# Probability - Couples seated round a table

1. Mar 11, 2010

### IniquiTrance

[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!

Last edited: Mar 11, 2010
2. Mar 11, 2010