2nd, 2014, 12:13 pm

The expansion is actually not too hard, the result is [$]2\frac{(w+m-1)!}{(w-1)!(m-1)!}[$] for the total number of pairs. The average is [$]\frac{2mw}{(m+w)}[$]. This is in agreement with Vanubis1 who posted correct answer before I could spot an error in my original recurrence relation. I find it hard to follow his logic though.
Last edited by EBal on July 1st, 2014, 10:00 pm, edited 1 time in total.
 
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Vanubis1
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Joined: February 21st, 2011, 7:41 am

Theatre Row Problem

July 2nd, 2014, 12:54 pm

I think you can do the average directly.A man has a probability of 1-1/(w+m) to have a left neighbour and in this case w/(w+m-1) it's a woman.Same number for right neighbour.Each man has an average number of women neighbours of 2*w/(w+m) By summing on all the men, we find 2*m*w/(w+m) = 7.46 with w=7 and m=8
Last edited by Vanubis1 on July 2nd, 2014, 10:00 pm, edited 1 time in total.
 
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EBal
Posts: 431
Joined: May 20th, 2005, 1:30 pm

Theatre Row Problem

July 2nd, 2014, 2:23 pm

Last edited by EBal on July 1st, 2014, 10:00 pm, edited 1 time in total.
 
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chocolatemoney
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Joined: October 8th, 2008, 6:50 am

Theatre Row Problem

July 5th, 2014, 8:58 am

 
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katastrofa
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Joined: August 16th, 2007, 5:36 am
Location: Alpha Centauri

Theatre Row Problem

July 6th, 2014, 10:54 am

Can you solve this skipping the word "marriageable" (i.e. when one pair is classified as a couple, they cannot be married to somebody else)? E.g. female male female -> 1 couplea
 
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Ultraviolet
Posts: 1655
Joined: August 15th, 2012, 9:46 am

Theatre Row Problem

July 6th, 2014, 6:07 pm

Previous solution - m * (m + n - 2) / (m + n) * n * (n - 1) / (n + m - 1) - analogous term for "beautiful models"?
 
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katastrofa
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Joined: August 16th, 2007, 5:36 am
Location: Alpha Centauri

Theatre Row Problem

July 7th, 2014, 6:16 am

nope...a
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