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one turdEight couples meet up at a restaurant. What is the probability that no one sits next to their mate? (assume they sit in a row)

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That will make sure nobody sits down.........one turd

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Are we supposed to also assume that the seats are assigned/chosen randomly?… assume they sit in a row …

If not, then is it a good exercise because it serves as practice for determining exercises that cannot be answered as given? :cool:

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Before retiring I loved to teach combinatorics at a simple level to in service teachers. They need renewal credits and a collection of problems giving "real world" applications. This was one I used to work on the concept ofEight couples meet up at a restaurant. What is the probability that no one sits next to their mate? (assume they sit in a row)

In order to find the number of ways that none of these four couples sits together we must subtract the number of ways at least on of the couples sit together from the total possible arrangements \(\displaystyle 8!\). To find the of ways at least one couple is together we use inclusion/exclusion.

We add \(\displaystyle ||A||+||B||+||C||+||D||-||AB||-||AC||-||AD||-||BC||-||BD||-||CD||+||ABC||+||ABD||+||ACD||+||BCD||-||ABCD||\)

The couples taken on at a time minus two at a time plus three at a time and finally minus four at a time.

Now lets consider each of the four types.

\(\displaystyle ||A||=2\cdot 7!\) is the number of ways couple \(\displaystyle A\) can sit together. The 2 comes from the fact the wife can sit on either side of the husband.

Then that one unit along with the six others makes seven.

\(\displaystyle ||AB||=4\cdot 6!\) is the number of ways couples \(\displaystyle A~\&~B\) can each sit together, although the couples are not necessarily together.

\(\displaystyle ||ABC||=8\cdot 5!~\&~||ABCD||=16\cdot 4!\)

\(\displaystyle \displaystyle\sum\limits_{k = 0}^4 {{{( - 1)}^k}\left( {\begin{array}{*{20}{c}} 4\\ k \end{array}} \right)\left[ {{2^k}(8 - k)!} \right]} \) is the number of ways that none of the couples sits together.

To see the probability that no couple is together SEE HERE

One can see how using a modified Mooore method, that can make a three hour night class.

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I thought it was a riddle.My 1/3 assumes completely at random …

\(\displaystyle \dfrac{12}{35} \Rightarrow \text{Pretty darn close to }\dfrac 1 3 \text{ but still }\Rightarrow \text{corner}\)My 1/3 assumes completely at random.

Plus 1/3 also solution to all similar cases > 1 couple.

Do I get the "corner" again?

Any combo containing one or more of pairs 12,21,34,43 are eliminated.

* = OK!

1234

1243

1324 *

1342

1423 *

1432

2134

2143

2314 *

2341

2413 *

2431

Similarly for 3??? and 4???

Clearly 2 out of 6 (or 8 out of 24)

Similarly did 3 couples and 4 couples (wrote program);

got similar results...so assumed it applied to all cases...

Where oh where did I goof?

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Did you read the solution? This is a well known and used problem in counting courses.2 couples: 1,2,3,4 ;

Similarly did 3 couples and 4 couples (wrote program);

got similar results...so assumed it applied to all cases...

Where oh where did I goof?

The "goof" as you call it is your model.