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IniquiTrance
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[SOLVED] Probability - Couples seated round a table
As per my textbook: (Ross, 8th ed.), the probability of 10 couples being seated around a table, where every guy's with his girl, or guy, is:
P\left(\bigcup^{10}_{1}E_{i}\right)
Where any:
P(E_{i_{1}}, E_{i_{2}}, E_{i_{3}}, E_{i_{4}}...E_{i_{n}}) = \frac{2^{n}(19 - n)!}{19!}
The book explains that it considers each of the 10 couples a single entity, and therefore calculates all possible outcomes of placing these entities around the table.
Yet why is it that for all the E_{i} intersections in the above equation, n is subtracted from 19, instead of 10?? When n = 1, there are now 18 other people that can be arranged in whatever way, which makes sense to me.
But when n = 2, we're permuting 17 others, when 2 couples, 4 people, have been removed from the table?? Shouldn't it be 16 for n = 2, 14 for n= 3... ?
Thanks!
As per my textbook: (Ross, 8th ed.), the probability of 10 couples being seated around a table, where every guy's with his girl, or guy, is:
P\left(\bigcup^{10}_{1}E_{i}\right)
Where any:
P(E_{i_{1}}, E_{i_{2}}, E_{i_{3}}, E_{i_{4}}...E_{i_{n}}) = \frac{2^{n}(19 - n)!}{19!}
The book explains that it considers each of the 10 couples a single entity, and therefore calculates all possible outcomes of placing these entities around the table.
Yet why is it that for all the E_{i} intersections in the above equation, n is subtracted from 19, instead of 10?? When n = 1, there are now 18 other people that can be arranged in whatever way, which makes sense to me.
But when n = 2, we're permuting 17 others, when 2 couples, 4 people, have been removed from the table?? Shouldn't it be 16 for n = 2, 14 for n= 3... ?
Thanks!
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